Three-Dimensional Geometry Exercise 11.1
Question 1. If a line makes angles 90°, 135°, and 45° with x, y, and z-axes respectively, find its direction cosines. Sol. Let the direction cosines of the line be l, m, and n.
Solution:
Let the direction cosines of the line be l, m, and n.
l = \(\cos 90^{\circ}=0 ; m=\cos \left(180^{\circ}-45^{\circ}\right)=-\cos 45^{\circ}=-\frac{1}{\sqrt{2}} ; \mathrm{n}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
Therefore, the direction cosines of the line are \(0,-\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\)
Therefore, the direction cosines of the line are 0,\(-\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\)
Question 2. Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let the line make an angle ‘α’ with each of the coordinate axes.
∴ l = \(\cos \alpha, \mathrm{m}=\cos \alpha, \mathrm{n}=\cos \alpha\)
∴ \(l^2+\mathrm{m}^2+\mathrm{n}^2=1\)
⇒ \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \Rightarrow 3 \cos ^2 \alpha=1 \Rightarrow \cos ^2 \alpha=\frac{1}{3} \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}\)
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are \(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\), and \(\pm \frac{1}{\sqrt{3}}\)
Question 3. If a line has the direction ratios -18, 12, – 4, then what are its direction cosines?
Solution:
If a line has direction ratios of -18, 12, and -4, then its direction cosines are \(\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}}, \frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}}, \frac{-4}{\sqrt{(-18)^2+(12)^2+(-4)^2}}\)
i.e., \(\frac{-18}{22}, \frac{12}{22}, \frac{-4}{22} \Rightarrow \frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\)
Thus, the direction cosines are \(\frac{-9}{11}, \frac{6}{11}\), and \(\frac{-2}{11}\)
Read and Learn More Class 12 Maths Chapter Wise with Solutions
Question 4. Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.
Solution:
The given points are A (2, 3, 4), B (- 1, – 2, 1), and C (5, 8, 7).
It is known that the direction ratios of lines joining the points, (x1, y1, z1) and (x2, y2, z2), are given by, (x2-x1, y2-y1, z2-z1).
The direction ratios of AB are a1= (-1 – 2), b1= (-2 – 3), and c1= (1 -4) i.e., a1 = -3, b1= -5, and c1= – 3.
The direction ratios of BC are a2= (5 – (- 1)), b2 = (8 – (- 2)), and c2= (7-1) i.e., a2 = 6, b2= 10, and c2= 6.
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{b}_1}{\mathrm{~b}_2}=\frac{\mathrm{c}_1}{\mathrm{c}_2}=-\frac{1}{2}\) i.e. they are proportional.
Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.
Question 5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (-1,1,2), and (-5,-5,-2)
Solution:
The vertices of ΔABC are A (3, 5, -4), B (-1, 1,2), and C (-5, -5, -2)
The direction ratios of side AB are (-1 -3), (1 -5), and (2 -(-4)) i.e., (- 4, -4, 6).
Therefore, the direction cosines of AB are \(\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}}, \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}}, \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}\)
⇒ \(\frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}}=\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\)
The direction ratios of BC are (-5-(-1)), (-5-1), and (-2-2) i.e., (-4,-6,-4). Therefore, the direction cosines of BC are \(\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}, \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}, \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}\)
= \(\frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}=\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\)
The direction ratios of AC are (-5-3),(-5-5), and (-2-(-4)) i.e., (-8,-10 and 2). Therefore, the direction cosines of AC are \(\frac{-8}{\sqrt{(-8)^2+(-10)^2+(2)^2}}, \frac{-10}{\sqrt{(-8)^2+(-10)^2+(2)^2}}, \frac{2}{\sqrt{(-8)^2+(-10)^2+(2)^2}}\)
= \(\frac{-8}{2 \sqrt{42}}, \frac{-10}{2 \sqrt{42}}, \frac{2}{2 \sqrt{42}}=\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\)
Three-Dimensional Geometry Exercise 11.2
Question 1. Show that the three lines with direction cosines \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) are mutually perpendicular.
Solution:
Two lines with direction cosines, l1, m1, n1, and l2, m2, n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0
1. For the lines with direction cosines, \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\) and \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\), we obtain
∴ \(l_1 l_2+m_1 m_2+n_1 n_2=\left(\frac{12}{13}\right) \times\left(\frac{4}{13}\right)+\left(\frac{-3}{13}\right) \times\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right) \times\left(\frac{3}{13}\right)\)
= \(\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=0\)
Therefore, the lines are perpendicular
2. For the lines with direction cosines, \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\) and \(\frac{3}{13}=\frac{-4}{13}, \frac{12}{13}\), we obtain
∴ \(l_1 l_2+m_1 m_2+n_1 n_2=\left(\frac{4}{13}\right) \times\left(\frac{3}{13}\right)+\left(\frac{12}{13}\right) \times\left(\frac{-4}{13}\right)+\left(\frac{3}{13}\right) \times\left(\frac{12}{13}\right)\)
= \(\frac{12}{169}-\frac{48}{169}+\frac{36}{169}=0\)
Therefore, the lines are perpendicular,
3. For the lines with direction cosines, \(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) and \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\), we obtain
∴ \(l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2=\left(\frac{3}{13}\right) \times\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right) \times\left(\frac{-3}{13}\right)+\left(\frac{12}{13}\right) \times\left(\frac{-4}{13}\right)\)
= \(\frac{36}{169}+\frac{12}{169}-\frac{48}{169}=0\)
Therefore, the lines are perpendicular.
Thus, all lines are mutually perpendicular.
Question 2. Show that the line through the points (1, -1,2) and (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
Let AB be the line joining the points, (1, -1, 2) and (3, 4, -2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios, a1, b∴ 1, c1, of AB are (3 -1), (4 – (-1)), and (-2 -2) i.e., (2, 5, – 4).
The direction ratios, a2, b2, c2, of CD are (3 – 0), (5 -3), and (6 -2) i.e., (3, 2, 4).
AB and CD will be perpendicular to each other if a1a2 + b1b2 + c1c2 = 0 ⇒ a1a2 + b1b2 + c1c2 = 2 x 3 + 5 x 2 + (-4) x 4= 6 + 10 -16 = 0
Therefore, AB and CD are perpendicular to each other.
Question 3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1,-2, 1), (1,2, 5).
Solution:
Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (-1, -2, 1) and (1, 2, 5).
The directions ratios, a1, b1, c1 of AB are (2 – 4), (3 -7), and (4 -8) i.e., (- 2, – 4, – 4).
The direction ratios, a2, b2, c2 of CD are (1 – (-1)), (2 – (-2)), and (5 – 1) i.e., (2, 4, 4).
AB will be parallel to CD, if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
⇒ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{-2}{2}=-1, \frac{\mathrm{b}_1}{\mathrm{~b}_2}=\frac{-4}{4}=-1\)
and \(\frac{\mathrm{c}_1}{\mathrm{c}_2}=\frac{-4}{4}=-1\)
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{b}_1}{\mathrm{~b}_2}=\frac{\mathrm{c}_1}{\mathrm{c}_2}=-1\)
Thus, AB is parallel to CD.
Question 4. Find the equation of the line that passes through the point (1, 2, 3) and is parallel to the vector \(3 \hat{i}+2 \hat{j}-2 \hat{k}\)
Solution:
It is given that the line passes through the point A(1,2,3). Therefore, the position vector point A is \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\).
Let, \(\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
It is known that the line which passes through point A and parallel to \(\vec{b}\) is given by \(\vec{r}=\vec{a}+\lambda \vec{b}\), where \(\lambda\) is a constant \(\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})\).
This is the required equation of the line.
Question 5. Find the equation of the line in vector and in Cartesian form’ that passes through the point with position vector \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) and is in the direction \(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\).
Solution:
⇒ \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
It is known that a line through a point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is given by the equation \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{i}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
This is the required equation of the line in vector form.
⇒ \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k} \Rightarrow x \hat{i}+y \hat{j}+z \hat{k}=(\lambda+2) \hat{i}+(2 \lambda-1) \hat{j}+(-\lambda+4) \hat{k}\)
Eliminating \(\lambda\), we obtain the Cartesian form equation as \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)
This is the required equation of the given time in Cartesian form.
Question 6. Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Solution:
It is given that the line passes through the point (-2, 4, -5) and is parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
The direction rations of the line, \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) are (3, 5, 6).
The required line is parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0
It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios, a, b, c is given by \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
Therefore the equation of the required line is \(\frac{x+2}{3 k}=\frac{y-4}{5 k}=\frac{z+5}{6 k} \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k\)
Question 7. The Cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its Vector form.
Solution:
The Cartesian equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)
The given line passes through the point (5, -4, 6), The position vector of this point is \(\overrightarrow{\mathrm{a}}=5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of the vector, \(\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
It is known that the line through position vector \(\vec{a}\) and in the direction of the vector \(\vec{b}\) is given by the equation, \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}, \lambda \in \mathrm{R} \Rightarrow \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)
This is the required equation of the given line in vector form.
Question 8. Find the angle between the following pairs of lines:
- \(\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) and \({\overrightarrow{\mathrm{r}}}=7 \hat{\mathrm{i}}-6 \hat{\mathrm{k}}+\mu(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)
- \(\overrightarrow{\mathrm{r}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\) and \(\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-56 \hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})\)
Solution:
1. Let θ be the angle between the given lines.
The angle between the given pairs of lines is given by, \(\cos \theta=\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}\right|\)
The given lines are parallel to the vectors, \(\vec{b}_1=3 \hat{i}+2 \hat{j}+6 \hat{k}\) and \(\vec{b}_2=\hat{i}+2 \hat{j}+2 \hat{k}\) respectively.
∴ \(\left|\vec{b}_1\right|=\sqrt{3^2+2^2+6^2}=7\)
∴ \(\left|\vec{b}_2\right|=\sqrt{(1)^2+(2)^2+(2)^2}=3\)
⇒ \(\vec{b}_1 \cdot \vec{b}_2=(3 \hat{i}+2 \hat{j}+6 \hat{k})(\hat{i}+2 \hat{j}+2 \hat{k})=3 \times 1+2 \times 2+6 \times 2=3+4+12=19\)
⇒ \(\cos \theta=\frac{19}{7 \times 3} \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)
2. The given lines are parallel to the vectors, \(\vec{b}_1=\hat{i}-\hat{j}-2 \hat{k}\) and \(\vec{b}_2=3 \hat{i}-5 \hat{j}-4 \hat{k}\) respectively
∴ \(\left|\vec{b}_1\right|=\sqrt{(1)^2+(-1)^2+(-2)^2}=\sqrt{6}\)
∴ \(\left|\vec{b}_2\right|=\sqrt{(3)^2+(-5)^2+(-4)^2}=\sqrt{50}=5 \sqrt{2}\)
∴ \(\vec{b}_1 \vec{b}_2=(\hat{i}-\hat{j}-2 \hat{k})(3 \hat{i}-5 \hat{j}-4 \hat{k})=1.3-1(-5)-2(-4)=3+5+8=16\)
⇒ \(\cos \theta=\left|\frac{\vec{b}_1, \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}\right|\)
⇒ \(\cos \theta=\frac{16}{\sqrt{6} \cdot 5 \sqrt{2}}=\frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5 \sqrt{2}}=\frac{16}{10 \sqrt{3}} \Rightarrow \cos \theta=\frac{8}{5 \sqrt{3}} \Rightarrow \theta=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)\)
Question 9. Find the angle between the following pairs of lines :
- \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
- \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution:
1. Let \(\vec{b}_1\) and \(\vec{b}_2\) be the vectors parallel to the pair of lines, \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\), respectively
⇒ \(\vec{b}_1=2 \hat{i}+5 \hat{j}-3 \hat{k}\) and \(\vec{b}_2=-\hat{i}+8 \hat{j}+4 \hat{k}\)
∴ \(\left|\vec{b}_1\right|=\sqrt{(2)^2+(5)^2+(-3)^2}=\sqrt{38}\)
∴ \(\left|\vec{b}_2\right|=\sqrt{(-1)^2+(8)^2+(4)^2}=\sqrt{81}=9\)
∴ \(\vec{b}_1 \vec{b}_2=(2 \hat{i}+5 \hat{j}-3 \hat{k}) \cdot(-\hat{i}+8 \hat{j}+4 \hat{k})=2(-1)+5 \times 8+(-3) \cdot 4=-2+40-12=26\)
The angle \(\theta\) between the given pair of lines is given by the relation,
⇒ \(\cos \theta=\left|\frac{\vec{b}_1, \vec{b}_2}{\left|\overrightarrow{\mathrm{b}}_1\right|\left|\overrightarrow{\mathrm{b}}_2\right|}\right| \Rightarrow \cos \theta=\frac{26}{9 \sqrt{38}} \Rightarrow \theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)\)
2. Let \(\vec{b}_1\) and \(\vec{b}_2\) be the vectors parallel to the given pair of lines, \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\), respectively.
⇒ \(\vec{b}_1=2 \hat{i}+2 \hat{j}+\hat{k} \text { and } \vec{b}_2=4 \hat{i}+\hat{j}+8 \hat{k}\)
⇒ \(\left|\vec{b}_1\right|=\sqrt{(2)^2+(2)^2+(1)^2}=\sqrt{9}=3\)
⇒ \(\left|\vec{b}_2\right|=\sqrt{4^2+1^2+8^2}=\sqrt{81}=9\)
⇒ \(\vec{b}_1, \vec{b}_2=(2 \hat{i}+2 \hat{j}+\hat{k}) \cdot(4 \hat{i}+\hat{j}+8 \hat{k})=2 \times 4+2 \times 1+1 \times 8=8+2+8=18\)
If \(\theta\) is the angle between the given pair of lines, then
∴ \(\cos \theta=\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right| \mid \vec{b}_2}\right| \Rightarrow \cos \theta=\frac{18}{3 \times 9}=\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
Question 10. Find the values of p so that the lines \(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} \text { and } \frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Solution:
The given equations can be written in the standard form as \(\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{\frac{-3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}\)
The direction ratios of the lines are – 3, 2p/7, 2, and -3p/7,1,-5 respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other, if a1a2 + b1b2 + c1c2 = 0
∴ (-3) \(\cdot\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right) \cdot(1)+2(-5)=0\)
⇒ \(\frac{9 p}{7}+\frac{2 p}{7}=10 \quad \Rightarrow 11 p=70 \quad \Rightarrow p=\frac{70}{11}\)
Thus, the value of p is \(\frac{70}{11}\)
Question 11. Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
Solution:
The equations of the given lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
The direction ratios of the given lines are 7, -5, 1, and 1, 2, and 3 respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other, if a1a2+ b1b2+ c1c2= 0
∴ 7 x 1 + (-5) x 2+1 x 3 = 7- 10 + 3 = 0
Therefore, the given lines are perpendicular to each other.
Question 12. Find the shortest distance between the lines: \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \text { and } \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)
Solution:
The equations of the given lines are \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \text { and } \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)
It is known that the shortest distance between the lines, \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\) and \(\vec{r}=\vec{a}_2+\mu \vec{b}_2\) is given by
d = \(\left|\frac{\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)……(1)
Comparing the given equations, we obtain
∴ \(\vec{a}_1=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}_1=\hat{i}-\hat{j}+\hat{k}, \vec{a}_2=2 \hat{i}-\hat{j}-\hat{k}, \vec{b}_2=2 \hat{i}+\hat{j}+2 \hat{k}\)
⇒ \(\vec{a}_2-\vec{a}_1=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k}\)
⇒ \(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\)
= \(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 1 \\
2 & 1 & 2
\end{array}\right|\)
⇒ \(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=(-2-1) \hat{\mathrm{i}}-(2-2) \hat{\mathrm{j}}+(1+2) \hat{\mathrm{k}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}\)
⇒ \(\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|=\sqrt{(-3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
Substituting all the values in equation (1), we obtain
d = \(\left|\frac{(-3 \hat{i}+3 \hat{k}) \cdot(\hat{i}-3 \hat{j}-2 \hat{k})}{3 \sqrt{2}}\right| \Rightarrow d=\left|\frac{-3 \cdot 1+3(-2)}{3 \sqrt{2}}\right| \Rightarrow d=\left|\frac{-9}{3 \sqrt{2}}\right|\)
⇒ d = \(\frac{3}{\sqrt{2}}=\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3 \sqrt{2}}{2}\)
Therefore, the shortest distance between the two lines is \(\frac{3 \sqrt{2}_2^{-}}{2}\) units.
Question 13. Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
The given lines are \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
It is known that the shortest distance between the two lines,
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) is given as :
d = \(\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\)…..(1)
Comparing the given equations, we obtain
⇒ \(x_1=-1, y_1=-1, z_1=-1 ; x_2=3, y_2=5, z_2=7\)
⇒ \(a_1=7, b_1=-6, c_1=1 ; a_2=1, b_2=-2, c_2=1\)
Then, \(\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|\)
= \(4(-6+2)-6(7-1)+8(-14+6)=-16 -36-64=-116\)
⇒ \(\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}=\sqrt{(-6+2)^2+(1-7)^2+(-14+6)^2}\)
= \(\sqrt{16+36+64}=\sqrt{116}=2 \sqrt{29}\)
Substituting all the values in equation (1), we obtain
d = \(\left|\frac{-116}{\sqrt{116}}\right|=\frac{116}{\sqrt{116}}=2 \sqrt{29}\)
Since distance is always non-negative, the distance between the given lines is \(2 \sqrt{29}\) units.
Question 14. Find the shortest distance between the lines whose vector equations are: \(\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\) and \(\vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})\)
Solution:
The given lines are \(\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\) and \(\vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})\)
It is known that the shortest distance between the lines, \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\) and \(\vec{r}=\vec{a}_2+\mu \vec{b}_2\) is given by,
d = \(\left|\frac{\left.\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right) \cdot\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right)}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}\right|\)…….(1)
Comparing the given equations with \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_1+\lambda \overrightarrow{\mathrm{b}}_1\) and \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_2+\mu \overrightarrow{\mathrm{b}}_2\); we have:
⇒ \(\vec{a}_1=\hat{i}+2 \hat{j}+3 \hat{k} ; \vec{a}_2=4 \hat{i}+5 \hat{j}+6 \hat{k}\)
⇒ \(\vec{b}_1=\hat{i}-3 \hat{j}+2 \hat{k} ; \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k}\)
⇒ \(\vec{a}_2-\vec{a}_1=(4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+3 \hat{j}+3 \hat{k}\)
⇒ \(\vec{b}_1 \times \vec{b}_2\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 2 \\
2 & 3 & 1
\end{array}\right|\)
= \((-3-6 \hat{i}-(1-4) \hat{j}+(3+6) \hat{k}=-9 \hat{i}+3 \hat{j}+9 \hat{k}\)
⇒ \(\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{81+9+81}=\sqrt{171}=3 \sqrt{19}\)
⇒ \(\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)=(-9 \hat{i}+3 \hat{j}+9 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+3 \hat{k})=-9 \times 3+3 \times 3+9 \times 3=9\)
Substituting all the values in equation (1), we obtain
d = \(\left|\frac{9}{3 \sqrt{19}}\right|=\frac{3}{\sqrt{19}}\)
Therefore, the shortest distance between the two given lines is \(\frac{3}{\sqrt{19}}\) units.
Question 15. Find the shortest distance between the lines whose vector equations are \(\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}\) and \(\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}\)
Solution:
The given lines are
⇒ \(\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}\)
⇒ \(\vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})\)….(1)
⇒ \(\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}\)
⇒ \(\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})\)…..(2)
It is known that the shortest distance between the lines, \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\) and \(\vec{r}=\vec{a}_2+\mu \vec{b}_2\) is given by
d = \(\left|\frac{\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)……..(3)
For the given equations,
⇒ \(\vec{a}_1=\hat{i}-2 \hat{j}+3 \hat{k} ; \vec{a}_2=\hat{i}-\hat{j}-\hat{k}\)
⇒ \(\vec{b}_1=-\hat{i}+\hat{j}-2 \hat{k} ; \vec{b}_2=\hat{i}+2 \hat{j}-2 \hat{k}\)
⇒ \(\vec{a}_2-\vec{a}_1=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\hat{j}-4 \hat{k}\)
⇒ \(\vec{b}_1 \times \vec{b}_2\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & -2 \\
1 & 2 & -2
\end{array}\right|\)
= \((-2+4) \hat{i}-(2+2) \hat{j}+(-2-1) \hat{k}=2 \hat{i}-4 \hat{j}-3 \hat{k}\)
⇒ \(\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{(2)^2+(-4)^2+(-3)^2}=\sqrt{4+16+9}=\sqrt{29}\)
∴ \(\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)=(2 \hat{i}-4 \hat{j}-3 \hat{k}) \cdot(\hat{j}-4 \hat{k})=-4+12=8\)
Substituting all the values in equation (3), we, obtain \(\mathrm{d}=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}}\)
Therefore, the shortest distance between the lines is \(\frac{8}{\sqrt{29}}\) units.
Three-Dimensional Geometry Miscellaneous Exercise
Question 1. Find the angle between the lines whose direction ratios are (a, b, c) and (b – c, c – a, a -b).
Solution:
The angle θ between the lines with direction cosines, (a, b, c) and (b – c, c – a, a -b), is given by,
cos \(\theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2}+\sqrt{a_2^2+b_2^2+c_2^2}}\right| \Rightarrow \cos \theta=\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}+\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}\right|\)
⇒cos \(\theta=0 \Rightarrow \theta=\cos ^{-1}(0) \Rightarrow \theta=90^{\circ}\)
Thus, the angle between the lines is 90°.
Question 2. Find the equation of a line parallel to the x-axis and passing through the origin.
Solution:
The line parallel to the x-axis and passing through the origin is the x-axis itself.
Let A be a point on the x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a ∈ R.
The direction ratios of OA are (a – 0), (0 – 0), (0 – 0) i.e. a, 0, 0
The equation of OA is given by \(\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0} \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\)
Thus, the equation of the line parallel to the x-axis and passing through the origin is \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\)
Question 3. If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution:
The direction ratios of the lines, \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\), are -3,2 k, 2 and 3 k, 1,-5 respectively.
It is known that two lines with direction ratios, \(\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1\) and \(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2\), are perpendicular if \(a_1 a_2+b_1 b_2+c_1 c_2=0\)
∴ -3(3k) + 2kx 1 + 2(-5) = 0 => ~9k + 2k-10 = 0 => 7k=-10 => k = -10/7
Question 4. Find the shortest distance between lines \(\overrightarrow{\mathrm{r}}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and \(\overrightarrow{\mathrm{r}}=-4 \hat{\mathrm{i}}-\hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
Solution:
The given lines are \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\)…..(1)
⇒ \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)….(2)
It is known that the shortest distance between two lines, \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\), and \(\vec{r}=\vec{a}_2+\mu \vec{b}_2\) is given by
d = \(\left|\frac{\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right) \cdot\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right)}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}\right|\)……(3)
Comparing \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_1+\lambda \overrightarrow{\mathrm{b}}_1\) and \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_2+\mu \overrightarrow{\mathrm{b}}_2\) to equations (1) and (2), we obtain
⇒ \(\vec{a}_1=6 \hat{i}+2 \hat{j}+2 \hat{k}, \vec{b}_1=\hat{i}-2 \hat{j}+2 \hat{k}, \vec{a}_2=-4 \hat{i}-\hat{k}, \vec{b}_2=3 \hat{i}-2 \hat{j}-2 \hat{k}\)
⇒ \(\vec{a}_2-\vec{a}_1=(-4 \hat{i}-\hat{k})-(6 \hat{i}+2 \hat{j}+2 \hat{k})=-10 \hat{i}-2 \hat{j}-3 \hat{k}\)
⇒ \(\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
i & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right|\)
= \((4+4) \hat{i}-(-2-6) \hat{j}+(-2+6) \hat{k}=8 \hat{i}+8 \hat{j}+4 \hat{k}\)
⇒ \(\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{(8)^2+(8)^2+(4)^2}=12\)
⇒ \(\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)=(8 \hat{i}+8 \hat{j}+4 \hat{k}) \cdot(-10 \hat{i}-2 \hat{j}-3 \hat{k})=-80-16-12=-108\)
Substituting all the values in equation (3), we obtain \(\mathrm{d}=\left|\frac{-108}{12}\right|=9\) units
Therefore, the shortest distance between the two given lines is 9 units.
Question 5. Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines: \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)
Solution:
Let the required line be parallel to the vector \(\vec{b}\) given by, \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
The position vector of the point (1,2,-4) is \(\vec{a}=\hat{i}+2 \hat{j}-4 \hat{k}\)
The equation of the line passing through (1,2,-4) and parallel to vector \(\vec{b}\) is \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})+\lambda\left(\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}+\mathrm{b}_3 \hat{\mathrm{k}}\right)\)……..(1)
The equations of the lines are \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\)…..(2)
and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)…..(3)
Lines (1) and line (2) are perpendicular to each other \(3 b_1-16 b_2+7 b_3=0\)…..(4)
Also, lines (1) and line (3) are perpendicular to each other. \(3 b_1+8 b_2-5 b_3=0\)…..(5)
From equations (4) and (5), we obtain
\(\frac{b_1}{(-16)(-5)-8 \times 7}=\frac{b_2}{7 \times 3-3(-5)}=\frac{b_3}{3 \times 8-3(-16)} \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72} \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\)∴ Direction ratios of \(\overrightarrow{\mathrm{b}}\) are 2,3, and 6.
∴ \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}\)
Substituting \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) in equation (1), we obtain \(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
This is the equation of the required line.