Vector Algebra Exercise 10.1
Question 1. Represent graphically a displacement of 40 km, 30° east of north.
Solution:
Here, \(\overrightarrow{\mathrm{OP}}\) vector represents the displacement of 40 km, 30° East of North.
Question 2. Classify the following measures as scalars and vectors.
- 10kg
- 2 metres north-west
- 40°
- 40 watt
- 10-19
- coulomb 20 m/s²
Solution:
- 10 kg is a scalar quantity because it involves only magnitude.
- 2 meters northwest is a vector quantity as it involves both magnitude and direction.
- 40° is a scalar quantity as it involves only magnitude.
- 40 watts is a scalar quantity as it involves only magnitude.
- 10-19 coulomb is a scalar quantity as it involves only magnitude.
- 20 m/s² is a vector quantity as it involves magnitude as well as direction.
Question 3. Classify the following as scalar and vector quantities.
- Time period
- Distance
- Force
- Velocity
- Work done
Solution:
- Time period is a scalar quantity as it involves only magnitude.
- Distance is a scalar quantity as it involves only magnitude.
- Force is a vector quantity as it involves both magnitude and direction.
- Velocity is a vector quantity as it involves both magnitude as well as direction.
- Work done is a scalar quantity as it involves only magnitude.
Question 4. Identify the following vectors,
- Coinitial
- Equal
- Collinear but not equal
Solution:
- Vectors \(\vec{a}\) and \(\vec{d}\) are coinitial because they have the same initial point.
- Vectors \(\vec{b}\) and \(\vec{d}\) are equal because they have the same magnitude and direction.
- Vectors \(\vec{a}\) and \(\vec{c}\) are collinear but not equal. This is because although they are parallel, their directions are not the same.
Read and Learn More Class 12 Maths Chapter Wise with Solutions
Question 5. Answer the following as true or false.
- \(\vec{\mathrm{a}}\) and –\(\vec{\mathrm{a}}\) are collinear.
- Two collinear vectors are always equal in magnitude.
- Two vectors having the same magnitude are collinear.
- Two collinear vectors having the same magnitude are equal.
Solution:
- True, Vectors \(\vec{\mathrm{a}}\) and –\(\vec{\mathrm{a}}\) are parallel to the same line.
- False, Collinear vectors are those vectors that are parallel to the same line.
- False, Two vectors having the same magnitude need not necessarily be parallel to the same line,
- False, Only if the magnitude and direction of two vectors are the same regardless of the positions of their initial points, the two vectors are said to be equal.
Vector Algebra Exercise 10.2
Question 1. Compute the magnitude of the following vectors: \(\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\)
Solution:
The given vectors are: \(\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\)
⇒ \(|\vec{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}\)
⇒ \(|\vec{b}|=\sqrt{(2)^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62}\)
⇒ \(|\vec{c}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(-\frac{1}{\sqrt{3}}\right)^2}=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=1\)
Question 2. Write two different vectors having the same magnitude.
Solution:
Consider \(\overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\) and \(\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})\).
It can be observed that \(|\overrightarrow{\mathrm{a}}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\)
and \(|\vec{b}|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}\)
Hence, \(\vec{a}\) and \(\vec{b}\) are two different vectors having the same magnitude. The vectors are different because they have different directions.
Question 3. Write two different vectors having the same direction.
Solution:
Consider \(\vec{a}=(\hat{i}+\hat{j}+\hat{k})\) and \(\vec{b}=(2 \hat{i}+2 \hat{j}+2 \hat{k})\).
The direction consines of \(\vec{a}\) are given by,
l = \(\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}, \mathrm{~m}=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}, \mathrm{n}=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}\).
The direction cosines of \(\vec{b}\) are given by
l = \(\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}, \mathrm{~m}=\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}, \mathrm{n}=\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
The direction cosines of \(\vec{a}\) and \(\vec{b}\) are the same. Hence, the two vectors have the same direction.
Question 4. Find the values of x and y so that the vectors \(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}} \text { and } x \hat{\mathrm{i}}+y \hat{\mathrm{j}}\) are equal.
Solution:
The two vectors \(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}} \text { and } x \hat{\mathrm{i}}+y \hat{\mathrm{j}}\) will be equal if their corresponding scalar components are equal. Hence, the required values of x and y are 2 and 3 respectively.
Question 5. Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5, 7).
Solution:
The vector with the initial point P (2, 1) and terminal point Q (-5, 7) can be given by, \(\overrightarrow{\mathrm{PQ}}=(-5-2) \hat{\mathrm{i}}+(7-1) \hat{\mathrm{j}} \Rightarrow \overrightarrow{\mathrm{PQ}}=-7 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}\)
Hence, the required scalar components are -7 and 6 while the vector components are \(-7 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}\)
Question 6. Find the sum of the vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \hat{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\).
Solution:
The given vectors are \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \hat{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\).
∴ \(\vec{a}+\vec{b}+\vec{c}=(1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7)) \hat{k}=0 \hat{i}-4 \hat{j}-1 \hat{k}=-4 \hat{j}-\hat{k}\)
Question 7. Find the unit vector in the direction of the vector \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\).
Solution:
The unit vector in the direction of the vector \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\) is given by \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\).
⇒ \(|\vec{a}|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}\)
∴ \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k}\)
Question 8. Find the unit vector in the direction of the vector \(\overrightarrow{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Solution:
The given points are P(l, 2, 3) and Q (4, 5, 6).
∴ \(\overrightarrow{P Q}=(4-1) \hat{i}+(5-2) \hat{j}+(6-3) \hat{k}=3 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ Magnitude of given vector, \(|\overrightarrow{\mathrm{PQ}}|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3 \sqrt{3}\)
Hence, the unit vector in the direction of \(\overrightarrow{\mathrm{PQ}}\) is \(\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}=\frac{3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}+\frac{1}{\sqrt{3}} \hat{\mathrm{k}}\)
Question 9. For given vectors, \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\), find the unit vector in the direction of the vector \(\vec{a}\) + \(\vec{b}\).
Solution:
The given vectors are \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\).
∴ \(\vec{a}+\vec{b}=(2-1) \hat{i}+(-1+1) \hat{j}+(2-1) \hat{k}=1 \hat{i}++0 \hat{j}+1 \hat{k}=\hat{i}+\hat{k}\)
⇒ \(|\vec{a}+\vec{b}|=\sqrt{1^2+1^2}=\sqrt{2}\)
Hence, the unit vector in the direction of \((\vec{a}+\vec{b})\) is \(\frac{(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\)
Question 10. Find a vector in the direction of the vector \(5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) which has magnitude 8 units.
Solution:
Let \(\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{5^2+(-1)^2+2^2}=\sqrt{25+1+4}=\sqrt{30}\)
⇒ \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\)
Hence, the vector in the direction of the vector \(5 \hat{i}-\hat{j}+2 \hat{k}\) which has magnitude of 8 units is given by,
8 \(\hat{\mathrm{a}}=8\left(\frac{5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{30}}\right)=\frac{40}{\sqrt{30}} \hat{\mathrm{i}}-\frac{8}{\sqrt{30}} \hat{\mathrm{j}}+\frac{16}{\sqrt{30}} \hat{\mathrm{k}}\)
Question 11. Show that the vectors \(2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are collinear.
Solution:
Let \(\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}\)
It is observed that \(\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}\)
∴ \(\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}}\)
where, λ=-2,
Hence, the given vectors are collinear.
Question 12. Find the direction cosines of the vector \(\hat{i}+2 \hat{j}+3 \hat{k}\)
Solution:
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
∴ \(|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\)
∴ \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\)
Hence, the direction cosines of a are \(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\).
Question 13. Find the direction cosines of the vector joining the points A (1, 2, -3) and B(-l, -2, 1), directed from A to B.
Solution:
The given points are A (1,2, -3) and B (-1, -2, 1).
∴ \(\overrightarrow{\mathrm{AB}}=(-1-1) \hat{\mathrm{i}}+(-2-2) \hat{\mathrm{j}}+\{1-(-3)\} \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Direction ratios of \(\overrightarrow{\mathrm{AB}}\) are a = -2, b = -4, c = 4
Now direction cosines of \(\overrightarrow{\mathrm{AB}}\) are :
l \(=\frac{-2}{\sqrt{(-2)^2+(-4)^2+(4)^2}}=\frac{-2}{6}=-\frac{1}{3}, \mathrm{~m}=\frac{-4}{\sqrt{(-2)^2+(-4)^2+(4)^2}}=\frac{-4}{6}=-\frac{2}{3}\)
n = \(\frac{4}{\sqrt{(-2)^2+(-4)^2+(4)^2}}=\frac{4}{6}=\frac{2}{3}\)
Hence, the direction cosines of \(\overrightarrow{\mathrm{AB}}\) are \(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\).
Question 14. Show that the vector \(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) is equally inclined to the axes, OX, OY, and OZ.
Solution:
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\). Direction ratios of \(\overrightarrow{\mathrm{a}}\) are \(\vec{a}=\mathrm{b}=\mathrm{c}=1\)
Now, direction cosines are
l = \(\frac{1}{\sqrt{(1)^2+(1)^2+(1)^2}}=\frac{1}{\sqrt{3}}=\mathrm{m}=\mathrm{n}\)
Therefore, the direction cosines of \(\vec{a}\) are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
Now, let \(\alpha, \beta, and \gamma\) be the angles formed by \(\vec{a}\) with the positive directions of x, y, and z axes.
Then, we have \(\cos \alpha=\frac{1}{\sqrt{3}}, \cos \beta=\frac{1}{\sqrt{3}}, \cos \gamma=\frac{1}{\sqrt{3}}, \alpha=\beta=\gamma=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Hence, the given vector is equally inclined to axes OX, OY, and OZ.
Question 15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \(\hat{i}+2 \hat{j}-\hat{k}\) and \(-\hat{i}+\hat{j}+\hat{k}\) respectively, in the ratio 2:1
- Internally
- Externally
Solution:
Here, \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}=\overrightarrow{\mathrm{a}}\) (let) and \(\overrightarrow{\mathrm{OQ}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}=\overrightarrow{\mathrm{b}}\) (let), also m=2, n=1 when R divides PQ internally in the ratio 2: 1, then
1. P . V. of \(R=\frac{m \vec{b}+n \vec{a}}{m+n}\)
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{(-2 \hat{i}+2 \hat{j}+2 \hat{k})+(\hat{i}+2 \hat{j}-\hat{k})}{3}\)
= \(\frac{-\hat{i}+4 \hat{j}+\hat{k}}{3}=-\frac{1}{3} \hat{i}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k}\)
2. when R divides PQ externally in the ratio 2: 1 then,
P.V. of R = \(\frac{m \vec{b}-n \vec{a}}{m-n}\)
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}\)
= \(-3 \hat{i}+3 \hat{k}\)
Question 16. Find the position vector of the midpoint of the vector joining the points P (2,3,4) and Q (4, 1, – 2).
Solution:
The position vector of P and Q are given by \(\vec{p}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{q}=4 \hat{i}+\hat{j}-2 \hat{k}\) respectively
∴ P.V. of midpoint of PQ = \(\frac{1}{2}(\vec{p}+\vec{q})\)
= \(\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}\)
Question 17. Show that the points A, B and C with position vectors, \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}\), respectively form the vertices of a right-angled triangle.
Solution:
Position vectors of points A, B, and C are respectively given as: \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\vec{i}-3 \hat{j}-5 \hat{k}\)
∴ \(\overrightarrow{A B}=\vec{b}-\vec{a}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k}\)
⇒ \(\overrightarrow{B C}=\vec{c}-\vec{b}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}\)
⇒ \(\overrightarrow{C A}=\vec{a}-\vec{c}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k}\)
Now, \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{k})+(-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})+(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=\overrightarrow{0}\)
∴ A, B, and C are vertices of the triangle
Now, \(|\overrightarrow{\mathrm{AB}}|^2=(-1)^2+3^2+5^2=1+9+25=35\)
⇒ \(|\overrightarrow{B C}|^2=(-1)^2+(-2)^2+(-6)^2=1+4+36=41\)
⇒ \(|\overrightarrow{\mathrm{CA}}|^2=2^2+(-1)^2+1^2=4+1+1=6\)
∴ \(|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{BC}}|^2=35+6=41\)
Hence, A, B, and C are vertices of a right-angled triangle.
Question 18. In triangle ABC which of the following is not true:
- \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
- \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\)
- \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
- \(\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
Solution:
On applying the triangle law of addition in the given triangle, we have:
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}\)….(1)
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}} \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)…..(2)
∴ The equation given in alternative A is true.
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}} \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\)
∴ The equation given in alternative 2 is true. From equation (2), we have: \(\overrightarrow{\mathrm{AB}}\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
∴ The equation given in alternative 4 is true.
Now, consider the equation given in alternative C: \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0} \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{CA}}[latex]
\)
From equation (1) and (3), we have: \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CA}}\)
∴ \(\overrightarrow{\mathrm{AC}}=-\overrightarrow{\mathrm{AC}} \Rightarrow \overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{0} \Rightarrow 2 \overrightarrow{\mathrm{AC}}=\overrightarrow{0} \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{0}\), which is not true.
Hence, the equation given in alternative C is incorrect.
The correct answer is (3).
Question 19. If \(\vec{a}\) and \(\vec{b}\) are two collinear vectors, then which of the following are incorrect:
- \(\vec{b}=\lambda \vec{a}\), for some scalar \(\lambda\)
- \(\overrightarrow{\mathrm{a}}= \pm \overrightarrow{\mathrm{b}}\)
- The respective components of \(\vec{a}\) and \(\vec{b}\) are proportional
- Both the vectors \(\vec{a}\) and \(\vec{b}\) have same direction, but different magnitudes
Solution:
If \(\vec{a}\) and \(\vec{b}\) are two collinear vectors, then they are parallel.
Therefore, we have: \(\vec{b}=\lambda \vec{a}\) (For some scalar \(\lambda\))
If \(\lambda= \pm 1\), then \(\vec{a}= \pm \vec{b}\).
If \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\) and \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\) and \(\vec{b}=\lambda \vec{a}\).
⇒ \(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}=\lambda\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)\)
⇒ \(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}=\left(\lambda a_1\right) \hat{i}+\left(\lambda a_2\right) \hat{j}+\left(\lambda a_3\right) \hat{k}\)
⇒ \(\mathrm{b}_1=\lambda \mathrm{a}_1, \mathrm{~b}_2=\lambda \mathrm{a}_2, \mathrm{~b}_3=\lambda \mathrm{a}_3\)
⇒ \(\frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}=\lambda\)
Thus, the respective scalar components of \(\vec{a}\) and \(\vec{b}\) are proportional.
However, vectors \(\vec{a}\) and \(\vec{b}\) can have different directions.
Hence, the statement given in 4 is incorrect.
The correct answer is 4.
Vector Algebra Exercise 10.3
Question 1. Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitude \(\sqrt{3}\) and 2, respectively having \(\vec{a} \cdot \vec{b}=\sqrt{6}\).
Solution:
It is given that, \(|\vec{a}|=\sqrt{3},|\vec{b}|=2\) and, \(\vec{a} \cdot \vec{b}=\sqrt{6}\)
Now, we know that \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\).
∴ \(\sqrt{6}=\sqrt{3} \times 2 \times \cos \theta \Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \times 2} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}\)
Hence, the angle between the given vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\).
Question 2. Find the angle between the vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\).
Solution:
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\).
⇒ \(|\vec{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\)
⇒ \(|\vec{b}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{9+4+1}=\sqrt{14}\)
Now, \(\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})=1 \cdot 3+(-2)(-2)+3 \cdot 1=3+4+3=10\)
Also, we know that \(\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta\)
∴ 10 = \(\sqrt{14} \sqrt{14} \cos \theta \Rightarrow \cos \theta=\frac{10}{14} \Rightarrow \theta=\cos ^{-1}\left(\frac{5}{7}\right)\)
Question 3. Find the projection of the vector \(\hat{i}-\hat{j}\) on the vector \(\hat{i}+\hat{j}\).
Solution;
Let \(\vec{a}=\hat{i}-\hat{j}\) and \(\vec{b}=\hat{i}+\hat{j}\)
Now, projection of vector \(\vec{a}\) on \(\vec{b}\) is given by, \(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{1+1}}\{1 \cdot 1+(-1)(1)\}=\frac{1}{\sqrt{2}}(1-1)=0\)
Hence, the projection of vector \(\vec{a}\) on \(\vec{b}\) is 0 .
Question 4. Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(7 \hat{i}-\hat{j}+8 \hat{k}\).
Solution:
Let \(\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}\) and \(\hat{b}=7 \hat{i}-\hat{j}+8 \hat{k}\).
Now, projection of vector \(\vec{a}\) on \(\vec{b}\) is given by-
⇒ \(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{7^2+(-1)^2+8^2}}\{1(7)+3(-1)+7(8)\}=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}\)
Hence, the projection of vector \(\vec{a}\) on \(\vec{b}\) is \(\frac{60}{\sqrt{114}}\).
Question 5. Show that each of the given three vectors is a unit vector: \(\frac{1}{7}(2 \hat{i}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot \frac{1}{7}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}), \frac{1}{7}(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\). Also, show that they are mutually perpendicular to each other.
Solution:
Let,
⇒ \(\overrightarrow{\mathrm{a}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=\frac{2}{7} \hat{\mathrm{i}}+\frac{3}{7} \hat{\mathrm{j}}+\frac{6}{7} \hat{\mathrm{k}}\)
∴ \(|\overrightarrow{\mathrm{a}}|=\sqrt{\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+\left(\frac{6}{7}\right)^2}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1\)
⇒ \(\overrightarrow{\mathrm{b}}=\frac{1}{7}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=\frac{3}{7} \hat{\mathrm{i}}-\frac{6}{7} \hat{\mathrm{j}}+\frac{2}{7} \hat{\mathrm{k}}\)
∴ \(|\overrightarrow{\mathrm{b}}|=\sqrt{\left(\frac{3}{7}\right)^2+\left(-\frac{6}{7}\right)^2+\left(\frac{2}{7}\right)^2}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=1\)
⇒ \({\mathrm{c}}=\frac{1}{7}(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=\frac{6}{7} \hat{\mathrm{i}}+\frac{2}{7} \hat{\mathrm{j}}-\frac{3}{7} \hat{\mathrm{k}}\)
∴ \(|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{6}{7}\right)^2+\left(\frac{2}{7}\right)_{-}^2+\left(-\frac{3}{7}\right)^2}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=1\)
Thus, each of the given three vectors is a unit vector.
⇒ \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{2}{7} \times \frac{3}{7}+\frac{3}{7} \times\left(\frac{-6}{7}\right)+\frac{6}{7} \times \frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0\)
⇒ \(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\frac{3}{7} \times \frac{6}{7}+\left(\frac{-6}{7}\right) \times \frac{2}{7}+\frac{2}{7} \times\left(\frac{-3}{7}\right)=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0\)
⇒ \(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\frac{6}{7} \times \frac{2}{7}+\frac{2}{7} \times \frac{3}{7}+\left(\frac{-3}{7}\right) \times \frac{6}{7}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0\)
Hence, the given three vectors are mutually perpendicular to each other.
Question 6. Find \(|\vec{a}|\) and \(|\vec{b}|\), if \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8\) and \(|\vec{a}|=8|\vec{b}|\).
Solution:
⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8 \Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=8\)
⇒ \(|\vec{a}|^2-|\vec{b}|^2=8\)
because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2\)
⇒ \((8 \mid \vec{b})^2-|\vec{b}|^2=8\)
because \(|\vec{a}|=8|\vec{b}|\)
⇒ \(64|\vec{b}|^2-|\vec{b}|^2=8 \Rightarrow 63|\vec{b}|^2=8 \Rightarrow|\vec{b}|^2=\frac{8}{63}\)
⇒ \(|\vec{b}|=\sqrt{\frac{8}{63}}\) Magnitude of a vector is non-negative
⇒ \(|\vec{b}|=\frac{2 \sqrt{2}}{3 \sqrt{7}}\)
⇒ \(|\vec{a}|=8|\vec{b}|=\frac{8 \times 2 \sqrt{2}}{3 \sqrt{7}}=\frac{16 \sqrt{2}}{3 \sqrt{7}}\)
Question 7. Evaluate the product \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\).
Solution:
⇒ \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})=6(\vec{a} \cdot \vec{a})+21(\vec{a} \cdot \vec{b})-10(\vec{b} \cdot \vec{a})-35(\vec{b} \cdot \vec{b})\)
= \(6(\vec{a} \cdot \vec{a})+21(\vec{a} \cdot \vec{b})-10(\vec{a} \cdot \vec{b})-35(\vec{b} \cdot \vec{b})\)
= \(6|\vec{a}|^2+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^2\) (because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2 \text { and } \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\))
Question 8. Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude and such that the angle between them is \(60^{\circ}\) and their scalar product is \(\frac{1}{2}\).
Solution:
Let \(\theta\) be the angle between the vectors \(\vec{a}\) and \(\vec{b}\).
It is given that \(|\vec{a}|=|\vec{b}|, \vec{a} \cdot \vec{b}=\frac{1}{2}\) and \(\theta=60^{\circ}\)…..(1)
We know that \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\).
∴ \(\frac{1}{2}=|\vec{a}||\vec{a}| \cos 60^{\circ}\) Using (1)
⇒ \(\frac{1}{2}=|\vec{a}|^2 \times \frac{1}{2} \Rightarrow|\vec{a}|^2=1 \Rightarrow|\vec{a}|=1 \Rightarrow|\vec{a}|=|\vec{b}|=1\)
Question 9. Find \(|\vec{x}|\), if for a unit vector \(\vec{a},(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12\).
Solution:
⇒ \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12 \Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{a}=12\)
⇒ \(|\vec{x}|^2-|\vec{a}|^2=12\) (because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2\))
and \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a})\)
⇒ \(|\vec{x}|^2-1=12 \quad |\vec{a}|=1\) as \(\vec{a}\) is a unit vector
⇒ \(|\overrightarrow{\mathrm{x}}|^2=13\)
∴ \(|\mathrm{x}|=\sqrt{13}\)
Question 10. If \(\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) are such that \(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) is perpendicular to \(\overrightarrow{\mathrm{c}}\), then find the value of \(\lambda\).
Solution:
The given vectors are \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and [/latex]\vec{c}=3 \hat{i}+\hat{j}[/latex].
Now, \(\vec{a}+\lambda \vec{b}=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})=(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}\)
If \((\vec{a}+\lambda \vec{b})\) is perpendicular to \(\vec{c}\), then \((\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0\)
⇒ \([(2-\lambda) \hat{\mathrm{i}}+(2+2 \lambda) \hat{\mathrm{j}}+(3+\lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})=0 \Rightarrow(2-\lambda) 3+(2+2 \lambda) 1+(3+\lambda) 0=0\)
⇒ 6 – \(3 \lambda+2+2 \lambda \Rightarrow 0 \Rightarrow-\lambda+8=0 \Rightarrow \lambda=8\)
Hence, the required value of \(\lambda\) is 8 .
Question 11. Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) is perpendicular to \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a}\), for any two nonzero vectors \(\vec{a}\) and \(\vec{b}\).
Solution:
⇒ \((|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a})\)
= \(|\vec{a}|^2(\vec{b} \cdot \vec{b})-|\vec{a}||\vec{b}|(\vec{b} \cdot \vec{a})+|\vec{b}||\vec{a}|(\vec{a} \cdot \vec{b})-|\vec{b}|^2(\vec{a} \cdot \vec{a})\)
= \(|\vec{a}|^2|\stackrel{\rightharpoonup}{b}|^2-|\vec{b}|^2|\vec{a}|^2=0\) (because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2\) and \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)
Hence, \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) and \(|\vec{a}| |\vec{b}|-|\vec{b}| \vec{a}\) are perpendicular to each other for any two non-zero vectors \(\vec{a}\) and \(\vec{b}\).
Question 12. If \(\vec{a} \cdot \vec{a}=0\) and \(\vec{a} \cdot \vec{b}=0\), then what can be concluded about the vector \(\vec{b}\)?
Solution:
It is given that \(\vec{a} \cdot \vec{a}=0\) and \(\vec{a} \cdot \vec{b}=0\).
Now, \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}=0 \Rightarrow|\overrightarrow{\mathrm{a}}|^2=0 \Rightarrow|\overrightarrow{\mathrm{a}}|=0\)
⇒ \(\overrightarrow{\mathrm{a}}\) is a zero vector.
Hence, vector \(\vec{b}\) satisfying \(\vec{a} \cdot \vec{b}=0\) can be any vector.
Question 13. If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\).
Solution;
Given \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors, therefore
⇒ \(|\vec{a}|=1,|\vec{b}|=1 \text { and }|\vec{c}|=1 \text {. }\)…….(1)
Again given \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}\)
⇒ \(|\vec{a}+\vec{b}+\vec{c}|=0 \quad \Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\)
⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)
⇒ \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\) (because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2)\))
⇒ \(1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\) [Using equation (1)]
⇒ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a})=-3 \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}\)
Question 14. If either vector \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}=0\). But the converse need not be true. Justify your answer with an example.
Solution:
Consider \(\vec{a}=2 \hat{i}+4 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+3 \hat{j}-6 \hat{k}\).
Then, \(\vec{a} \cdot \vec{b}=2 \cdot 3+4 \cdot 3+3(-6)=6+12-18=0\)
We now observe that: \(|\vec{a}|=\sqrt{2^2+4^2+3^2}=\sqrt{29}\)
⇒ \(\vec{a} \neq \overrightarrow{0}\)
⇒ \(|\overrightarrow{\mathrm{b}}|=\sqrt{3^2+3^2+(-6)^2}=\sqrt{54}\)
⇒ \(\overrightarrow{\mathrm{b}} \neq \overrightarrow{0}\)
Hence, the converse of the given statement need not be true.
Question 15. If the vertices A, B, and C of a triangle ABC are (1,2,3),(-1,0,0),(0,1,2), respectively, then find \(\angle \mathrm{ABC}\). \(\angle \mathrm{ABC}\) is the angle between the vectors \(\overrightarrow{\mathrm{BA}}\) and \(\overrightarrow{\mathrm{BC}}\).
Solution:
The vertices of \(\triangle \mathrm{ABC}\) are given as A(1,2,3), B}(-1,0,0), and C(0,1,2). Also, it is given that \(\angle \mathrm{ABC}\) is the angle between the vectors \(\overrightarrow{\mathrm{BA}}\) and \(\overrightarrow{\mathrm{BC}}\).
⇒ \(\overrightarrow{B A}=(1-(-1)) \hat{i}+(2-0) \hat{j}+(3-0) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
⇒ \(\overrightarrow{\mathrm{BC}}=(0-(-1)) \hat{\mathrm{i}}+(1-0) \hat{\mathrm{j}}+(2-0) \hat{k}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
∴ \(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=2 \times 1+2 \times 1+3 \times 2=2+2+6=10\)
⇒ \(|\overrightarrow{\mathrm{BA}}|=\sqrt{2^2+2^2+3^2}=\sqrt{4+4+9}=\sqrt{17}\)
⇒ \(|\overrightarrow{\mathrm{BC}}|=\sqrt{1+1+2^2}=\sqrt{6}\)
Now, it is known that: \(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}| \cos (\angle \mathrm{ABC})\)
⇒ 10 = \(\sqrt{17} \times \sqrt{6} \cos (\angle \mathrm{ABC}) \Rightarrow \cos (\angle \mathrm{ABC})=\frac{10}{\sqrt{17} \times \sqrt{6}} \Rightarrow \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)\)
Question 16. Show that the points A(1,2,7), B(2,6,3) and C(3,10,-1) are collinear.
Solution:
The given points are A(1,2,7), B(2,6,3), and C(3,10,-1).
Position vectors of points A, B, and C are \(\vec{a}=\hat{i}+2 \hat{j}+7 \hat{k}, \vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\) and \(\vec{c}=3 \hat{i}+10 \hat{j}-\hat{k}\) respectively.
∴ \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=(2-1) \hat{i}+(6-2) \hat{j}+(3-7) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}}=(3-2) \hat{i}+(10-6) \hat{j}+(-1-3) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}\)
Since \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{BC}} \Rightarrow \overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{BC}}\)
Here, point B is common in \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BC}}\)
So, \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BC}}\) are collinear.
Hence, A, B, and C are collinear
Question 17. Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) and \(3 \hat{i}-4 \hat{j}-4 \hat{k}\) form the vertices of right angled triangle.
Solution:
Let vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) and \(3 \hat{i}-4 \hat{j}-4 \hat{k}\) be position vectors of points A, B and C respectively.
i.e., \(\overline{O A}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{O B}=\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\overline{O C}=3 \hat{i}-4 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{A B}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}\)
∴ \(\overrightarrow{\mathrm{BC}}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k}\)
∴ \(\overrightarrow{\mathrm{AC}}=(3-2) \hat{\mathrm{i}}+(-4+1) \hat{\mathrm{j}}+(-4-1) \hat{k}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
Now, \(\overrightarrow{A B}+\overrightarrow{B C}=(-\hat{i}-2 \hat{j}+6 \hat{k})+(2 \hat{i}-\hat{j}+\hat{k})=\hat{i}-3 \hat{j}-5 \hat{k}=\overrightarrow{A C}\)
A, B, and C are vertices of a triangle.
Now, \(|\overrightarrow{\mathrm{AB}}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{1+4+36}=\sqrt{41}\)
∴ \(|\overrightarrow{\mathrm{BC}}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{4+1+1}=\sqrt{6}\)
∴ \(|\overrightarrow{\mathrm{AC}}|=\sqrt{(1)^2+(-3)^2+(-5)^2}=\sqrt{1+9+25}=\sqrt{35}\)
∴ \(|\overrightarrow{\mathrm{BC}}|^2+|\overrightarrow{\mathrm{AC}}|^2=6+35=41=|\overrightarrow{\mathrm{AB}}|^2\)
Hence, A, B, and C are vertices of a right-angle triangle.
Question 18. If \(\vec{a}\) is a nonzero vector of magnitude ‘a’ and \(\lambda\) a nonzero scalar, then \(\lambda \vec{a}\) is unit vector if
- \(\lambda=1\)
- \(\lambda=-1\)
- \(a=|\lambda|\)
- \(a=\frac{1}{|\lambda|}\)
Solution: 4. \(a=\frac{1}{|\lambda|}\)
Given \(\lambda \vec{a}\) is a unit vector.
∴ \(|\lambda \vec{a}|=1 \quad \Rightarrow \quad|\lambda||\vec{a}|=1\)
⇒ \(|\vec{a}|=\frac{1}{|\lambda|} \quad[\lambda \neq 0]\)
⇒ \(a=\frac{1}{|\lambda|}\) (because \(|\vec{a}|\)=a)
Hence, vector \(\lambda \vec{a}\) is a unit vector if \(\mathrm{a}=\frac{1}{|\vec{\lambda}|}, \quad(\lambda \neq 0)\)
The correct answer is (4).
Vector Algebra Exercise 10.4
Question 1. Find \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\), if \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\).
Solution:
We have, \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
⇒ \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)
= \(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\hat{\mathrm{i}}(-14+14)-\hat{\mathrm{j}}(2-21)+\hat{\mathrm{k}}(-2+21)=0 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}+19 \hat{\mathrm{k}}\)
∴ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{(19)^2+(19)^2}=\sqrt{2 \times(19)^2}=19 \sqrt{2}\)
Question 2. Find a unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\).
Solution:
We have, \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
∴ \(\vec{a}+\vec{b}=4 \hat{i}+4 \hat{j}, \vec{a}-\vec{b}=2 \hat{i}+4 \hat{k}\)
∴ \((\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})\)
= \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
4 & 4 & 0 \\
2 & 0 & 4
\end{array}\right|\)
= \(\hat{i}(16)-\hat{j}(16)+\hat{k}(-8)=16 \hat{i}-16 \hat{j}-8 \hat{k}\)
∴ \(|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\sqrt{16^2+(-16)^2+(-8)^2}=8 \sqrt{2^2+2^2+1}=8 \sqrt{9}=8 \times 3=24\)
Hence, the unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) is given by the relation,
± \(\frac{(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|}= \pm \frac{(16 \hat{i}-16 \hat{j}-8 \hat{k})}{24}= \pm \frac{(2 \hat{i}-2 \hat{j}-\hat{k})}{3}\)
Required vector is \(\frac{2}{3} \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}-\frac{1}{3} \hat{\mathrm{k}}\) or \(-\frac{2}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\)
Question 3. If a unit vector a makes an angle \(\frac{\pi}{3}\) with \(\hat{i}, \frac{\pi}{4}\) with \(\hat{j}\) and an acute angle \(\theta\) with \(\hat{k}\), then find \(\theta\) and hence, the components of \(\vec{a}\).
Solution:
Let unit vector \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\)
Since \(\vec{a}\) is a unit vector, \(|\vec{a}|=1\).
Also, it is given that \(\overrightarrow{\mathrm{a}}\) makes angle \(\frac{\pi}{3}\) with \(\hat{\mathrm{i}}, \frac{\pi}{4}\) with \(\hat{\mathrm{j}}\) and an acute angle \(\theta\) with \(\hat{k}\).
Then, we have: \(\cos \frac{\pi}{3}=\frac{a_1}{|\vec{a}|} \Rightarrow \frac{1}{2}=a_1 \quad[|\vec{a}|=1]\)
Also, \(\cos \frac{\pi}{4}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|} \Rightarrow \frac{1}{\sqrt{2}}=\mathrm{a}_2\) (\(|\vec{a}|=1\))
Also, \(\cos \theta=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|} \Rightarrow \mathrm{a}_3=\cos \theta\)
Now, \(|\overrightarrow{\mathrm{a}}|=1 \Rightarrow \sqrt{\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2}=1\)
⇒ \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \theta=1 \Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^2 \theta=1\)
∴ \(\frac{3}{4}+\cos ^2 \theta=1 \Rightarrow \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4} \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\) (because \(\theta\) is acute angle)
∴ \(a_3=\cos \frac{\pi}{3}=\frac{1}{2}\)
Hence, \(\theta=\frac{\pi}{3}\) and the components of a are \(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\).
Question 4. Show that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Solution:
L.H.S. \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})\)
= \((\vec{a}-\vec{b}) \times \vec{a}+(\vec{a}-\vec{b}) \times \vec{b}\) [By distributivity of vector product over addition]
= \(\vec{a} \times \vec{a}-\vec{b} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{b}\) [Again, by distributivity of vector product over addition]
= \(\overrightarrow{0}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-\overrightarrow{0}\)
= \(2(\vec{a} \times \vec{b})\) R.H.S.
Question 5. Find \(\lambda\) and \(\mu\) if \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\).
Solution:
⇒ \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)
⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|\)
= \(0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \Rightarrow \hat{\mathrm{i}}(6 \mu-27 \lambda)-\hat{\mathrm{j}}(2 \mu-27)+\hat{\mathrm{k}}(2 \lambda-6)\)
=0 \(\hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\)
On comparing the corresponding components, we have :
6 \(\mu-27 \lambda=0\)….(1)
2 \(\mu-27=0\)….(2)
2 \(\lambda-6=0\)….(3)
Now, from equation (3), \(2 \lambda-6=0 \Rightarrow \lambda=3\)
from equation (2), \(2 \mu-27=0 \Rightarrow \mu=\frac{27}{2}\)
Here, values of \(\lambda\) and \(\mu\) satisfy to equation (1)
Hence, \(\lambda=3\) and \(\mu=\frac{27}{2}\)
Question 6. Given that \(\vec{a} \cdot \vec{b}=0\) and \(\vec{a} \times \vec{b}=\overrightarrow{0}\). What can you conclude about the vectors \(\vec{a}\) and \(\vec{b}\)?
Solution:
⇒ \(\vec{a}, \vec{b}=0\) Either \(|\vec{a}|=0\) or \(|\vec{b}|=0\) or \(\vec{a} \perp \vec{b}\)
⇒ \(\vec{a} \times \vec{b}=0\) Then, either \(|\vec{a}|=0\) or \(|\vec{b}|=0\) or \(\vec{a} \| \vec{b}\)
But, \(\vec{a}\) and \(\vec{b}\) cannot be perpendicular and parallel simultaneously.
Hence, \(|\vec{a}|=0\) or \(|\vec{b}|=0\). i.e. either \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\)
Question 7. Let the vectors \(\vec{a}, \vec{b}, \vec{c}\) are given as \(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\) respectively.
Then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\).
Solution:
We have \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)
⇒ \((\vec{b}+\vec{c})=\left(b_1+c_1\right) \hat{i}+\left(b_2+c_2\right) \hat{j}+\left(b_3+c_3\right) \hat{k}\)
Now, \(\vec{a} \times(\vec{b}+\vec{c})\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3\end{array}\right|\)
= \(\hat{i}\left[a_2\left(b_3+c_3\right)-a_3\left(b_2+c_2\right)\right]-\hat{j}\left[a_1\left(b_3+c_3\right)-a_3\left(b_1+c_1\right)\right]+\hat{k}\left[a_1\left(b_2+c_2\right)-a_2\left(b_1+c_1\right)\right]\)
= \(\hat{i}\left[a_2 b_3+a_2 c_3-a_3 b_2-a_3 c_2\right]+\hat{j}\left[-a_1 b_3-a_1 c_3+a_3 b_1+a_3 c_1\right]\)+\(\hat{k}\left[a_1 b_2+a_1 c_2-a_2 b_1-a_2 c_1\right]\)…..(1)
⇒ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)
= \(\hat{i}\left[a_2 b_3-a_3 b_2\right]+\hat{j}\left[a_3 b_1-a_1 b_3\right]+\hat{k}\left[a_1 b_2-a_2 b_1\right]\)….(2)
⇒ \(\vec{a} \times \vec{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)
= \(\hat{i}\left[a_2 c_3-a_3 c_2\right]+\hat{j}\left[a_3 c_1-a_1 c_3\right]+\hat{k}\left[a_1 c_2-a_2 c_1\right]\)….(3)
On adding (2) and (3), we get :
(\(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})=\hat{i}\left[a_2 b_3+a_2 c_3-a_3 b_2-a_3 c_2\right]+\hat{j}\left[b_1 a_3+a_3 c_1-a_1 b_3-a_1 c_3\right]\)
+ \(\hat{k}\left[a_1 b_2+a_1 c_2-a_2 b_1-a_2 c_1\right]\)….(4)
Now, from (1) and (4), we have: \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)
Question 8. If either \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\), then \(\vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example.
Solution:
Take any parallel non-zero vectors so that \(\vec{a} \times \vec{b}=\overrightarrow{0}\).
Let \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}\).
Then, \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)
= \(\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right|=\hat{\mathrm{i}}(24-24)-\hat{\mathrm{j}}(16-16)+\hat{\mathrm{k}}(12-12)=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\)
= \(\overrightarrow{0}\)
It can now be observed that : \(|\vec{a}|=\sqrt{2^2+3^2+4^2}=\sqrt{29}\)
⇒ \(\vec{a} \neq \overrightarrow{0}\)
⇒ \(|\vec{b}|=\sqrt{4^2+6^2+8^2}=\sqrt{116}\)
⇒ \(\vec{b} \neq \overrightarrow{0}\)
Hence, the converse of the given statement need not be true.
Question 9. Find the area of the triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Solution:
The vertices of triangle ABC are given as A(1,1,2), B(2,3,5), and C(1,5,5).
Position vector of points A, B and C are \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{i}+5 \hat{j}+5 \hat{k}\) respectively.
The adjacent sides A, B and \(\overrightarrow{B C}\) of \(\triangle A B C\) are given as:
⇒ \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \Rightarrow \overrightarrow{\mathrm{AB}}=(2-1) \hat{\mathrm{i}}+(3-1) \hat{\mathrm{j}}+(5-2) \hat{\mathrm{k}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
and \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{BC}}=(1-2) \hat{\mathrm{i}}+(5-3) \hat{\mathrm{j}}+(5-5) \hat{k}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}\)
∴ \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
-1 & 2 & 0
\end{array}\right|\)
= \(\hat{i}(-6)-\hat{j}(3)+\hat{k}(2+2)=-6 \hat{i}-3 \hat{j}+4 \hat{k}\)
∴ \(|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=\sqrt{(-6)^2+(-3)^2+4^2}=\sqrt{36+9+16}=\sqrt{61}\)
Area of \(\triangle \mathrm{ABC}\)=\(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2} \sqrt{61}\)
Area of \(\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2} \sqrt{61}\)
Hence, the area of \(\triangle \mathrm{ABC}\) is \(\frac{\sqrt{61}}{2}\) square units.
Question 10. Find the area of the parallelogram whose adjacent sides are determined by the vector \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}\).
Solution:
The area of the parallelogram whose adjacent sides are \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\).
Adjacent sides are given as: \(\vec{a}=\hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}\)
∴ \(\vec{a} \times \vec{b}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 3 \\
2 & -7 & 1
\end{array}\right|\)
= \(\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20 \hat{i}+5 \hat{j}-5 \hat{k}\)
⇒ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{20^2+5^2+5^2}=\sqrt{400+25+25}=15 \sqrt{2}\)
Hence, the area of the given parallelogram is \(15 \sqrt{2}\) square units.
Choose The Correct Answer
Question 11. Let the vectors \(\vec{a}\) and \(\vec{b}\) be such that \(|\vec{a}|=3\) and \(|\vec{b}|=\frac{\sqrt{2}}{3}\), then \(\vec{a} \times \vec{b}\) is a unit vector, if the angle between \(\vec{a}\) and \(\vec{b}\) is
- \(\frac{\pi}{6}\)
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{3}\)
- \(\frac{\pi}{2}\)
Solution: 2. \(\frac{\pi}{4}\)
It is given that \(|\vec{a}|=3\) and \(|\vec{b}|=\frac{\sqrt{2}}{3}\).
We know that \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\), where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).
Now, \(\vec{a} \times \vec{b}\) is a unit vector if \(|\vec{a} \times \vec{b}|=1\)
⇒ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=1 \Rightarrow||\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta|=1 \Rightarrow 3 \times \frac{\sqrt{2}}{3} \times \sin \theta=1 \Rightarrow \sin \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}\)
Hence, \(\vec{a} \times \vec{b}\) is a unit vector if the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\).
The correct answer is 2.
Question 12. Area of a rectangle having vertices A, B, C, and D with position vectors \(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\) \(\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k},-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\) respectively is-
- \(\frac{1}{2}\)
- 1
- 2
- 4
Solution: 3. 2
The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:
⇒ \(\overrightarrow{O A}=-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{O B}=\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{O C}=\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{O D}=-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\)
The adjacent sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BC}}\) of the given rectangle are given as:
⇒ \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(1+1) \hat{\mathrm{i}}+\left(\frac{1}{2}-\frac{1}{2}\right) \hat{\mathrm{j}}+(4-4) \hat{\mathrm{k}}=2 \hat{\mathrm{i}}\)
⇒ \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(1-1) \hat{\mathrm{i}}+\left(-\frac{1}{2}-\frac{1}{2}\right) \hat{\mathrm{j}}+(4-4) \hat{\mathrm{k}}=-\hat{\mathrm{j}}\)
∴ \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}\)
= \(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 0 & 0 \\
0 & -1 & 0
\end{array}\right|\)
= \(\hat{\mathrm{k}}(-2)=-2 \hat{\mathrm{k}} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=\sqrt{(-2)^2}=2\)
Now, it is known that the area of a rectangle whose adjacent sides are \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\)
Hence, the area of the given rectangle is \(|\overrightarrow{A B} \times \overrightarrow{B C}|=2\) square units.
The correct answer is 3.
Vector Algebra Miscellaneous Exercise
Question 1. Write down a unit vector in XY-plane, making an angle of 30n with the positive direction of the x-axis.
Solution:
If \(\overrightarrow{\mathrm{r}}\) is a unit vector in the \(\mathrm{XY}\)-plane, then \(\overrightarrow{\mathrm{r}}=\cos \theta \hat{\mathrm{i}}+\sin \theta \hat{\mathrm{j}}\).
Here, \(\theta\) is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for \(\theta=30^{\circ}\):
⇒ \(\overrightarrow{\mathrm{r}}=\cos 30^{\circ} \hat{\mathrm{i}}+\sin 30^{\circ} \hat{\mathrm{j}}=\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\)
Hence, the required unit vector is \(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\)
Question 2. Find the scalar components and magnitude of the vector joining the points \(P\left(x_1, y_1, z_1\right)\) and \(\mathrm{Q}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)\).
Solution:
The vector joining the points \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) and \(\mathrm{Q}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)\) can be obtained by,
⇒ \(\overrightarrow{P Q}\) = P.V. of Q-P.V. of P =\(\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)
⇒ \(|\overrightarrow{P Q}|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)
Hence, the scalar components and the magnitude of the vector joining the given points are \(\left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)\) and \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\) respectively
Question 3. A girl walks \(4 \mathrm{~km}\) towards west, then she walks \(3 \mathrm{~km}\) in a direction \(30^{\circ}\) east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution:
Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:
OA= \(4 \mathrm{~km}, \mathrm{AB}=3 \mathrm{~km}, \overrightarrow{\mathrm{OA}}=-4 \hat{\mathrm{i}}\) and \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CB}} \)
⇒ \(\overrightarrow{\mathrm{AB}}=\left(|\overrightarrow{\mathrm{AB}}| \cos 60^{\circ}\right) \hat{\mathrm{i}}+\left(|\overrightarrow{\mathrm{AB}}| \sin 60^{\circ}\right) \hat{\mathrm{j}}\)
= \(3 \times \frac{1}{2} \hat{\mathrm{i}}+3 \times \frac{\sqrt{3}}{2} \hat{\mathrm{j}}=\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)
By the triangle law of vector addition, we have: \(\overrightarrow{\mathrm{OB}} =\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=(-4 \hat{\mathrm{i}})+\left(\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\right)=\left(-4+\frac{3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)
= \(\left(\frac{-8+3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}=\left(\frac{-5}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)
Hence, the girl’s displacement from her initial point of departure is \(\left(\frac{-5}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)
Question 4. If \(\vec{a}=\vec{b}+\vec{c}\), then is it true that \(|\vec{a}|=|\vec{b}|+|\vec{c}|\)? Justify your answer.
Solution:
In \(\triangle \mathrm{ABC}\), let \(\overrightarrow{\mathrm{CB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}\), and \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}}\) (as shown in the following figure).
Now, by the triangle law of vector addition, we have \(\vec{a}=\vec{b}+\vec{c}\).
It is clearly known that \(|\vec{a}|,|\vec{b}|\) and \(|\vec{c}|\) represent the sides of \(\triangle \mathrm{ABC}\).
Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.
∴ \(|\vec{a}|<|\vec{b}|+|\vec{c}|\)
Hence, it is not true that \(|\vec{a}|=|\vec{b}|+|\vec{c}|\).
Question 5. Find the value of x for which \(x(\hat{i}+\hat{j}+\hat{k})\) is a unit vector.
Solution:
If \(x(\hat{i}+\hat{j}+\hat{k})\) is a unit vector, then \(|x(\hat{i}+\hat{j}+\hat{k})|=1\)
⇒ \(\sqrt{\mathrm{x}^2+\mathrm{x}^2+\mathrm{x}^2}=1 \Rightarrow \sqrt{3 \mathrm{x}^2}=1 \Rightarrow \pm \sqrt{3} \mathrm{x}=1 \Rightarrow \mathrm{x}= \pm \frac{1}{\sqrt{3}}\)
Hence, the required value of x is \(\pm \frac{1}{\sqrt{3}}\).
Question 6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\).
Solution:
We have, \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
Let \(\vec{c}\) be the resultant of \(\vec{a}\) and \(\vec{b}\).
Then, \(\vec{c}=\vec{a}+\vec{b}=(2+1) \hat{i}+(3-2) \hat{j}+(-1+1) \hat{k}=3 \hat{i}+\hat{j}\)
⇒ \(|\overrightarrow{\mathrm{c}}|=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\)
⇒ \(\hat{\mathrm{c}}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}=\frac{(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})}{\sqrt{10}}\)
Hence, the vector of magnitude 5 units and parallel to the resultant of vectors \(\vec{a}\) and \(\vec{b}\) is \(\pm 5\). \(\hat{\mathrm{c}}= \pm 5 \cdot \frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})= \pm \frac{3 \sqrt{10}}{2} \hat{\mathrm{i}} \pm \frac{\sqrt{10}}{2} \hat{\mathrm{j}}\).
Question 7. If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{c}=\hat{i}-2 \hat{j}+3 \hat{k}\), find a unit vector parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
We have, \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{c}=\hat{i}-2 \hat{j}+3 \hat{k}\)
⇒ \(2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k})\)
= \(2 \hat{i}+2 \hat{j}+2 \hat{k}-2 \hat{i}+\hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k}=3 \hat{i}-3 \hat{j}+2 \hat{k}\)
⇒ \(|2 \vec{a}-\vec{b}+3 \vec{c}|=\sqrt{3^2+(-3)^2+2^2}=\sqrt{9+9+4}=\sqrt{22}\)
Hence, the unit vector parallel to \(2 \vec{a}-\vec{b}+3 \vec{c}\) is.
± \(\frac{(2 \vec{a}-\vec{b}+3 \vec{c})}{2 \vec{a}-\vec{b}+3 \vec{c}}= \pm \frac{3 \hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{22}}= \pm \frac{3}{\sqrt{22}} \hat{i} \mp \frac{3}{\sqrt{22}} \hat{j} \pm \frac{2}{\sqrt{22}} \hat{k}\)
Question 8. Show that the points A(1,-2,-8), B(5,0,-2) and C(11,3,7) are collinear, and find the ratio in which B divides AC.
Solution:
The given points are A(1,-2,-8), B(5,0,-2), and C(11,3,7).
P.V. of point A is \(\vec{a}=\hat{i}-2 \hat{j}-8 \hat{k}\)
P.V. of point B is \(\vec{b}=5 \hat{i}-2 \hat{k}\)
P.V. of point C is \(\overrightarrow{\mathrm{c}}=11 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
⇒ \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} =(5-1) \hat{\mathrm{i}}+(0+2) \hat{\mathrm{j}}+(-2+8) \hat{k}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)
⇒ \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} =(11-5) \hat{\mathrm{i}}+(3-0) \hat{\mathrm{j}}+(7+2) \hat{k}=6 \hat{i}+3 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}\)
= \(\frac{3}{2}(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=\frac{3}{2} \overrightarrow{\mathrm{AB}}\)
⇒\(\overrightarrow{\mathrm{BC}}=\frac{3}{2} \overrightarrow{\mathrm{AB}} \text { i.e. } \overrightarrow{\mathrm{BC}} \| \overrightarrow{\mathrm{AB}}\)
Here, B is common in \(\overrightarrow{\mathrm{BC}}\) and \(\overrightarrow{\mathrm{AB}}\)
So, \(\overrightarrow{\mathrm{BC}}\) and \(\overrightarrow{\mathrm{AB}}\) are collinear
Hence, A, B, and C are collinear
Now, let point B divide AC in the ratio \(\lambda: 1\). Then, we have:
⇒ \(\overrightarrow{\mathrm{b}}=\frac{\lambda \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}}{\lambda+1}\)
⇒ \(5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}}=\frac{\lambda(11 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{\lambda+1}\)
⇒ \((\lambda+1)(5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}})=(11 \lambda \hat{\mathrm{i}}+3 \lambda \hat{\mathrm{j}}+7 \lambda \hat{\mathrm{k}})+(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})\)
⇒ \(5(\lambda+1) \hat{\mathrm{i}}-2(\lambda+1) \hat{\mathrm{k}}=(11 \lambda+1) \hat{\mathrm{i}}+(3 \lambda-2) \hat{\mathrm{j}}+(7 \lambda-8) \hat{\mathrm{k}}\)
On equating the corresponding components, we get: 5\((\lambda+1)=11 \lambda+1\)
⇒ 5\(\lambda+5=11 \lambda+1\)
⇒ 6\(\lambda=4 \Rightarrow \lambda=\frac{4}{6}=\frac{2}{3}\)
Hence, point B divides AC in the ratio 2: 3, internally.
Question 9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \((2 \vec{a}+\vec{b})\) and \((\vec{a}-3 \vec{b})\) externally in the ratio 1: 2. Also, show that P is the midpoint of the line segment RQ.
Solution:
It is given that \(\overrightarrow{O P}=2 \vec{a}+\vec{b}, \overrightarrow{O Q}=\vec{a}-3 \vec{b}\).
It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2.
Then, on using the section formula, we get: \(\overrightarrow{\mathrm{OR}}=\frac{2(2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}})}{2-1}=\frac{4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}}{1}=3 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}}\)
Therefore, the position vector of point R is \(3 \vec{a}+5 \vec{b}\).
Position vector of the mid-point of RQ = \(\frac{\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}}}{2}=\frac{(\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}})+(3 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})}{-2}=2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{OP}}\)
Hence, P is the mid-point of the line segment RQ.
Question 10. The two adjacent sides of a parallelogram are \(2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to its diagonal. Also, find its area.
Solution:
Adjacent sides of a parallelogram are given as: \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\).
Then, the diagonal of a parallelogram is given by \(\vec{a}+\vec{b}\).
= \((2+1) \hat{i}+(-4-2) \hat{j}+(5-3) \hat{k}=3 \hat{i}-6 \hat{j}+2 \hat{k}\)
Thus, the unit vector parallel to the diagonal is
± \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}= \pm \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{3^2+(-6)^2+2^2}}= \pm \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{9+36+4}}= \pm \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{7}\)
= \(\pm\left(\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}\right)\).
Now, Area of parallelogram ABCD = \(|\vec{a} \times \vec{b}|\)
⇒ \(\vec{a} \times \vec{b}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -4 & 5 \\
1 & -2 & -3
\end{array}\right|\)
= \(\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)=22 \hat{i}+11 \hat{j}\)
⇒ \(\vec{a} \times \vec{b}=11(2 \hat{i}+\hat{j})\)
⇒ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=11 \sqrt{2^2+1^2}=11 \sqrt{5}\)
Hence, the area of the parallelogram is \(11 \sqrt{5}\) square units.
Question 11. Show that the direction cosines of a vector equally inclined to the axes OX, OY, and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
Solution:
Let a vector be equally inclined to axes OX, OY, and OZ at angle \(\alpha\).
Then, the direction cosines of the vector are cos\(\alpha\), cos\(\alpha\), and cos\(\alpha\).
Now, \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \Rightarrow 3 \cos ^2 \alpha=1 \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}\)
Hence, the direction cosines of the vector which are equally inclined to the axes are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\) or \(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\).
Question 12. Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\). Find a vector \(\overrightarrow{\mathrm{d}}\) which is perpendicular to both \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\), and \(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=15\).
Solution:
Vector \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b} \Rightarrow \vec{d}=\lambda(\vec{a} \times \vec{b})\)
Now, \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)
= \(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & 4 & 2 \\
3 & -2 & 7
\end{array}\right|\)
= \(32 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}}\)
∴ \(\overrightarrow{\mathrm{d}}=\lambda(32 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}})\)
Now, \(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=15 \Rightarrow(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(32 \lambda \hat{\mathrm{i}}-\lambda \hat{\mathrm{j}}-14 \lambda \hat{\mathrm{k}})=15\)
⇒ 64 \(\lambda+\lambda-56 \lambda=15 \Rightarrow \lambda=\frac{5}{3} \Rightarrow \overrightarrow{\mathrm{d}}=\frac{5}{3}(32 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}})\)
Question 13. The scalar product of the vector \(\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of the vector \(2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equal to one. Find the value of \(\lambda\).
Solution:
Let, \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\)
⇒ \(\vec{b}+\vec{c}=(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})=(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}\)
Therefore, unit vector along \(\vec{b}+\vec{c}\) is given as:
± \(\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}= \pm \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{4+4 \lambda+\lambda^2+36+4}}= \pm \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}\)
Scalar product of (\(\vec{a}\)) with this unit vector is 1 .
⇒ \((\hat{i}+\hat{j}+\hat{k}),\left\{ \pm \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}\right\}=1 \Rightarrow \frac{(2+\lambda)+6-2}{\sqrt{\lambda^2+4 \lambda+44}}= \pm 1\)
⇒ \(\sqrt{\lambda^2+4 \lambda+44}= \pm(\lambda+6) \Rightarrow \lambda^2+4 \lambda+44=(\lambda+6)^2\)
⇒ \(\lambda^2+4 \lambda+44=\lambda^2+12 \lambda+36 \Rightarrow 8 \lambda=8 \Rightarrow \lambda=1\)
Hence, the value of \(\lambda\) is 1 .
Question 14. If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular vectors of equal magnitudes, show that the vector \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}, \vec{b}\) and \(\vec{c}\).
Solution:
Since \(\vec{a}, \vec{b}\) and \(\vec{c}\) are mutually perpendicular vectors, we have \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0\)
It is given that: \(|\vec{a}|=|\vec{b}|=|\vec{c}|\)
Let vector \(\vec{a}+\vec{b}+\vec{c}\) be inclined to \(\vec{a}, \vec{b}\) and \(\vec{c}\) at angles \(\theta_1, \theta_2\) and \(\theta_3\) respectively.
Then, we have:
cos\(\theta_1=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}=\frac{\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}\) (\(\vec{b} \cdot \vec{a}=\vec{c} \cdot \vec{a}=0\))
= \(\frac{|\vec{a}|^2}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}=\frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}\)
cos\(\theta_2=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}=\frac{\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{ba}+\vec{c} \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}\) (\(\vec{b} \cdot \vec{a}=\vec{c} \cdot \vec{b}=0\))
= \(\frac{|\vec{b}|^2}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}=\frac{|\vec{b}|}{|\vec{a}+\vec{b}+\vec{c}|}\)
cos\(\theta_3=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{c}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|}=\frac{\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{c}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|}\)
= \(\frac{|\vec{c}|^2}{\mid \vec{a}+\vec{b}+\vec{c}=0]}=\frac{|\vec{c}|}{|\vec{a}+\vec{b}+\vec{c}|}\)
Now, as \(|\vec{a}|=|\vec{b}|=|\vec{c}|\)
∴ \(\cos \theta_1=\cos \theta_2=\cos \theta_3 \Rightarrow \theta_1=\theta_2=\theta_3\)
Hence, the vector \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined to \(\vec{a}, \vec{b}\) and \(\vec{c}\).
Question 15. Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\), if and only if \(\vec{a}, \vec{b}\) are perpendicular, given \(\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}\).
Solution:
Let \((\vec{a}+\vec{b})(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\)
⇒ \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^{-2}\) [Distributivity of scalar products over addition]
⇒ \(|\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2 \quad[\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) (Scalar product is commutative)
⇒ \(2 \vec{a} \cdot \vec{b}=0 \quad \Rightarrow \vec{a} \cdot \vec{b}=0\)
∴ \(\vec{a}\) and \(\vec{b}\) are perpendicular. \(\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}\) (Given)
Further, let \(\vec{a} \perp \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0\)
Now \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b}) \cdot\)
= \(|\vec{a}|^2+|\vec{b}|^2\) (because \(\vec{a} \cdot \vec{b}=0\))
Hence, \((\vec{a}+\vec{b})(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\)
Choose The Correct Answer
Question 16. If \(\theta\) is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a} \cdot \vec{b} \geq 0\) only when:
- \(0<\theta<\frac{\pi}{2}\)
- \(0 \leq \theta \leq \frac{\pi}{2}\)
- \(0<\theta<\pi\)
- \(0 \leq \theta \leq \pi\)
Solution: 2. \(0 \leq \theta \leq \frac{\pi}{2}\)
Let \(\theta\) be the angle between two vectors \(\vec{a}\) and \(\vec{b}\), if \(\vec{a} \cdot \vec{b} \geq 0\)
⇒ \(|\vec{a}||\vec{b}| \cos \theta \geq 0 \Rightarrow \cos \theta \geq 0\)
⇒ \([|\vec{a}| and |\vec{b}|\) are positive]
⇒ \(0 \leq \theta \leq \frac{\pi}{2}\)
Hence, \(\vec{a} \cdot \vec{b} \geq 0\) when \(0 \leq \theta \leq \frac{\pi}{2}\).
The correct answer is (B).
Question 17. Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and \(\theta\) is the angle between them. Then \(\vec{a}+\vec{b}\) is a unit vector if :
- \(\theta=\frac{\pi}{4}\)
- \(\theta=\frac{\pi}{3}\)
- \(\theta=\frac{\pi}{2}\)
- \(\theta=\frac{2 \pi}{3}\)
Solution: \(\theta=\frac{2 \pi}{3}\)
Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and \(\theta\) be the angle between them. Then, \(|\vec{a}|=|\vec{b}|=1\).
Now, \(\vec{a}+\vec{b}\) is a unit vector then \(|\vec{a}+\vec{b}|=1\)
⇒ \(|\vec{a}+\vec{b}|^2=1 \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1\)
⇒ \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=1\) (because \(\vec{a} \cdot \vec{a}=|\vec{a}|^2)\))
⇒ \(|\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2=1 \Rightarrow 1+2|\vec{a}||\vec{b}| \cos \theta+1=1 \Rightarrow \cos \theta=-\frac{1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\)
Hence, \(\vec{a}+\vec{b}\) is a unit vector if \(\theta=\frac{2 \pi}{3}\).
The correct answer is (4).
Question 18. The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})\) is
- 0
- -1
- 1
- 3
Solution: 3. 1
⇒ \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})=\hat{i} \cdot \hat{i}+\hat{j} \cdot(-\hat{j})+\hat{k} \cdot \hat{k}=1-1+1=1\)
The correct answer is (3).
Question 19. If \(\theta\) is the angle between any two vectors \(\vec{a}\) and \(\vec{b}\), then \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\) when \(\theta\) is equal to
- 0
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
- \(\pi\)
Solution: 2. \(\frac{\pi}{4}\)
Let \(\theta\) be the angle between two vectors \(\vec{a}\) and \(\vec{b}\).
⇒ \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}| \Rightarrow|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta\)
cos \(\theta=\sin \theta \Rightarrow \tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}\)
Hence, \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\) when \(\theta\) is equal to \(\frac{\pi}{4}\).
The correct answer is (2).