NCERT Exemplar Solutions Class 12 Maths Three Dimensional Geometry — Long Answer Q16–28

NCERT Exemplar Problems, Class 12 Mathematics — Three Dimensional Geometry, Long Answer questions 16 to 28. All 13 questions solved.

The long answers are where the unit’s techniques combine. Questions 16 to 18 find the foot of a perpendicular, to a line and to a plane; Questions 21 and 26 are the shortest distance between skew lines, first as a number and then as the two actual points; Questions 22 and 24 use the family of planes through a line of intersection. The proofs, 23, 27 and 28, each turn on one short observation.

A note on the syllabus. The current NCERT textbook chapter teaches lines only; the plane was removed in the rationalised edition. Questions 16, 17, 21, 26, 27 and 28 are about lines and sit squarely inside it. Questions 18 to 20 and 22 to 25 use planes; they remain in the JEE Advanced syllabus.

Key insight. The foot of a perpendicular is always “general point, then one dot product equal to zero”. Write the general point on the line with a parameter, make the joining vector perpendicular to the line’s direction, and solve one linear equation. Questions 16, 17 and 26 are that recipe, once, once and twice.

Question 16

Find the foot of perpendicular from the point $(2, 3, -8)$ to the line $\dfrac{4 – x}{2} = \dfrac{y}{6} = \dfrac{1 – z}{3}$. Also, find the perpendicular distance from the given point to the line.

Solution. First put the line in standard form. The numerators must read “$x$ minus something”, so rewrite $\frac{4 – x}{2} = \frac{x – 4}{-2}$ and $\frac{1 – z}{3} = \frac{z – 1}{-3}$:

$$\frac{x – 4}{-2} = \frac{y}{6} = \frac{z – 1}{-3} = t$$

The direction is $(-2, 6, -3)$, and a general point is $P = (4 – 2t,\; 6t,\; 1 – 3t)$. With A $(2, 3, -8)$,

$$\overrightarrow{AP} = (2 – 2t,\; 6t – 3,\; 9 – 3t)$$

For P to be the foot, $\overrightarrow{AP}$ must be perpendicular to the line:

$$-2(2 – 2t) + 6(6t – 3) – 3(9 – 3t) = 49t – 49 = 0 \quad\Longrightarrow\quad t = 1$$

So the foot is $(2, 6, -2)$, and $\overrightarrow{AP} = (0, 3, 6)$ has length $\sqrt{45} = 3\sqrt5$.

Question 17

Find the distance of a point $(2, 4, -1)$ from the line $\dfrac{x + 5}{1} = \dfrac{y + 3}{4} = \dfrac{z – 6}{-9}$.

Solution. The same recipe. A general point is $P = (-5 + t,\; -3 + 4t,\; 6 – 9t)$ and, with A $(2, 4, -1)$,

$$\overrightarrow{AP} = (t – 7,\; 4t – 7,\; 7 – 9t)$$

Perpendicular to the direction $(1, 4, -9)$:

$$(t – 7) + 4(4t – 7) – 9(7 – 9t) = 98t – 98 = 0 \quad\Longrightarrow\quad t = 1$$

The foot is $(-4, 1, -3)$ and $\overrightarrow{AP} = (-6, -3, -2)$, of length $\sqrt{36 + 9 + 4} = 7$.

Question 18

Find the length and the foot of perpendicular from the point $\left(1, \frac32, 2\right)$ to the plane $2x – 2y + 4z + 5 = 0$.

Solution. The length comes straight from the distance formula:

$$d = \frac{|2(1) – 2\left(\frac32\right) + 4(2) + 5|}{\sqrt{4 + 4 + 16}} = \frac{12}{2\sqrt6} = \sqrt6$$

For the foot, move from the point along the normal $(2, -2, 4)$: the foot is $\left(1 + 2k,\; \frac32 – 2k,\; 2 + 4k\right)$ for the $k$ that puts it on the plane.

$$2(1 + 2k) – 2\left(\tfrac32 – 2k\right) + 4(2 + 4k) + 5 = 12 + 24k = 0 \quad\Longrightarrow\quad k = -\frac12$$

The key prints only the length.

Question 19

Find the equations of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x + 2y = 0$ and $3y – z = 0$.

Solution. A line parallel to a plane is perpendicular to its normal. Parallel to both planes, it is perpendicular to both normals $(1, 2, 0)$ and $(0, 3, -1)$, so it runs along their cross product:

$$\begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & 2 & 0\\ 0 & 3 & -1\end{vmatrix} = -2\hat i + \hat j + 3\hat k$$

The key’s “$(x – 3)\hat j$” should read $(x – 3)\hat i$; its direction vector agrees.

Question 20

Find the equation of the plane through the points $(2, 1, -1)$ and $(-1, 3, 4)$, and perpendicular to the plane $x – 2y + 4z = 10$.

Solution. The required plane contains the vector joining the two points, $(-3, 2, 5)$. Being perpendicular to the given plane, it also contains that plane’s normal $(1, -2, 4)$. Its own normal is perpendicular to both:

$$\begin{vmatrix}\hat i & \hat j & \hat k\\ -3 & 2 & 5\\ 1 & -2 & 4\end{vmatrix} = 18\hat i + 17\hat j + 4\hat k$$

Through $(2, 1, -1)$: $18(x – 2) + 17(y – 1) + 4(z + 1) = 0$.

Check the second point: $-18 + 51 + 16 = 49$.

Question 21

Find the shortest distance between the lines given by $\vec r = (8 + 3\lambda)\hat i – (9 + 16\lambda)\hat j + (10 + 7\lambda)\hat k$ and $\vec r = 15\hat i + 29\hat j + 5\hat k + \mu\left(3\hat i + 8\hat j – 5\hat k\right)$.

Solution. Separate the first line into point and direction: $\vec a_1 = 8\hat i – 9\hat j + 10\hat k$, $\vec b_1 = 3\hat i – 16\hat j + 7\hat k$. The second has $\vec a_2 = 15\hat i + 29\hat j + 5\hat k$, $\vec b_2 = 3\hat i + 8\hat j – 5\hat k$.

The common perpendicular is along $\vec b_1\times\vec b_2$:

$$\begin{vmatrix}\hat i & \hat j & \hat k\\ 3 & -16 & 7\\ 3 & 8 & -5\end{vmatrix} = 24\hat i + 36\hat j + 72\hat k = 12\left(2\hat i + 3\hat j + 6\hat k\right)$$

The shortest distance is the projection of $\vec a_2 – \vec a_1 = 7\hat i + 38\hat j – 5\hat k$ on that direction, whose unit vector is $\frac17(2, 3, 6)$:

$$d = \frac{|14 + 114 – 30|}{7} = \frac{98}{7}$$

Question 22

Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z – 4 = 0$ and $2x + y – z + 5 = 0$.

Solution. Every plane through the line of intersection has the form

$$(x + 2y + 3z – 4) + k(2x + y – z + 5) = 0$$

with normal $(1 + 2k,\; 2 + k,\; 3 – k)$. Perpendicular planes have perpendicular normals, so

$$5(1 + 2k) + 3(2 + k) + 6(3 – k) = 29 + 7k = 0 \quad\Longrightarrow\quad k = -\frac{29}{7}$$

Multiply the family equation by $7$ and substitute: $7(x + 2y + 3z – 4) – 29(2x + y – z + 5) = 0$, which is $-51x – 15y + 50z – 173 = 0$.

Question 23

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm\left(\sqrt{a^2 + b^2}\tan\alpha\right)z = 0$.

Solution. The line about which the plane turns lies on both $ax + by = 0$ and $z = 0$. Every plane through that line is $(ax + by) + kz = 0$, so the new position has this form for some $k$, with normal $(a, b, k)$.

The angle between the old and new positions is the angle between their normals $(a, b, 0)$ and $(a, b, k)$:

$$\cos\alpha = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}\,\sqrt{a^2 + b^2 + k^2}} = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2 + k^2}}$$

Squaring and rearranging, $(a^2 + b^2 + k^2)\cos^2\alpha = a^2 + b^2$, so $k^2\cos^2\alpha = (a^2 + b^2)\sin^2\alpha$ and

$$k = \pm\sqrt{a^2 + b^2}\,\tan\alpha$$

The two signs are the two directions of rotation.

Question 24

Find the equation of the plane through the intersection of the planes $\vec r\cdot\left(\hat i + 3\hat j\right) – 6 = 0$ and $\vec r\cdot\left(3\hat i – \hat j – 4\hat k\right) = 0$, whose perpendicular distance from origin is unity.

Solution. The family through the intersection is

$$\vec r\cdot\left[(1 + 3k)\hat i + (3 – k)\hat j – 4k\hat k\right] = 6$$

Its distance from the origin is $6$ divided by the length of the normal, and that must be $1$:

$$(1 + 3k)^2 + (3 – k)^2 + 16k^2 = 36 \quad\Longrightarrow\quad 10 + 26k^2 = 36 \quad\Longrightarrow\quad k = \pm 1$$

$k = 1$ gives the normal $(4, 2, -4)$ and $k = -1$ gives $(-2, 4, 4)$.

Equivalently $2x + y – 2z = 3$ and $x – 2y – 2z + 3 = 0$; each has normal of length $3$ and constant $3$, so each is at distance $1$.

Question 25

Show that the points $\left(\hat i – \hat j + 3\hat k\right)$ and $3\left(\hat i + \hat j + \hat k\right)$ are equidistant from the plane $\vec r\cdot\left(5\hat i + 2\hat j – 7\hat k\right) + 9 = 0$ and lies on opposite side of it.

Solution. Substitute each point into the left-hand side, $5x + 2y – 7z + 9$:

$$(1, -1, 3):\; 5 – 2 – 21 + 9 = -9, \qquad (3, 3, 3):\; 15 + 6 – 21 + 9 = 9$$

The distance of a point from the plane is the absolute value divided by $|\vec n| = \sqrt{25 + 4 + 49} = \sqrt{78}$, so both points are $\frac{9}{\sqrt{78}}$ away. The sign of the expression tells which side of the plane a point is on, and the signs here are opposite.

Question 26

$\overrightarrow{AB} = 3\hat i – \hat j + \hat k$ and $\overrightarrow{CD} = -3\hat i + 2\hat j + 4\hat k$ are two vectors. The position vectors of the points A and C are $6\hat i + 7\hat j + 4\hat k$ and $-9\hat j + 2\hat k$, respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{CD}$ both.

Solution. This is the foot-of-perpendicular recipe applied to both lines at once. General points are

$$P = (6 + 3s,\; 7 – s,\; 4 + s), \qquad Q = (-3t,\; -9 + 2t,\; 2 + 4t)$$

$$\overrightarrow{PQ} = (-6 – 3s – 3t,\; -16 + s + 2t,\; -2 – s + 4t)$$

Two conditions, two unknowns:

$$\overrightarrow{PQ}\cdot(3, -1, 1) = -11s – 7t – 4 = 0, \qquad \overrightarrow{PQ}\cdot(-3, 2, 4) = 7s + 29t – 22 = 0$$

Solving, $s = -1$ and $t = 1$.

As a check, $\overrightarrow{PQ} = (-6, -15, 3)$, and its dot products with $(3, -1, 1)$ and $(-3, 2, 4)$ are $-18 + 15 + 3 = 0$ and $18 – 30 + 12 = 0$. Its length $\sqrt{270} = 3\sqrt{30}$ is the shortest distance between the two lines.

Question 27

Show that the straight lines whose direction cosines are given by $2l + 2m – n = 0$ and $mn + nl + lm = 0$ are at right angles.

Solution. From the first equation $n = 2(l + m)$. Substituting into the second,

$$n(l + m) + lm = 2(l + m)^2 + lm = 2l^2 + 5lm + 2m^2 = 0$$

This factorises as $(2l + m)(l + 2m) = 0$.

• $m = -2l$: then $n = 2(l – 2l) = -2l$, direction ratios $(1, -2, -2)$.
• $l = -2m$: then $n = 2(-2m + m) = -2m$, direction ratios $(-2, 1, -2)$.

Their dot product is $-2 – 2 + 4 = 0$.

Question 28

If $l_1, m_1, n_1$; $l_2, m_2, n_2$; $l_3, m_3, n_3$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_1 + l_2 + l_3$, $m_1 + m_2 + m_3$, $n_1 + n_2 + n_3$ makes equal angles with them.

Solution. Let $\hat u_1, \hat u_2, \hat u_3$ be the unit vectors $(l_1, m_1, n_1)$ and so on. They are mutually perpendicular, so $\hat u_i\cdot\hat u_j = 0$ for $i \ne j$, and each has length $1$. The new line is along $\vec s = \hat u_1 + \hat u_2 + \hat u_3$, whose length is

$$|\vec s|^2 = 1 + 1 + 1 + 2(0 + 0 + 0) = 3$$

The angle $\theta_1$ it makes with the first line:

$$\cos\theta_1 = \frac{\vec s\cdot\hat u_1}{|\vec s|} = \frac{1 + 0 + 0}{\sqrt3} = \frac{1}{\sqrt3}$$

By exactly the same working, $\cos\theta_2 = \cos\theta_3 = \frac{1}{\sqrt3}$.

Geometrically, the three lines are a rotated set of coordinate axes, and $\vec s$ is the long diagonal of the unit cube they span.

Common mistakes

• Reading direction ratios off a line not in standard form. In Question 16 the line is written $\frac{4 – x}{2}$ and $\frac{1 – z}{3}$; its direction is $(-2, 6, -3)$, not $(2, 6, 3)$. Rewrite every numerator as “variable minus constant” first.
• Treating the shortest distance as the distance between the two given points in Question 21. It is that join projected on the common perpendicular $\vec b_1\times\vec b_2$.
• Using only one condition in Question 26. P and Q have two unknown parameters, so $\overrightarrow{PQ}$ needs to be perpendicular to both lines, giving two equations.
• Forgetting the second member of the family in Question 24. The distance condition is quadratic in $k$, so two planes satisfy it.
• Taking the absolute value too early in Question 25. The distances are equal either way; it is the signs of $-9$ and $+9$ that show the points are on opposite sides.
• Dividing by $l$ or $m$ when it may be zero in Question 27. Factorising $2l^2 + 5lm + 2m^2$ avoids dividing at all.

Practise next

• Exemplar Short Answer Q1–15, the first band — direction cosines, lines that meet, and planes from a point and a normal.
• Exemplar Objective, Fill in the Blanks and True or False Q29–49, the rest of the unit, including one statement the printed key gets wrong.
• NCERT textbook Exercise 11.2 for the shortest distance between skew lines, as in Questions 21 and 26.
• NCERT textbook Miscellaneous Exercise on Three Dimensional Geometry for more feet of perpendiculars, as in Questions 16 and 17.