NCERT Class 12 Mathematics — Three Dimensional Geometry, Exercise 11.2. All 15 questions solved.

Everything in this exercise comes out of one object: a line described by a point on it and a vector along it. In vector form,

$$\vec r = \vec a + \lambda\vec b$$

and in Cartesian form, with $\vec a = (x_1, y_1, z_1)$ and $\vec b = (a, b, c)$,

$$\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$$

From those two, three formulas do the rest of the work:

$$\cos\theta = \left|\frac{\vec b_1 \cdot \vec b_2}{\left|\vec b_1\right|\left|\vec b_2\right|}\right| \qquad \text{(angle between two lines)}$$

$$d = \left|\frac{\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right)}{\left|\vec b_1 \times \vec b_2\right|}\right| \qquad \text{(skew lines)}$$

$$d = \left|\frac{\vec b \times \left(\vec a_2 – \vec a_1\right)}{\left|\vec b\right|}\right| \qquad \text{(parallel lines)}$$

Key insight. Read a Cartesian equation by matching it against the template. The numbers subtracted from $x$, $y$, $z$ give a point on the line; the numbers dividing give the direction. So $\tfrac{x-5}{3} = \tfrac{y+4}{7} = \tfrac{z-6}{2}$ passes through $(5, -4, 6)$ — note the sign flip on the $-4$ — with direction $(3, 7, 2)$.

The dangerous cases are questions 10 and 15, where the equation is not in template form. $\tfrac{1-x}{3}$ must first be rewritten as $\tfrac{x-1}{-3}$, and $\tfrac{7y-14}{2p}$ as $\tfrac{y-2}{2p/7}$, before any direction can be read off. Skipping that step is the single commonest error in this exercise.

Question 1

Show that the three lines with direction cosines $\tfrac{12}{13}, \tfrac{-3}{13}, \tfrac{-4}{13}$; $\tfrac{4}{13}, \tfrac{12}{13}, \tfrac{3}{13}$; $\tfrac{3}{13}, \tfrac{-4}{13}, \tfrac{12}{13}$ are mutually perpendicular.

Solution. Two lines are perpendicular exactly when $l_1l_2 + m_1m_2 + n_1n_2 = 0$. Take the three pairs in turn; the common denominator $169$ can be ignored since we only need the sum to vanish.

Lines 1 and 2: $12(4) + (-3)(12) + (-4)(3) = 48 – 36 – 12 = 0$ ✓

Lines 2 and 3: $4(3) + 12(-4) + 3(12) = 12 – 48 + 36 = 0$ ✓

Lines 1 and 3: $12(3) + (-3)(-4) + (-4)(12) = 36 + 12 – 48 = 0$ ✓

All three pairs are perpendicular, so the lines are mutually perpendicular.

Question 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 $(3, 5, 6)$.

Solution. Direction ratios of the first line: $3-1, 4+1, -2-2$, that is $2, 5, -4$. Direction ratios of the second: $3-0, 5-3, 6-2$, that is $3, 2, 4$.

$$2(3) + 5(2) + (-4)(4) = 6 + 10 – 16 = 0$$

The dot product vanishes, so the lines are perpendicular.

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. Direction ratios of the first: $2-4, 3-7, 4-8$, that is $-2, -4, -4$. Direction ratios of the second: $1+1, 2+2, 5-1$, that is $2, 4, 4$.

The second is exactly $-1$ times the first, so the ratios are proportional and the lines are parallel. (They are distinct lines, not the same one, since $(4,7,8)$ does not satisfy the second line’s equation.)

Question 4

Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat i + 2\hat j – 2\hat k$.

Solution. Substitute straight into $\vec r = \vec a + \lambda\vec b$ with $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = 3\hat i + 2\hat j – 2\hat k$.

Question 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 $\hat i + 2\hat j – \hat k$.

Solution. The vector form is immediate. For the Cartesian form, read off the point $(2, -1, 4)$ and the direction $(1, 2, -1)$ and substitute into the template — noting that $y_1 = -1$ makes the numerator $y – (-1) = y + 1$.

Question 6

Find the Cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the line given by $\dfrac{x+3}{3} = \dfrac{y-4}{5} = \dfrac{z+8}{6}$.

Solution. Parallel lines share direction ratios, so read $(3, 5, 6)$ off the given line and pair it with the new point $(-2, 4, -5)$:

Question 7

The Cartesian equation of a line is $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$. Write its vector form.

Solution. The point is $(5, -4, 6)$ — the $+4$ in the numerator means $y_1 = -4$ — and the direction is $(3, 7, 2)$.

Question 8

Find the angle between the following pairs of lines:

(i) $\vec r = 2\hat i – 5\hat j + \hat k + \lambda\left(3\hat i + 2\hat j + 6\hat k\right)$ and $\vec r = 7\hat i – 6\hat k + \mu\left(\hat i + 2\hat j + 2\hat k\right)$

(ii) $\vec r = 3\hat i + \hat j – 2\hat k + \lambda\left(\hat i – \hat j – 2\hat k\right)$ and $\vec r = 2\hat i – \hat j – 56\hat k + \mu\left(3\hat i – 5\hat j – 4\hat k\right)$

Solution. Only the direction vectors matter; the points are irrelevant to the angle.

(i) $\vec b_1 = (3, 2, 6)$ and $\vec b_2 = (1, 2, 2)$.

$$\vec b_1 \cdot \vec b_2 = 3 + 4 + 12 = 19, \qquad \left|\vec b_1\right| = 7, \qquad \left|\vec b_2\right| = 3$$

(ii) $\vec b_1 = (1, -1, -2)$ and $\vec b_2 = (3, -5, -4)$.

$$\vec b_1 \cdot \vec b_2 = 3 + 5 + 8 = 16, \qquad \left|\vec b_1\right| = \sqrt6, \qquad \left|\vec b_2\right| = \sqrt{50} = 5\sqrt2$$

$$\cos\theta = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt3} = \frac{8}{5\sqrt3}$$

Question 9

Find the angle between the following pair of lines:

(i) $\dfrac{x-2}{2} = \dfrac{y-1}{5} = \dfrac{z+3}{-3}$ and $\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{z-5}{4}$

(ii) $\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}$ and $\dfrac{x-5}{4} = \dfrac{y-2}{1} = \dfrac{z-3}{8}$

Solution. Read the direction ratios from the denominators.

(i) $(2, 5, -3)$ and $(-1, 8, 4)$.

$$\vec b_1 \cdot \vec b_2 = -2 + 40 – 12 = 26, \qquad \left|\vec b_1\right| = \sqrt{38}, \qquad \left|\vec b_2\right| = \sqrt{81} = 9$$

(ii) $(2, 2, 1)$ and $(4, 1, 8)$.

$$\vec b_1 \cdot \vec b_2 = 8 + 2 + 8 = 18, \qquad \left|\vec b_1\right| = 3, \qquad \left|\vec b_2\right| = 9$$

$$\cos\theta = \frac{18}{27} = \frac23$$

Question 10

Find the values of $p$ so that the lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2p} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3p} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$ are at right angles.

Solution. Neither equation is in standard form, so rewrite both before reading any direction. The coefficient of $x$, $y$ or $z$ in the numerator must be $1$.

First line. $\tfrac{1-x}{3} = \tfrac{x-1}{-3}$, and $\tfrac{7y-14}{2p} = \tfrac{7(y-2)}{2p} = \tfrac{y-2}{2p/7}$. So the direction ratios are

$$\left(-3,\ \frac{2p}{7},\ 2\right)$$

Second line. $\tfrac{7-7x}{3p} = \tfrac{-7(x-1)}{3p} = \tfrac{x-1}{-3p/7}$, and $\tfrac{6-z}{5} = \tfrac{z-6}{-5}$. So the direction ratios are

$$\left(\frac{-3p}{7},\ 1,\ -5\right)$$

Setting the dot product to zero:

$$(-3)\left(\frac{-3p}{7}\right) + \frac{2p}{7}(1) + 2(-5) = \frac{9p}{7} + \frac{2p}{7} – 10 = 0$$

$$\frac{11p}{7} = 10$$

Question 11

Show that the lines $\dfrac{x-5}{7} = \dfrac{y+2}{-5} = \dfrac{z}{1}$ and $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ are perpendicular to each other.

Solution. Direction ratios $(7, -5, 1)$ and $(1, 2, 3)$:

$$7(1) + (-5)(2) + 1(3) = 7 – 10 + 3 = 0$$

The dot product is zero, so the lines are perpendicular.

Question 12

Find the shortest distance between the lines $\vec r = \left(\hat i + 2\hat j + \hat k\right) + \lambda\left(\hat i – \hat j + \hat k\right)$ and $\vec r = 2\hat i – \hat j – \hat k + \mu\left(2\hat i + \hat j + 2\hat k\right)$.

Solution. The direction vectors $(1, -1, 1)$ and $(2, 1, 2)$ are not proportional, so the lines are skew and the general formula applies.

$$\vec a_2 – \vec a_1 = (2-1,\ -1-2,\ -1-1) = (1, -3, -2)$$

$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 2 & 1 & 2\end{vmatrix} = (-2-1)\hat i – (2-2)\hat j + (1+2)\hat k = (-3, 0, 3)$$

$$\left|\vec b_1 \times \vec b_2\right| = \sqrt{9 + 0 + 9} = 3\sqrt2$$

$$\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right) = -3 + 0 – 6 = -9$$

$$d = \frac{|-9|}{3\sqrt2} = \frac{3}{\sqrt2}$$

Question 13

Find the shortest distance between the lines $\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}$.

Solution. Points $(-1, -1, -1)$ and $(3, 5, 7)$; directions $(7, -6, 1)$ and $(1, -2, 1)$.

$$\vec a_2 – \vec a_1 = (4, 6, 8)$$

$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{vmatrix} = (-6+2)\hat i – (7-1)\hat j + (-14+6)\hat k = (-4, -6, -8)$$

$$\left|\vec b_1 \times \vec b_2\right| = \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}$$

$$\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right) = -16 – 36 – 64 = -116$$

$$d = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$$

The last step uses $58 = 2 \times 29$, so $\tfrac{58}{\sqrt{29}} = 2\sqrt{29}$.

Question 14

Find the shortest distance between the lines whose vector equations are $\vec r = \left(\hat i + 2\hat j + 3\hat k\right) + \lambda\left(\hat i – 3\hat j + 2\hat k\right)$ and $\vec r = 4\hat i + 5\hat j + 6\hat k + \mu\left(2\hat i + 3\hat j + \hat k\right)$.

Solution.

$$\vec a_2 – \vec a_1 = (3, 3, 3)$$

$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & -3 & 2 \\ 2 & 3 & 1\end{vmatrix} = (-3-6)\hat i – (1-4)\hat j + (3+6)\hat k = (-9, 3, 9)$$

$$\left|\vec b_1 \times \vec b_2\right| = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}$$

$$\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right) = -27 + 9 + 27 = 9$$

$$d = \frac{9}{3\sqrt{19}}$$

Question 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 $\vec r = (s+1)\hat i + (2s-1)\hat j – (2s+1)\hat k$.

Solution. Neither equation is in $\vec a + \lambda\vec b$ form, so regroup first — collect the constant terms and the parameter terms separately.

$$\vec r = \left(\hat i – 2\hat j + 3\hat k\right) + t\left(-\hat i + \hat j – 2\hat k\right)$$

$$\vec r = \left(\hat i – \hat j – \hat k\right) + s\left(\hat i + 2\hat j – 2\hat k\right)$$

Now $\vec a_1 = (1, -2, 3)$, $\vec b_1 = (-1, 1, -2)$, $\vec a_2 = (1, -1, -1)$, $\vec b_2 = (1, 2, -2)$.

$$\vec a_2 – \vec a_1 = (0, 1, -4)$$

$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ -1 & 1 & -2 \\ 1 & 2 & -2\end{vmatrix} = (-2+4)\hat i – (2+2)\hat j + (-2-1)\hat k = (2, -4, -3)$$

$$\left|\vec b_1 \times \vec b_2\right| = \sqrt{4+16+9} = \sqrt{29}$$

$$\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right) = 0 – 4 + 12 = 8$$

Common mistakes

• Reading direction ratios from an equation not in standard form. Questions 10 and 15 both punish this. $\tfrac{7y-14}{2p}$ is not direction ratio $2p$ for $y$; factor the $7$ out first.
• Getting the sign of the point wrong. In $\tfrac{y+4}{7}$ the point’s $y$-coordinate is $-4$, not $4$ — the template has $y – y_1$.
• Forgetting the modulus in the angle formula. The angle between two lines is taken acute, so $\cos\theta$ carries an absolute value. Without it a negative dot product gives an obtuse answer.
• Using the skew-lines formula on parallel lines. If $\vec b_1 \times \vec b_2 = \vec 0$ the formula divides by zero; parallel lines need the other formula.
• Not simplifying the surd. $\tfrac{58}{\sqrt{29}}$ and $2\sqrt{29}$ are the same number, and NCERT prints the second. Rationalise before comparing with the key.
• Sign errors in the cross product’s middle term. The $\hat j$ component carries a minus sign in the cofactor expansion. Every shortest-distance question here depends on getting it right.

Practise next

• Miscellaneous Exercise on Chapter 11 — the same formulas with a parameter to determine, plus a line perpendicular to two given lines.
• Exercise 10.4 — the cross product that every shortest-distance calculation here relies on.