Three Dimensional Geometry — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 11, Three Dimensional Geometry (ত্ৰিবিমিতীয় জ্যামিতি). It covers Exercise 11.1, Exercise 11.2 and the Miscellaneous Exercise on Chapter 11, working out direction cosines and direction ratios, equations of lines in space, the angle between two lines and the shortest distance between skew and parallel lines, so you can follow every step and prepare confidently for your examination.
Summary
The direction cosines of a line are the cosines $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$ of the angles the line makes with the positive $x$, $y$ and $z$-axes. They always satisfy $l^2 + m^2 + n^2 = 1$. Any three numbers $a$, $b$, $c$ proportional to $l$, $m$, $n$ are called the direction ratios of the line; from them the direction cosines are recovered as $l = \pm\frac{a}{\sqrt{a^2 + b^2 + c^2}}$, $m = \pm\frac{b}{\sqrt{a^2 + b^2 + c^2}}$, $n = \pm\frac{c}{\sqrt{a^2 + b^2 + c^2}}$. The direction ratios of the segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are $x_2 – x_1$, $y_2 – y_1$, $z_2 – z_1$, and dividing by $PQ$ gives its direction cosines.
A line through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ has vector equation $\vec{r} = \vec{a} + \lambda\vec{b}$, and, taking $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and direction ratios $a$, $b$, $c$, the Cartesian equation $\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$. If $\theta$ is the acute angle between two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$, then $\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}$; the lines are perpendicular when $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ and parallel when $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Skew lines are lines in space that are neither parallel nor intersecting. The shortest distance between $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ is $d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right|$, and for two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ the distance is $d = \frac{|\vec{b} \times (\vec{a}_2 – \vec{a}_1)|}{|\vec{b}|}$.
Summary: ASSEB Class 12 Mathematics Chapter 11 Three Dimensional Geometry explains direction cosines and direction ratios, the vector and Cartesian equations of a line in space, the angle between two lines, and the shortest distance between skew and parallel lines, with complete worked answers to Exercise 11.1, Exercise 11.2 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 11.1
A directed line makes angles $\alpha$, $\beta$, $\gamma$ with the positive $x$, $y$, $z$-axes; its direction cosines are $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$, and they always satisfy $l^2 + m^2 + n^2 = 1$.
1. If a line makes angles $90°$, $135°$, $45°$ with the $x$, $y$ and $z$-axes respectively, find its direction cosines.
Answer: The direction cosines are the cosines of the given angles:
$$l = \cos 90° = 0, \qquad m = \cos 135° = -\frac{1}{\sqrt{2}}, \qquad n = \cos 45° = \frac{1}{\sqrt{2}}$$
Hence the direction cosines are $0,\ -\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}$. As a check, $l^2 + m^2 + n^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$, as required.
2. Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer: If the line makes the same angle $\alpha$ with each axis, then $l = m = n = \cos\alpha$. Using $l^2 + m^2 + n^2 = 1$,
$$\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \implies 3\cos^2\alpha = 1 \implies \cos\alpha = \pm\frac{1}{\sqrt{3}}$$
Therefore the direction cosines are $\pm\frac{1}{\sqrt{3}},\ \pm\frac{1}{\sqrt{3}},\ \pm\frac{1}{\sqrt{3}}$ (the same sign throughout, depending on the direction chosen).
3. If a line has the direction ratios $-18$, $12$, $-4$, then what are its direction cosines?
Answer: With direction ratios $a = -18$, $b = 12$, $c = -4$, first find $\sqrt{a^2 + b^2 + c^2}$:
$$\sqrt{(-18)^2 + 12^2 + (-4)^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$$
Dividing each direction ratio by $22$ gives the direction cosines:
$$l = \frac{-18}{22} = -\frac{9}{11}, \qquad m = \frac{12}{22} = \frac{6}{11}, \qquad n = \frac{-4}{22} = -\frac{2}{11}$$
So the direction cosines are $-\frac{9}{11},\ \frac{6}{11},\ -\frac{2}{11}$.
4. Show that the points $(2, 3, 4)$, $(-1, -2, 1)$, $(5, 8, 7)$ are collinear.
Answer: Let $A(2, 3, 4)$, $B(-1, -2, 1)$ and $C(5, 8, 7)$. The direction ratios of a segment joining two points are the differences of their coordinates.
Direction ratios of $AB$: $-1 – 2,\ -2 – 3,\ 1 – 4 = -3,\ -5,\ -3$.
Direction ratios of $BC$: $5 – (-1),\ 8 – (-2),\ 7 – 1 = 6,\ 10,\ 6$.
Since $6,\ 10,\ 6 = -2\,(-3,\ -5,\ -3)$, the direction ratios of $AB$ and $BC$ are proportional, so $AB$ is parallel to $BC$. As the point $B$ is common to both, the three points $A$, $B$, $C$ lie on the same straight line, i.e. they are collinear.
5. Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4)$, $(-1, 1, 2)$ and $(-5, -5, -2)$.
Answer: Let the vertices be $A(3, 5, -4)$, $B(-1, 1, 2)$ and $C(-5, -5, -2)$. For each side, find the direction ratios (coordinate differences) and divide by the length of that side.
Side $AB$: direction ratios $-1 – 3,\ 1 – 5,\ 2 – (-4) = -4,\ -4,\ 6$, and
$$AB = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$$
so its direction cosines are $\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}$.
Side $BC$: direction ratios $-5 – (-1),\ -5 – 1,\ -2 – 2 = -4,\ -6,\ -4$, and
$$BC = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$$
so its direction cosines are $\frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{3}{\sqrt{17}},\ -\frac{2}{\sqrt{17}}$.
Side $CA$: direction ratios $3 – (-5),\ 5 – (-5),\ -4 – (-2) = 8,\ 10,\ -2$, and
$$CA = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$$
so its direction cosines are $\frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}} = \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ -\frac{1}{\sqrt{42}}$.
Exercise 11.2
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.
Answer: Two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are perpendicular when $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$. Take the three lines as $L_1, L_2, L_3$ in the order given.
$$L_1 \cdot L_2 = \frac{12 \times 4 + (-3) \times 12 + (-4) \times 3}{169} = \frac{48 – 36 – 12}{169} = 0$$
$$L_2 \cdot L_3 = \frac{4 \times 3 + 12 \times (-4) + 3 \times 12}{169} = \frac{12 – 48 + 36}{169} = 0$$
$$L_3 \cdot L_1 = \frac{3 \times 12 + (-4) \times (-3) + 12 \times (-4)}{169} = \frac{36 + 12 – 48}{169} = 0$$
Each pair gives $0$, so the three lines are mutually perpendicular.
2. Show that the line through the points $(1, -1, 2)$, $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.
Answer: Direction ratios of the first line: $3 – 1,\ 4 – (-1),\ -2 – 2 = 2,\ 5,\ -4$. Direction ratios of the second line: $3 – 0,\ 5 – 3,\ 6 – 2 = 3,\ 2,\ 4$. The lines are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$:
$$2 \times 3 + 5 \times 2 + (-4) \times 4 = 6 + 10 – 16 = 0$$
Since the sum is $0$, the two lines are perpendicular to each other.
3. Show that the line through the points $(4, 7, 8)$, $(2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1)$, $(1, 2, 5)$.
Answer: Direction ratios of the first line: $2 – 4,\ 3 – 7,\ 4 – 8 = -2,\ -4,\ -4$. Direction ratios of the second line: $1 – (-1),\ 2 – (-2),\ 5 – 1 = 2,\ 4,\ 4$. For parallel lines the ratios must be proportional:
$$\frac{-2}{2} = \frac{-4}{4} = \frac{-4}{4} = -1$$
All three ratios equal $-1$, so the direction ratios are proportional and the two lines are parallel.
4. Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat{i} + 2\hat{j} – 2\hat{k}$.
Answer: Here $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}$. The vector equation $\vec{r} = \vec{a} + \lambda\vec{b}$ is
$$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} – 2\hat{k})$$
Taking the direction ratios $3, 2, -2$ and the point $(1, 2, 3)$, the Cartesian equation is
$$\frac{x – 1}{3} = \frac{y – 2}{2} = \frac{z – 3}{-2}$$
5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{i} – \hat{j} + 4\hat{k}$ and is in the direction $\hat{i} + 2\hat{j} – \hat{k}$.
Answer: Here $\vec{a} = 2\hat{i} – \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} – \hat{k}$, so the vector equation is
$$\vec{r} = (2\hat{i} – \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} – \hat{k})$$
The point is $(2, -1, 4)$ with direction ratios $1, 2, -1$, so the Cartesian equation is
$$\frac{x – 2}{1} = \frac{y + 1}{2} = \frac{z – 4}{-1}$$
6. Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}$.
Answer: The given line has direction ratios $3, 5, 6$. A parallel line has the same direction ratios, and it passes through $(-2, 4, -5)$, so its Cartesian equation is
$$\frac{x + 2}{3} = \frac{y – 4}{5} = \frac{z + 5}{6}$$
7. The cartesian equation of a line is $\frac{x – 5}{3} = \frac{y + 4}{7} = \frac{z – 6}{2}$. Write its vector form.
Answer: Comparing with $\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$, the line passes through the point $(5, -4, 6)$ and has direction ratios $3, 7, 2$. Hence $\vec{a} = 5\hat{i} – 4\hat{j} + 6\hat{k}$ and $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$, and the vector form is
$$\vec{r} = (5\hat{i} – 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$$
8. Find the angle between the following pairs of lines:
(i) $\vec{r} = 2\hat{i} – 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$ and $\vec{r} = 7\hat{i} – 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$
Answer: The direction vectors are $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$. The angle $\theta$ satisfies $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}$.
$$\vec{b}_1 \cdot \vec{b}_2 = 3 + 4 + 12 = 19, \quad |\vec{b}_1| = \sqrt{9 + 4 + 36} = 7, \quad |\vec{b}_2| = \sqrt{1 + 4 + 4} = 3$$
$$\cos\theta = \frac{19}{7 \times 3} = \frac{19}{21} \implies \theta = \cos^{-1}\!\left(\frac{19}{21}\right)$$
(ii) $\vec{r} = 3\hat{i} + \hat{j} – 2\hat{k} + \lambda(\hat{i} – \hat{j} – 2\hat{k})$ and $\vec{r} = 2\hat{i} – \hat{j} – 56\hat{k} + \mu(3\hat{i} – 5\hat{j} – 4\hat{k})$
Answer: The direction vectors are $\vec{b}_1 = \hat{i} – \hat{j} – 2\hat{k}$ and $\vec{b}_2 = 3\hat{i} – 5\hat{j} – 4\hat{k}$.
$$\vec{b}_1 \cdot \vec{b}_2 = 3 + 5 + 8 = 16, \quad |\vec{b}_1| = \sqrt{1 + 1 + 4} = \sqrt{6}, \quad |\vec{b}_2| = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$$
$$\cos\theta = \frac{16}{\sqrt{6} \times 5\sqrt{2}} = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}} = \frac{8\sqrt{3}}{15}$$
Hence $\theta = \cos^{-1}\!\left(\frac{8\sqrt{3}}{15}\right)$.
9. Find the angle between the following pair of lines:
(i) $\frac{x – 2}{2} = \frac{y – 1}{5} = \frac{z + 3}{-3}$ and $\frac{x + 2}{-1} = \frac{y – 4}{8} = \frac{z – 5}{4}$
Answer: Direction ratios are $\vec{b}_1 = (2, 5, -3)$ and $\vec{b}_2 = (-1, 8, 4)$.
$$\vec{b}_1 \cdot \vec{b}_2 = 2(-1) + 5(8) + (-3)(4) = -2 + 40 – 12 = 26$$
$$|\vec{b}_1| = \sqrt{4 + 25 + 9} = \sqrt{38}, \quad |\vec{b}_2| = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$$
$$\cos\theta = \frac{26}{9\sqrt{38}} \implies \theta = \cos^{-1}\!\left(\frac{26}{9\sqrt{38}}\right)$$
(ii) $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x – 5}{4} = \frac{y – 2}{1} = \frac{z – 3}{8}$
Answer: Direction ratios are $\vec{b}_1 = (2, 2, 1)$ and $\vec{b}_2 = (4, 1, 8)$.
$$\vec{b}_1 \cdot \vec{b}_2 = 2(4) + 2(1) + 1(8) = 8 + 2 + 8 = 18$$
$$|\vec{b}_1| = \sqrt{4 + 4 + 1} = 3, \quad |\vec{b}_2| = \sqrt{16 + 1 + 64} = 9$$
$$\cos\theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3} \implies \theta = \cos^{-1}\!\left(\frac{2}{3}\right)$$
10. Find the values of $p$ so that the lines $\frac{1 – x}{3} = \frac{7y – 14}{2p} = \frac{z – 3}{2}$ and $\frac{7 – 7x}{3p} = \frac{y – 5}{1} = \frac{6 – z}{5}$ are at right angles.
Answer: First put each line in the standard form $\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$.
Line 1: $\frac{1 – x}{3} = \frac{x – 1}{-3}$ and $\frac{7y – 14}{2p} = \frac{y – 2}{2p/7}$, so its direction ratios are $-3,\ \frac{2p}{7},\ 2$.
Line 2: $\frac{7 – 7x}{3p} = \frac{x – 1}{-3p/7}$ and $\frac{6 – z}{5} = \frac{z – 6}{-5}$, so its direction ratios are $-\frac{3p}{7},\ 1,\ -5$.
For right angles, $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$:
$$(-3)\!\left(-\frac{3p}{7}\right) + \frac{2p}{7}(1) + 2(-5) = \frac{9p}{7} + \frac{2p}{7} – 10 = \frac{11p}{7} – 10 = 0$$
$$\frac{11p}{7} = 10 \implies p = \frac{70}{11}$$
11. Show that the lines $\frac{x – 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.
Answer: Direction ratios of the lines are $7, -5, 1$ and $1, 2, 3$. Then
$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 7(1) + (-5)(2) + 1(3) = 7 – 10 + 3 = 0$$
Since the sum is $0$, the two lines are perpendicular to each other.
12. Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} – \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} – \hat{j} – \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})$.
Answer: Compare with $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$: here $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b}_1 = \hat{i} – \hat{j} + \hat{k}$, $\vec{a}_2 = 2\hat{i} – \hat{j} – \hat{k}$, $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$. So $\vec{a}_2 – \vec{a}_1 = \hat{i} – 3\hat{j} – 2\hat{k}$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$$
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1) = (-3)(1) + 0(-3) + 3(-2) = -9$$
$$d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right| = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$$
The shortest distance is $\frac{3\sqrt{2}}{2}$ units.
13. Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$ and $\frac{x – 3}{1} = \frac{y – 5}{-2} = \frac{z – 7}{1}$.
Answer: The first line passes through $\vec{a}_1 = (-1, -1, -1)$ with direction $\vec{b}_1 = (7, -6, 1)$; the second through $\vec{a}_2 = (3, 5, 7)$ with direction $\vec{b}_2 = (1, -2, 1)$. So $\vec{a}_2 – \vec{a}_1 = (4, 6, 8)$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -4\hat{i} – 6\hat{j} – 8\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{116} = 2\sqrt{29}$$
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1) = (-4)(4) + (-6)(6) + (-8)(8) = -16 – 36 – 64 = -116$$
$$d = \frac{|-116|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$$
The shortest distance is $2\sqrt{29}$ units.
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 $\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu(2\hat{i} + 3\hat{j} + \hat{k})$.
Answer: Here $\vec{a}_1 = (1, 2, 3)$, $\vec{b}_1 = (1, -3, 2)$, $\vec{a}_2 = (4, 5, 6)$, $\vec{b}_2 = (2, 3, 1)$, so $\vec{a}_2 – \vec{a}_1 = (3, 3, 3)$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = -9\hat{i} + 3\hat{j} + 9\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{171} = 3\sqrt{19}$$
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1) = (-9)(3) + 3(3) + 9(3) = -27 + 9 + 27 = 9$$
$$d = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$$
The shortest distance is $\frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$ units.
15. Find the shortest distance between the lines whose vector equations are $\vec{r} = (1 – t)\hat{i} + (t – 2)\hat{j} + (3 – 2t)\hat{k}$ and $\vec{r} = (s + 1)\hat{i} + (2s – 1)\hat{j} – (2s + 1)\hat{k}$.
Answer: Group each equation into a fixed part plus a parameter multiple.
$$\vec{r} = (\hat{i} – 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} – 2\hat{k}), \qquad \vec{r} = (\hat{i} – \hat{j} – \hat{k}) + s(\hat{i} + 2\hat{j} – 2\hat{k})$$
So $\vec{a}_1 = (1, -2, 3)$, $\vec{b}_1 = (-1, 1, -2)$, $\vec{a}_2 = (1, -1, -1)$, $\vec{b}_2 = (1, 2, -2)$, giving $\vec{a}_2 – \vec{a}_1 = (0, 1, -4)$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = 2\hat{i} – 4\hat{j} – 3\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$$
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1) = 2(0) + (-4)(1) + (-3)(-4) = -4 + 12 = 8$$
$$d = \frac{|8|}{\sqrt{29}} = \frac{8}{\sqrt{29}} = \frac{8\sqrt{29}}{29}$$
The shortest distance is $\frac{8}{\sqrt{29}} = \frac{8\sqrt{29}}{29}$ units.
Miscellaneous Exercise on Chapter 11
1. Find the angle between the lines whose direction ratios are $a$, $b$, $c$ and $b – c$, $c – a$, $a – b$.
Answer: If $\theta$ is the angle between the lines, then $\cos\theta$ is proportional to $a_1 a_2 + b_1 b_2 + c_1 c_2$. Here
$$a(b – c) + b(c – a) + c(a – b) = ab – ac + bc – ab + ac – bc = 0$$
Since the sum of the products of corresponding direction ratios is $0$, we get $\cos\theta = 0$, so $\theta = 90°$. The two lines are perpendicular.
2. Find the equation of a line parallel to $x$-axis and passing through the origin.
Answer: A line parallel to the $x$-axis is parallel to the unit vector $\hat{i}$, so its direction ratios are $1, 0, 0$. It passes through the origin $\vec{a} = \vec{0}$, so its vector equation is
$$\vec{r} = \vec{0} + \lambda\hat{i} = \lambda\hat{i}$$
In Cartesian form, using the point $(0, 0, 0)$ and direction ratios $1, 0, 0$, the line is $\frac{x – 0}{1} = \frac{y – 0}{0} = \frac{z – 0}{0}$, i.e. $y = 0$ and $z = 0$. This is precisely the $x$-axis itself.
3. If the lines $\frac{x – 1}{-3} = \frac{y – 2}{2k} = \frac{z – 3}{2}$ and $\frac{x – 1}{3k} = \frac{y – 1}{1} = \frac{z – 6}{-5}$ are perpendicular, find the value of $k$.
Answer: Direction ratios of the two lines are $-3, 2k, 2$ and $3k, 1, -5$. For perpendicular lines, $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$:
$$(-3)(3k) + (2k)(1) + (2)(-5) = -9k + 2k – 10 = -7k – 10 = 0$$
$$-7k = 10 \implies k = -\frac{10}{7}$$
4. Find the shortest distance between lines $\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} – 2\hat{j} + 2\hat{k})$ and $\vec{r} = -4\hat{i} – \hat{k} + \mu(3\hat{i} – 2\hat{j} – 2\hat{k})$.
Answer: Here $\vec{a}_1 = (6, 2, 2)$, $\vec{b}_1 = (1, -2, 2)$, $\vec{a}_2 = (-4, 0, -1)$, $\vec{b}_2 = (3, -2, -2)$, so $\vec{a}_2 – \vec{a}_1 = (-10, -2, -3)$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = 8\hat{i} + 8\hat{j} + 4\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{144} = 12$$
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1) = 8(-10) + 8(-2) + 4(-3) = -80 – 16 – 12 = -108$$
$$d = \frac{|-108|}{12} = \frac{108}{12} = 9$$
The shortest distance is $9$ units.
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 $\frac{x – 15}{3} = \frac{y – 29}{8} = \frac{z – 5}{-5}$.
Answer: The required line is perpendicular to both given lines, so its direction vector is the cross product of their direction vectors $\vec{b}_1 = 3\hat{i} – 16\hat{j} + 7\hat{k}$ and $\vec{b}_2 = 3\hat{i} + 8\hat{j} – 5\hat{k}$.
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = 24\hat{i} + 36\hat{j} + 72\hat{k}$$
Dividing by the common factor $12$, a convenient direction vector is $2\hat{i} + 3\hat{j} + 6\hat{k}$. The line passes through $(1, 2, -4)$, i.e. $\vec{a} = \hat{i} + 2\hat{j} – 4\hat{k}$, so its vector equation is
$$\vec{r} = (\hat{i} + 2\hat{j} – 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$
Additional Questions and Answers
Multiple Choice Questions
1. If a line makes angles $90°$, $60°$ and $30°$ with the positive $x$, $y$ and $z$-axes respectively, its direction cosines are
(A) $0,\ \frac{1}{2},\ \frac{\sqrt{3}}{2}$ (B) $1,\ \frac{1}{2},\ \frac{\sqrt{3}}{2}$ (C) $0,\ \frac{1}{2},\ \frac{1}{2}$ (D) $\frac{1}{2},\ \frac{1}{2},\ \frac{1}{2}$
Answer: (A) $0,\ \frac{1}{2},\ \frac{\sqrt{3}}{2}$. The direction cosines are $\cos 90° = 0$, $\cos 60° = \frac{1}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$.
2. The direction cosines of the $x$-axis are
(A) $1, 0, 0$ (B) $0, 1, 0$ (C) $0, 0, 1$ (D) $1, 1, 1$
Answer: (A) $1, 0, 0$. The $x$-axis makes angles $0°$, $90°$, $90°$ with the axes, giving $\cos 0°, \cos 90°, \cos 90° = 1, 0, 0$.
3. If $l$, $m$, $n$ are the direction cosines of a line, then
(A) $l^2 + m^2 + n^2 = 0$ (B) $l + m + n = 1$ (C) $l^2 + m^2 + n^2 = 1$ (D) $l^2 + m^2 + n^2 = 3$
Answer: (C) $l^2 + m^2 + n^2 = 1$. This is the fundamental identity satisfied by direction cosines.
4. A line has direction ratios $1, -2, 2$. Its direction cosines are
(A) $\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$ (B) $1, -2, 2$ (C) $\frac{1}{9}, -\frac{2}{9}, \frac{2}{9}$ (D) $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$
Answer: (A) $\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$. Here $\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$, so divide each ratio by $3$.
5. If the lines with direction ratios $k, -3, 2$ and $1, 2, 3$ are perpendicular, then $k$ equals
(A) $0$ (B) $1$ (C) $-1$ (D) $6$
Answer: (A) $0$. Perpendicularity requires $k(1) + (-3)(2) + 2(3) = k – 6 + 6 = k = 0$.
6. The shortest distance between two intersecting lines is
(A) $0$ (B) $1$ (C) infinite (D) equal to their length
Answer: (A) $0$. Two lines that meet at a point have zero shortest distance between them.
7. Skew lines are lines in space that are
(A) parallel (B) intersecting (C) neither parallel nor intersecting (D) coincident
Answer: (C) neither parallel nor intersecting. Skew lines lie in different planes.
8. The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is
(A) $\vec{r} = \vec{a} \cdot \vec{b}$ (B) $\vec{r} = \vec{a} + \lambda\vec{b}$ (C) $\vec{r} = \vec{a} \times \vec{b}$ (D) $\vec{r} = \lambda(\vec{a} + \vec{b})$
Answer: (B) $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\lambda$ is a real parameter.
Fill in the Blanks
1. If $l$, $m$, $n$ are the direction cosines of a line, then $l^2 + m^2 + n^2 = $ ________.
Answer: $1$
2. The direction cosines of the $y$-axis are ________.
Answer: $0, 1, 0$
3. Two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are perpendicular when $a_1 a_2 + b_1 b_2 + c_1 c_2 = $ ________.
Answer: $0$
4. Lines in space which are neither parallel nor intersecting are called ________ lines.
Answer: skew
5. The direction ratios of the line joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are ________.
Answer: $x_2 – x_1,\ y_2 – y_1,\ z_2 – z_1$
True or False
1. The direction cosines of a line in space are unique.
Answer: False. A line has two sets of direction cosines that differ in sign; they become unique only when the line is taken as a directed line.
2. If $a$, $b$, $c$ are direction ratios of a line, then $ka$, $kb$, $kc$ (with $k \neq 0$) are also direction ratios of the same line.
Answer: True. A line has infinitely many sets of direction ratios, all proportional to one another.
3. The lines whose direction ratios are $a, b, c$ and $b – c, c – a, a – b$ are perpendicular.
Answer: True. $a(b – c) + b(c – a) + c(a – b) = 0$, so the angle between them is $90°$.
4. Two parallel lines have proportional direction ratios.
Answer: True. Parallel lines share the same direction, so their direction ratios are proportional.
5. The shortest distance between two parallel lines is always zero.
Answer: False. For parallel lines it equals the perpendicular distance between them, which is generally non-zero; it is zero only when the lines coincide.
Short Answer Questions
1. Find the direction cosines of a line whose direction ratios are $1, 2, -2$.
Answer: Here $\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$, so the direction cosines are $\frac{1}{3},\ \frac{2}{3},\ -\frac{2}{3}$.
2. Find the vector equation of the line passing through the point $(2, -1, 3)$ and parallel to the vector $2\hat{i} – \hat{j} + 4\hat{k}$.
Answer: With $\vec{a} = 2\hat{i} – \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} – \hat{j} + 4\hat{k}$, the vector equation is $\vec{r} = (2\hat{i} – \hat{j} + 3\hat{k}) + \lambda(2\hat{i} – \hat{j} + 4\hat{k})$.
3. Find the angle between two lines whose direction ratios are $1, 1, 2$ and $2, 1, 1$.
Answer: $\vec{b}_1 \cdot \vec{b}_2 = 1(2) + 1(1) + 2(1) = 5$, and $|\vec{b}_1| = |\vec{b}_2| = \sqrt{1 + 1 + 4} = \sqrt{6}$. Hence $\cos\theta = \frac{5}{\sqrt{6}\,\sqrt{6}} = \frac{5}{6}$, so $\theta = \cos^{-1}\!\left(\frac{5}{6}\right)$.
4. Find the direction cosines of the line joining the points $(1, 2, 3)$ and $(4, 5, 6)$.
Answer: Direction ratios are $4 – 1,\ 5 – 2,\ 6 – 3 = 3, 3, 3$, and the length is $\sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$. Dividing gives the direction cosines $\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}$.
Key Terms
| Term | Meaning |
|---|---|
| Direction angles | The angles $\alpha$, $\beta$, $\gamma$ that a directed line makes with the positive $x$, $y$ and $z$-axes. |
| Direction cosines | The cosines $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$ of the direction angles; they satisfy $l^2 + m^2 + n^2 = 1$. |
| Direction ratios | Any three numbers $a$, $b$, $c$ proportional to the direction cosines of a line; also called direction numbers. |
| Position vector | The vector $\vec{a}$ from the origin to a given point, used to locate a point on a line in space. |
| Vector equation of a line | $\vec{r} = \vec{a} + \lambda\vec{b}$, the position vector of a general point on the line through $\vec{a}$ parallel to $\vec{b}$. |
| Cartesian equation of a line | $\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$, obtained from the vector equation. |
| Angle between two lines | The acute angle $\theta$ with $\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}$. |
| Perpendicular lines | Lines for which $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, so that $\theta = 90°$. |
| Parallel lines | Lines whose direction ratios are proportional, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. |
| Skew lines | Lines in space that are neither parallel nor intersecting; they lie in different planes. |
| Shortest distance | For skew lines the length of the common perpendicular, $d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right|$. |