NCERT Exemplar Solutions Class 12 Maths Three Dimensional Geometry — Short Answer Q1–15

NCERT Exemplar Problems, Class 12 Mathematics — Three Dimensional Geometry, Short Answer questions 1 to 15. All 15 questions solved.

Fifteen short questions that cover the whole unit in miniature: direction cosines (1, 12, 13, 14), the equation of a line and where two lines meet (2, 3, 5, 11), the angle between lines (4, 6), and the equation of a plane from a point and a normal (7, 8, 9, 10). Questions 13 and 15 are proofs, and both are shorter than they look.

A note on the syllabus. The current NCERT textbook chapter on three dimensional geometry teaches lines only; the sections on the plane were removed in the rationalised edition. Questions 7 to 10, 14 and 15 are about planes. They remain in the JEE Advanced syllabus, so work them if you are preparing for it; for the board examination, the line questions are the ones to prioritise.

Key insight. Nearly every question here reduces to one vector. A line is a point plus a direction vector; a plane is a point plus a normal vector. Find that vector first, and the equation, the angle or the distance follows in one line.

Question 1

Find the position vector of a point A in space such that $\overrightarrow{OA}$ is inclined at $60^\circ$ to OX and at $45^\circ$ to OY and $|\overrightarrow{OA}| = 10$ units.

Solution. The direction cosines of $\overrightarrow{OA}$ are $l = \cos 60^\circ = \frac12$ and $m = \cos 45^\circ = \frac{1}{\sqrt2}$. The third comes from $l^2 + m^2 + n^2 = 1$:

$$n^2 = 1 – \frac14 – \frac12 = \frac14 \quad\Longrightarrow\quad n = \pm\frac12$$

The question fixes the angles with OX and OY only, so both signs of $n$ are allowed: A can lie above or below the $xy$-plane. Multiplying the unit vector $(l, m, n)$ by the length $10$:

The printed key gives only $5\hat i + 5\sqrt2\,\hat j + 5\hat k$; the vector with $-5\hat k$ satisfies every condition too.

Question 2

Find the vector equation of the line which is parallel to the vector $3\hat i – 2\hat j + 6\hat k$ and which passes through the point $(1, -2, 3)$.

Solution. A line through the point with position vector $\vec a$ parallel to $\vec b$ is $\vec r = \vec a + \lambda\vec b$. Here $\vec a = \hat i – 2\hat j + 3\hat k$:

The key writes the same line as $(x – 1)\hat i + (y + 2)\hat j + (z – 3)\hat k = \lambda(3\hat i – 2\hat j + 6\hat k)$; its “$3\hat j$” in the bracket is a misprint for $3\hat i$.

Question 3

Show that the lines $\dfrac{x – 1}{2} = \dfrac{y – 2}{3} = \dfrac{z – 3}{4}$ and $\dfrac{x – 4}{5} = \dfrac{y – 1}{2} = z$ intersect. Also, find their point of intersection.

Solution. Write a general point on each line with its own parameter:

$$(1 + 2s,\; 2 + 3s,\; 3 + 4s) \qquad\text{and}\qquad (4 + 5t,\; 1 + 2t,\; t)$$

The lines meet if one pair $(s, t)$ makes all three coordinates equal. Use two equations to find $s$ and $t$, then test the third:

$$1 + 2s = 4 + 5t, \qquad 2 + 3s = 1 + 2t$$

From the first, $2s – 5t = 3$; from the second, $3s – 2t = -1$. Solving, $s = -1$ and $t = -1$. Now the third coordinate: $3 + 4(-1) = -1$ and $t = -1$. They agree, so the lines do intersect. (Two lines in space usually do not meet, which is why the third equation has to be checked rather than assumed.)

Question 4

Find the angle between the lines $\vec r = 3\hat i – 2\hat j + 6\hat k + \lambda\left(2\hat i + \hat j + 2\hat k\right)$ and $\vec r = \left(2\hat j – 5\hat k\right) + \mu\left(6\hat i + 3\hat j + 2\hat k\right)$.

Solution. The angle between two lines is the angle between their direction vectors; the points they pass through play no part. With $\vec b_1 = 2\hat i + \hat j + 2\hat k$ and $\vec b_2 = 6\hat i + 3\hat j + 2\hat k$:

$$\cos\theta = \frac{\vec b_1\cdot\vec b_2}{|\vec b_1|\,|\vec b_2|} = \frac{12 + 3 + 4}{3 \times 7} = \frac{19}{21}$$

Question 5

Prove that the line through A $(0, -1, -1)$ and B $(4, 5, 1)$ intersects the line through C $(3, 9, 4)$ and D $(-4, 4, 4)$.

Solution. The direction of AB is $(4, 6, 2)$ and of CD is $(-7, -5, 0)$. General points on the two lines are

$$(4s,\; -1 + 6s,\; -1 + 2s) \qquad\text{and}\qquad (3 – 7t,\; 9 – 5t,\; 4)$$

The $z$-coordinate is the quickest place to start, because the second line has constant $z = 4$: $-1 + 2s = 4$ gives $s = \frac52$, and the point on AB is $(10, 14, 4)$. Now match $x$: $3 – 7t = 10$ gives $t = -1$. Check $y$ with that $t$: $9 – 5(-1) = 14$. All three coordinates agree, so the lines meet.

Question 6

Prove that the lines $x = py + q$, $z = ry + s$ and $x = p’y + q’$, $z = r’y + s’$ are perpendicular if $pp’ + rr’ + 1 = 0$.

Solution. Put the first line in symmetric form by making $y$ the parameter. From $x = py + q$ and $z = ry + s$,

$$\frac{x – q}{p} = \frac{y}{1} = \frac{z – s}{r}$$

so its direction ratios are $p, 1, r$. In the same way the second line has direction ratios $p’, 1, r’$. Two lines are perpendicular exactly when the dot product of their direction ratios is zero:

$$pp’ + 1\cdot 1 + rr’ = 0$$

Question 7

Find the equation of a plane which bisects perpendicularly the line joining the points A $(2, 3, 4)$ and B $(4, 5, 8)$ at right angles.

Solution. The plane passes through the midpoint of AB, which is $(3, 4, 6)$, and is perpendicular to AB, so its normal is $\overrightarrow{AB} = (2, 2, 4)$, or more simply $(1, 1, 2)$:

$$1(x – 3) + 1(y – 4) + 2(z – 6) = 0$$

Question 8

Find the equation of a plane which is at a distance $3\sqrt3$ units from origin and the normal to which is equally inclined to coordinate axis.

Solution. A normal equally inclined to the three axes has equal direction cosines, so it is along $\hat i + \hat j + \hat k$ and the plane is $x + y + z = d$. Its distance from the origin is $\frac{|d|}{\sqrt3}$, so

$$\frac{|d|}{\sqrt3} = 3\sqrt3 \quad\Longrightarrow\quad |d| = 9$$

Two parallel planes fit, one on each side of the origin. The key prints only $x + y + z = 9$.

Question 9

If the line drawn from the point $(-2, -1, -3)$ meets a plane at right angle at the point $(1, -3, 3)$, find the equation of the plane.

Solution. The line meets the plane at right angles, so it is along the normal. The normal is therefore the vector from $(-2, -1, -3)$ to $(1, -3, 3)$, which is $(3, -2, 6)$, and the plane passes through $(1, -3, 3)$:

$$3(x – 1) – 2(y + 3) + 6(z – 3) = 0$$

Question 10

Find the equation of the plane through the points $(2, 1, 0)$, $(3, -2, -2)$ and $(3, 1, 7)$.

Solution. Call the points A, B, C. Two vectors lying in the plane are $\overrightarrow{AB} = (1, -3, -2)$ and $\overrightarrow{AC} = (1, 0, 7)$; their cross product is perpendicular to both, so it is a normal:

$$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & -3 & -2\\ 1 & 0 & 7\end{vmatrix} = -21\hat i – 9\hat j + 3\hat k$$

Dividing by $-3$, the normal is $(7, 3, -1)$. Through A $(2, 1, 0)$: $7(x – 2) + 3(y – 1) – z = 0$.

The key writes it as $21x + 9y – 3z – 51 = 0$, which is the same plane multiplied by $3$. Checking B: $21 – 6 + 2 = 17$; checking C: $21 + 3 – 7 = 17$.

Question 11

Find the equations of the two lines through the origin which intersect the line $\dfrac{x – 3}{2} = \dfrac{y – 3}{1} = \dfrac{z}{1}$ at angles of $\dfrac{\pi}{3}$ each.

Solution. Each required line runs from the origin O to some point P on the given line, so its direction is $\overrightarrow{OP}$. A general point is $P = (3 + 2t,\; 3 + t,\; t)$, and the given line has direction $\vec d = (2, 1, 1)$ with $|\vec d| = \sqrt6$.

$$\overrightarrow{OP}\cdot\vec d = 2(3 + 2t) + (3 + t) + t = 9 + 6t, \qquad |\overrightarrow{OP}|^2 = 6t^2 + 18t + 18$$

The angle is $\frac\pi3$, so $\cos\frac\pi3 = \frac12$:

$$\frac{9 + 6t}{\sqrt6\,\sqrt{6t^2 + 18t + 18}} = \frac12$$

Squaring, $4(9 + 6t)^2 = 6(6t^2 + 18t + 18)$, which simplifies to $108t^2 + 324t + 216 = 0$, that is $t^2 + 3t + 2 = 0$. So $t = -1$ or $t = -2$, giving $P = (1, 2, -1)$ or $P = (-1, 1, -2)$. (At $t = -2$ the dot product $9 + 6t$ is negative, so the vectors make $\frac{2\pi}{3}$; but a line has no preferred direction, and the angle between the lines is then $\pi – \frac{2\pi}{3} = \frac\pi3$. Both values count.)

Question 12

Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0$, $l^2 + m^2 – n^2 = 0$.

Solution. Eliminate $n$. From the first equation $n = -(l + m)$; substituting in the second,

$$l^2 + m^2 – (l + m)^2 = 0 \quad\Longrightarrow\quad -2lm = 0$$

So $l = 0$ or $m = 0$.

• If $l = 0$, then $n = -m$: direction ratios $(0, 1, -1)$.
• If $m = 0$, then $n = -l$: direction ratios $(1, 0, -1)$.

The angle between them:

$$\cos\theta = \frac{0 + 0 + 1}{\sqrt2\,\sqrt2} = \frac12$$

Question 13

If a variable line in two adjacent positions has direction cosines $l, m, n$ and $l + \delta l,\; m + \delta m,\; n + \delta n$, show that the small angle $\delta\theta$ between the two positions is given by $\delta\theta^2 = \delta l^2 + \delta m^2 + \delta n^2$.

Solution. Both sets are direction cosines, so each satisfies the sum-of-squares rule:

$$l^2 + m^2 + n^2 = 1, \qquad (l + \delta l)^2 + (m + \delta m)^2 + (n + \delta n)^2 = 1$$

Expanding the second and subtracting the first,

$$2(l\,\delta l + m\,\delta m + n\,\delta n) = -\left(\delta l^2 + \delta m^2 + \delta n^2\right) \tag{1}$$

The angle between the two positions is given by the dot product of the two unit vectors:

$$\cos\delta\theta = l(l + \delta l) + m(m + \delta m) + n(n + \delta n) = 1 + (l\,\delta l + m\,\delta m + n\,\delta n)$$

Using (1),

$$1 – \cos\delta\theta = \frac12\left(\delta l^2 + \delta m^2 + \delta n^2\right)$$

For a small angle, $1 – \cos\delta\theta = 2\sin^2\frac{\delta\theta}{2} \approx 2\cdot\frac{\delta\theta^2}{4} = \frac{\delta\theta^2}{2}$. Equating the two expressions:

The geometric picture: the two unit vectors have tips $\delta\theta$ apart along an arc of the unit sphere, and for a small angle that arc is the straight-line gap $\sqrt{\delta l^2 + \delta m^2 + \delta n^2}$ between the tips.

Question 14

O is the origin and A is $(a, b, c)$. Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.

Solution. The direction ratios of OA are $a, b, c$. Dividing by the length $\sqrt{a^2 + b^2 + c^2}$ gives the direction cosines. The plane at right angles to OA has OA as its normal, and passes through A:

$$a(x – a) + b(y – b) + c(z – c) = 0$$

Question 15

Two systems of rectangular axis have the same origin. If a plane cuts them at distances $a, b, c$ and $a’, b’, c’$, respectively, from the origin, prove that

$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a’^2} + \frac{1}{b’^2} + \frac{1}{c’^2}.$$

Solution. In the first system the plane has intercepts $a, b, c$, so its equation is $\frac xa + \frac yb + \frac zc = 1$. Its perpendicular distance from the origin is

$$p = \frac{1}{\sqrt{\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}}}$$

In the second system the same argument gives $p’ = 1\Big/\sqrt{\frac{1}{a’^2} + \frac{1}{b’^2} + \frac{1}{c’^2}}$. But it is the same plane and the same origin: the perpendicular distance from a point to a plane is a length, and rotating the axes cannot change it. So $p = p’$, and squaring and inverting:

Common mistakes

• Dropping one sign when two answers exist. Question 1 fixes only two angles, so $n = \pm\frac12$; Question 8 fixes only a distance, so $d = \pm 9$. The printed key gives one answer for each.
• Assuming two lines meet because two coordinates match. In Questions 3 and 5 the third coordinate is the actual test; skip it and you would “prove” that skew lines intersect.
• Using a point on the line in the angle formula. In Question 4 only the direction vectors matter; $3\hat i – 2\hat j + 6\hat k$ and $2\hat j – 5\hat k$ are positions, not directions.
• Taking the normal from the wrong pair of points in Question 9. The normal is the vector between the two given points, and the plane passes through the foot $(1, -3, 3)$, not through $(-2, -1, -3)$.
• Forgetting that a symmetric form needs the coefficient of $y$ as well in Question 6. Writing $y = \frac{y}{1}$ is what makes the middle direction ratio $1$, and that $1$ is where the “$+1$” in the condition comes from.

Practise next

• Exemplar Long Answer Q16–28, the next band — feet of perpendiculars, shortest distance between skew lines and planes through a line.
• NCERT textbook Exercise 11.1 for direction cosines and direction ratios, as in Questions 1 and 12.
• NCERT textbook Exercise 11.2 for the equation of a line and the angle between two lines, as in Questions 2, 4 and 6.
• NCERT textbook Miscellaneous Exercise on Three Dimensional Geometry, which has more questions on lines meeting at a given angle, like Question 11.