NCERT Exemplar Solutions Class 12 Maths Three Dimensional Geometry — Objective, Fill in the Blanks and True or False Q29–49

NCERT Exemplar Problems, Class 12 Mathematics — Three Dimensional Geometry, Objective Type questions 29 to 36, Fill in the Blanks 37 to 41 and True or False 42 to 49. All 21 questions solved.

Twenty-one quick questions, each a single formula used correctly: the distance from an axis, the sum of squares of direction cosines, the angle between a line and a plane, the intercept form of a plane, and the equation of a line in vector form. The true-or-false statements are mostly formulas with one detail changed, so check every number.

One answer in the printed key is wrong, Question 46, and the reason is explained where it occurs.

A note on the syllabus. The current NCERT textbook chapter teaches lines only; the plane was removed in the rationalised edition. Questions 29, 30, 33, 38 to 40, 47 and 48 are inside it; the rest use planes, which remain in the JEE Advanced syllabus.

Key insight. For a line and a plane, the line’s direction and the plane’s normal decide everything. Direction perpendicular to normal means the line is parallel to the plane (and may or may not lie in it); otherwise the sine of the angle between line and plane is the cosine of the angle between direction and normal.

Question 29

Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is

Solution. The foot of the perpendicular from $(\alpha, \beta, \gamma)$ to the $y$-axis is $(0, \beta, 0)$: same height along the axis, zero in the other two directions. The distance between them is $\sqrt{\alpha^2 + 0 + \gamma^2}$. Option (B) is the distance from the $xz$-plane, a different thing.

Question 30

If the directions cosines of a line are $k, k, k$, then

Solution. Direction cosines satisfy $l^2 + m^2 + n^2 = 1$, so $3k^2 = 1$ and $k = \pm\frac{1}{\sqrt3}$. The negative value is the same line pointing the other way.

Question 31

The distance of the plane $\vec r\cdot\left(\frac27\hat i + \frac37\hat j – \frac67\hat k\right) = 1$ from the origin is

Solution. The normal has length $\sqrt{\frac{4 + 9 + 36}{49}} = 1$, so the equation is already in normal form $\vec r\cdot\hat n = d$, and $d = 1$ is the distance. The $7$s are a distraction: they make the vector a unit vector.

Question 32

The sine of the angle between the straight line $\dfrac{x – 2}{3} = \dfrac{y – 3}{4} = \dfrac{z – 4}{5}$ and the plane $2x – 2y + z = 5$ is

Solution. The angle $\theta$ between a line and a plane is the complement of the angle between the line and the normal, so $\sin\theta$ uses the dot product:

$$\sin\theta = \frac{|(3, 4, 5)\cdot(2, -2, 1)|}{\sqrt{50}\cdot 3} = \frac{|6 – 8 + 5|}{15\sqrt2} = \frac{1}{5\sqrt2} = \frac{\sqrt2}{10}$$

Question 33

The reflection of the point $(\alpha, \beta, \gamma)$ in the $xy$-plane is

Solution. The $xy$-plane is $z = 0$. Reflecting in it keeps $x$ and $y$ and reverses the distance from the plane, so $z$ changes sign. Option (A) is the foot of the perpendicular, halfway to the reflection; (C) is the reflection in the $z$-axis.

Question 34

The area of the quadrilateral ABCD, where A$(0, 4, 1)$, B$(2, 3, -1)$, C$(4, 5, 0)$ and D$(2, 6, 2)$, is equal to

Solution. Look at the sides first. $\overrightarrow{AB} = (2, -1, -2)$ and $\overrightarrow{DC} = (2, -1, -2)$ are equal, so ABCD is a parallelogram. $\overrightarrow{BC} = (2, 2, 1)$, and $\overrightarrow{AB}\cdot\overrightarrow{BC} = 4 – 2 – 2 = 0$, so it is a rectangle. Both sides have length $\sqrt{4 + 1 + 4} = 3$: it is a square of side $3$.

The general method gives the same: area $= \frac12\left|\overrightarrow{AC}\times\overrightarrow{BD}\right| = \frac12|(4, 1, -1)\times(0, 3, 3)| = \frac12|(6, -12, 12)| = 9$.

Question 35

The locus represented by $xy + yz = 0$ is

Solution. Factorise: $y(x + z) = 0$, so $y = 0$ or $x + z = 0$. Each linear equation in $x, y, z$ is a plane, not a line. Their normals $(0, 1, 0)$ and $(1, 0, 1)$ have dot product $0$, so the planes are perpendicular.

Question 36

The plane $2x – 3y + 6z – 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with x-axis. The value of $\alpha$ is equal to

Solution. The $x$-axis is a line with direction $\hat i$. As in Question 32, the sine of the angle between a line and a plane is

$$\sin\theta = \frac{|\hat i\cdot(2, -3, 6)|}{1\cdot\sqrt{4 + 9 + 36}} = \frac27$$

Question 37

A plane passes through the points $(2, 0, 0)$, $(0, 3, 0)$ and $(0, 0, 4)$. The equation of plane is ______.

Solution. The three points are the intercepts on the axes, so the intercept form applies directly: $\frac x2 + \frac y3 + \frac z4 = 1$. Multiplying by $12$:

Question 38

The direction cosines of the vector $\left(2\hat i + 2\hat j – \hat k\right)$ are ______.

Solution. The length is $\sqrt{4 + 4 + 1} = 3$; divide each component by it.

Question 39

The vector equation of the line $\dfrac{x – 5}{3} = \dfrac{y + 4}{7} = \dfrac{z – 6}{2}$ is ______.

Solution. The line passes through $(5, -4, 6)$, read from the numerators with the signs reversed, and has direction $(3, 7, 2)$ from the denominators.

Question 40

The vector equation of the line through the points $(3, 4, -7)$ and $(1, -1, 6)$ is ______.

Solution. The direction is the difference of the two points, $(1 – 3, -1 – 4, 6 + 7) = (-2, -5, 13)$, and the line passes through $(3, 4, -7)$.

Question 41

The cartesian equation of the plane $\vec r\cdot\left(\hat i + \hat j – \hat k\right) = 2$ is ______.

Solution. Put $\vec r = x\hat i + y\hat j + z\hat k$ and take the dot product.

Question 42

State True or False: The unit vector normal to the plane $x + 2y + 3z – 6 = 0$ is $\dfrac{1}{\sqrt{14}}\hat i + \dfrac{2}{\sqrt{14}}\hat j + \dfrac{3}{\sqrt{14}}\hat k$.

Solution. The normal is $(1, 2, 3)$, of length $\sqrt{14}$, and dividing by it gives the stated vector. (Its negative is also a unit normal; the statement names one of the two correctly.)

Question 43

State True or False: The intercepts made by the plane $2x – 3y + 5z + 4 = 0$ on the co-ordinate axis are $-2, \dfrac43, -\dfrac45$.

Solution. Set two coordinates to zero each time. $y = z = 0$: $2x + 4 = 0$, $x = -2$. $x = z = 0$: $-3y + 4 = 0$, $y = \frac43$. $x = y = 0$: $5z + 4 = 0$, $z = -\frac45$.

Question 44

State True or False: The angle between the line $\vec r = \left(5\hat i – \hat j – 4\hat k\right) + \lambda\left(2\hat i – \hat j + \hat k\right)$ and the plane $\vec r\cdot\left(3\hat i – 4\hat j – \hat k\right) + 5 = 0$ is $\sin^{-1}\left(\dfrac{5}{2\sqrt{91}}\right)$.

Solution.

$$\sin\theta = \frac{|(2, -1, 1)\cdot(3, -4, -1)|}{\sqrt6\,\sqrt{26}} = \frac{|6 + 4 – 1|}{\sqrt{156}} = \frac{9}{2\sqrt{39}}$$

That is about $0.72$; the stated $\frac{5}{2\sqrt{91}}$ is about $0.26$.

Question 45

State True or False: The angle between the planes $\vec r\cdot\left(2\hat i – 3\hat j + \hat k\right) = 1$ and $\vec r\cdot\left(\hat i – \hat j\right) = 4$ is $\cos^{-1}\left(\dfrac{-5}{\sqrt{58}}\right)$.

Solution. The angle between planes is the angle between their normals:

$$\cos\theta = \frac{|(2, -3, 1)\cdot(1, -1, 0)|}{\sqrt{14}\,\sqrt2} = \frac{5}{\sqrt{28}} = \frac{5}{2\sqrt7}$$

The stated value has the wrong denominator ($\sqrt{58}$ is not $\sqrt{14}\cdot\sqrt2$), and a negative cosine would describe the obtuse angle, not the angle between the planes.

Question 46

State True or False: The line $\vec r = 2\hat i – 3\hat j – \hat k + \lambda\left(\hat i – \hat j + 2\hat k\right)$ lies in the plane $\vec r\cdot\left(3\hat i + \hat j – \hat k\right) + 2 = 0$.

Solution. A line lies in a plane when two things hold: it is parallel to the plane, and one of its points is on the plane.

1. Parallel: the direction $(1, -1, 2)$ and the normal $(3, 1, -1)$ give $3 – 1 – 2 = 0$. So the line is parallel to the plane.
2. On the plane: the point $(2, -3, -1)$ gives $3(2) + 1(-3) – 1(-1) + 2 = 6 – 3 + 1 + 2 = 6 \ne 0$.

The second test fails. The line runs parallel to the plane at a distance $\frac{6}{\sqrt{11}}$ from it, and never touches it.

The printed key says True, and it is wrong. The key’s answer is what you get by stopping after the first test: parallel is necessary for “lies in”, but not sufficient.

Question 47

State True or False: The vector equation of the line $\dfrac{x – 5}{3} = \dfrac{y + 4}{7} = \dfrac{z – 6}{2}$ is $\vec r = 5\hat i – 4\hat j + 6\hat k + \lambda\left(3\hat i + 7\hat j + 2\hat k\right)$.

Solution. This is Question 39 again: the point is $(5, -4, 6)$ and the direction $(3, 7, 2)$, exactly as stated.

Question 48

State True or False: The equation of a line, which is parallel to $2\hat i + \hat j + 3\hat k$ and which passes through the point $(5, -2, 4)$, is $\dfrac{x – 5}{2} = \dfrac{y + 2}{-1} = \dfrac{z – 4}{3}$.

Solution. The point is right, but the denominators $(2, -1, 3)$ are not the direction $(2, 1, 3)$. The correct line is $\frac{x – 5}{2} = \frac{y + 2}{1} = \frac{z – 4}{3}$.

Question 49

State True or False: If the foot of perpendicular drawn from the origin to a plane is $(5, -3, -2)$, then the equation of plane is $\vec r\cdot\left(5\hat i – 3\hat j – 2\hat k\right) = 38$.

Solution. The perpendicular from the origin is along the normal, so the normal is $(5, -3, -2)$ and the plane passes through $(5, -3, -2)$. The constant is $5(5) – 3(-3) – 2(-2) = 25 + 9 + 4 = 38$.

Common mistakes

• Stopping at “parallel” in Question 46. A line parallel to a plane either lies in it or never meets it; substituting one point decides which. The printed key stops one step early.
• Using cosine for the angle between a line and a plane in Questions 32, 36 and 44. The dot product with the normal gives the cosine of the angle with the normal, which is the sine of the angle with the plane.
• Distance from an axis versus distance from a plane in Question 29. From the $y$-axis it is $\sqrt{\alpha^2 + \gamma^2}$; from the $xz$-plane it is $|\beta|$.
• Calling $xy + yz = 0$ a pair of lines in Question 35. In three dimensions a linear equation is a plane; only in two dimensions is it a line.
• Copying the point’s coordinates with the wrong signs in Questions 39 and 48. In $\frac{y + 4}{7}$ the point has $y = -4$; the sign flips when reading it off.

Practise next

• Exemplar Short Answer Q1–15, the first band — direction cosines, lines that meet, and planes from a point and a normal.
• Exemplar Long Answer Q16–28 — feet of perpendiculars, skew lines and the family of planes through a line.
• NCERT textbook Exercise 11.1 for direction cosines, as in Questions 30 and 38.
• NCERT textbook Exercise 11.2 for the vector and cartesian equations of a line, as in Questions 39, 40, 47 and 48.