The availability of step-by-step ML Aggarwal Class 12 Solutions ISC Chapter 2 Three Dimensional Geometry MCQs can make challenging problems more manageable.
ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry MCQs
Choose the correct answer from the given four options in questions (1 to 42) :
Question 1.
The equations of the x-axis are
(a) x = 0, y = 0
(b) y = 0, z = 0
(c) z = 0, x = 0
(d) x = 0
Solution:
(b) y = 0, z = 0
Since x-axis be the intersection of XOZ plane and XOY plane
i.e. y = 0 and z = 0
Question 2.
The coordinates of the foot of perpendicular drawn from the point P (2, – 3, 4) on the y-axis are
(a) (2, 3, 4)
(b) (- 2, – 3, – 4)
(c) (0, – 3, 0)
(d) (2, 0, 4)
Solution:
(c) (0, – 3, 0)
Let M be the foot of ⊥ drawn from P (2, – 3, 4) on y-axis.
∴ Coordinates of M are (0, – 3, 0)
Question 3.
The distance of the point P (α, β, γ) from x-axis is
(a) |α|
(b) \(\sqrt{\alpha^2+\beta^2}\)
(c) \(\sqrt{\beta^2+\gamma^2}\)
(d) \(\sqrt{\gamma^2+\alpha^2}\)
Solution:
(c) \(\sqrt{\beta^2+\gamma^2}\)
Let M be the foot of J drawn from point P (α, β, γ) to x-axis then coordinates of M are (α, 0. 0)
∴ required ⊥ distance = |MP|
= \(\sqrt{(\alpha-\alpha)^2+(\beta-0)^2+(\gamma-0)^2}\)
= \(\sqrt{\beta^2+\gamma^2}\)
Question 4.
The length of perpendicular drawn from the point (4, – 7, 3) on they-axis is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 7 units
Solution:
(c) 5 units
Let M be the foot of ⊥ drawn from point P (4, – 7, 3) on y-axis
i.e. x = 0 and z = 0.
Then coordinates of M are (0, – 7, 0).
∴ required distance = |PM|
= \(\sqrt{(4-0)^2+(-7+7)^2+(3-0)^2}\)
= \(\sqrt{16+9}\)
= 5 units
Question 5.
A rectangular parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes. The length of a diagonal of the parallelopiped is
(a) 7 units
(b) \(\sqrt{38}\) units
(c) \(\sqrt{155}\) units
(d) none of these
Solution:
(a) 7 units
required length of diagonal of rectangular || piped = |PQ|
= \(\sqrt{(5-2)^2+(9-3)^2+(7-5)^2}\)
= \(\sqrt{9+36+4}\)
= 7 units
Question 6.
If the direction cosines of a line are < k, k, k >, then
(a) k > 0
(b) 0 < k < 1
(c) k = 1
(d) k = – \(\frac{1}{\sqrt{3}}\) or k = + \(\frac{1}{\sqrt{3}}\)
Solution:
(d) k = – \(\frac{1}{\sqrt{3}}\) or k = + \(\frac{1}{\sqrt{3}}\)
direction cosines of line are < k, k, k >.
∴ k2 + k2 + k2 = 1
⇒ 3k2 = 1
⇒ k = ± \(\frac{1}{\sqrt{3}}\)
Question 7.
If a line is equally inclined with the coordinate axes, then its direction cosines are
(a) ± < 1, 1, 1 >
(b) ± \(\left\langle\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right\rangle\)
(c) ± \(\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle\)
(d) ± \(\left\langle-\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right\rangle\)
Solution:
(c) ± \(\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle\)
Let α be the angle made by line with coordinate axis.
Then direction cosines of line are < cos α, cos α, cos α >
since cos2 α + cos2 α + cos2 α = 1
⇒ 3 cos2 α = 1
⇒ cos2 α = \(\frac{1}{3}\)
⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)
∴ required direction cosines of line are ;
< ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >
Question 8.
The image of the point (2, – 1, 5) in the plane \(\vec{r} \cdot \hat{i}\) = 0 is
(a) (- 2, – 1, 5)
(b) (2, 1, – 5)
(c) (- 2, 1, – 5)
(d) (2, 0, 0)
Solution:
(a) (- 2, – 1, 5)
eqn. of given plane be \(\vec{r} \cdot \hat{i}\) = 0
⇒ x = 0
i.e. YOZ-plane.
Let M be the foot of ⊥ drawn from P (2, – 1, 5) to plane x = 0.
Then coordinates of foot of ⊥ are (0,- 1, 5).
Let Q (α, β, γ) be the required image of point P in plane x = 0.
then M be the mid-point of PQ.
∴ \(\frac{\alpha+2}{2}\) = 0
\(\frac{\beta-1}{2}\) = – 1
\(\frac{\gamma+5}{2}\) = 5
⇒ α = – 2 ;
β = – 1;
γ = 5
Question 9.
O is the origin and P is a point at a distance of 3 units from origin. If direction ratios of OP are < 1, – 2, – 2 >, then the coordinates of P are
(a) (1, – 2, – 2)
(b) (3, – 6, – 6)
(c) \(\left(\frac{1}{3},-\frac{2}{3},-\frac{2}{3}\right)\)
(d) \(\left(\frac{1}{9},-\frac{2}{9},-\frac{2}{9}\right)\)
Solution:
(a) (1, – 2, – 2)
The eqn. of line OP be given by
\(\frac{x-0}{1}=\frac{y-0}{-2}=\frac{z-0}{-2}\) = t (say)
So any point on line OP be given by P (t, – 2t, – 2t)
Also |OP| = 3
⇒ \(\sqrt{t^2+4 t^2+4 t^2}\) = 3
⇒ 9t2 = 9
⇒ t = ± 1
Thus coordinates of P are (1, – 2, – 2) or (- 1, 2, 2)
Question 10.
A point P lies on the line segment joining the points (- 1, 3, 2) and (5, 0, 6). If x coordinate of P is 2 then its z-coordinates is [CBSE 2020]
(a) – 1
(b) 4
(c) \(\frac{3}{2}\)
(d) 8
Solution:
(b) 4
The eqn. of line QR which pass through Q (- 1, 3, 2) and R (5, 0, 6) be given by
\(\frac{x+1}{5+1}=\frac{y-3}{0-3}=\frac{z-2}{6-2}\)
i.e. \(\frac{x+1}{6}=\frac{y-3}{-3}=\frac{z-2}{4}\) ……………….(1)
Let z coordinate of point P be a and given x-coordinate of point P be 2 and point P lies on line (1).
∴ \(\frac{2+1}{6}=\frac{\alpha-2}{4}\)
⇒ \(\frac{1}{2}\) × 4 = α – 2
⇒ α = 4
Question 11.
If a line makes angle α, β and γ with the axes respectively, then cos 2α + cos 2β + cos 2γ =
(a) – 2
(b) – 1
(c) 1
(d) 2
Solution:
(b) – 1
Given line l makes an angle α, β, γ with coordinates axes.
∴ l = cos α ;
m = cos β ;
n = cos γ
where l, m, n are direction cosines of line.
∴ l2 + m2 + n2 = 1
⇒ cos2 α + cos2 β + cos2 γ = 1
⇒ \(\frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}+\frac{1+\cos 2 \gamma}{2}\) = 1
⇒ cos 2α + cos 2β + cos 2γ = – 1
Question 12.
If a line makes angle \(\frac{\pi}{3}\) and \(\frac{\pi}{4}\) with x-axis and y-axis respectively, then the angle made by the line with z-axis is
(a) π/2
(b) π/3
(c) π/4
(d) 5π/12
Solution:
(b) π/3
Let θ be the angle made by line with z-axis.
Since line makes an angle \(\frac{\pi}{3}\) and \(\frac{\pi}{4}\) with x-axis andy-axis.
Thus direction cosines of line are < cos \(\frac{\pi}{3}\), cos \(\frac{\pi}{3}\), cos θ >
Also l2 + m2 + n2 = 1 ; where < l, m, n > be
the D’ cosines of line.
∴ cos2 \(\frac{\pi}{3}\) + cos2 \(\frac{\pi}{4}\) + cos2 θ = 1
⇒ \(\frac{1}{4}+\frac{1}{2}\) + cos2 θ = 1
⇒ cos2 θ = \(\frac{1}{4}\)
cos θ = ± \(\frac{1}{2}\)
Since θ is acute
∴ cos θ > 0
∴ cos θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
Question 13.
The acute angle between the planes 2x – y + z = 6 and x + y + 2z = 3 is
(a) 45°
(b) 60°
(c) 30°
(d) 75°
Solution:
(b) 60°
We know that angle e between given planes
a1x + b1y + c1z + d1 = 0
and a2x + b2y + c2z + d2 = 0 is given by
cos θ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
Here a1 = 2;
b1 = – 1 ;
c1 = 1 ;
a2 = 1 ;
b2 = 1;
c2 = 2
∴ cos θ = \(\frac{2(1)-1(1)+1(2)}{\sqrt{4+1+1} \sqrt{1+1+4}}\)
= \(\frac{3}{6}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
Question 14.
The equation of the plane which cuts equal intercepts of unit length on coordinate axes is
(a) x + y + z – 1 = 0
(b) x + y + z + 1 = 0
(c) x + y – z = 1
(d) x – y – z = 1
Solution:
(a) x + y + z – 1 = 0
The eqn. of plane having intercepts a, b, e on coordinates axes be given by
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 …………………(1)
Here a = b = c = 1
∴ eqn. (1) reduces to ; x + y + z – 1 = 0
Question 15.
The distance of the plane \(\vec{r} \cdot\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right)\) = 1 from origin is
(a) 1
(b) 7
(c) \(\frac{1}{7}\)
(d) None of these
Solution:
(a) 1
The given equation of plane is cartesian form is given by
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right)\) = 1
⇒ \(\frac{2}{7} x+\frac{3}{7} y-\frac{6}{7} z\) = 1
⇒2x + 3y – 6z – 7 = 0 …………………(1)
∴ ⊥ distance of the plane (1) from O (0, 0, 0)
= \(\frac{|2 \times 0+3 \times 0-6 \times 0-7|}{\sqrt{2^2+3^2+(-6)^2}}\)
= \(\frac{7}{7}\)
= 1 units
Question 16.
The plane 2x – 3y + 6z – 11 = 0 makesan angle sin-1 (α) with x-axis. The value of α is equal to
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{\sqrt{2}}{3}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{3}{7}\)
Solution:
(c) \(\frac{2}{7}\)
The given plane be 2x – 3y + 6z – 11 = 0
∴ \(\vec{n}=2 \hat{i}-3 \hat{j}+6 \hat{k}\)
equation of given line be x-axis and direction ratios of x-axis are < 1, 0, 0 >
i.e., \(\vec{b}=\hat{i}+0 \hat{j}+0 \hat{k}\)
Since θ be the angle between given plane and given line.
∴ sin θ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k})}{\sqrt{2^2+(-3)^2+6^2} \sqrt{1^2+0^2+0^2}}\)
= \(\frac{2(1)-3(0)+6(0)}{7 \times 1}=\frac{2}{7}\)
⇒ θ = sin-1 \(\frac{2}{7}\)
Also given plane makes an angle sin-1 α with given line.
∴ α = \(\frac{2}{7}\)
Question 17.
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and the plane 2x – 2y + z = 5 is
(a) \(\frac{10}{6 \sqrt{5}}\)
(b) \(\frac{4}{5 \sqrt{2}}\)
(c) \(\frac{2 \sqrt{3}}{5}\)
(d) \(\frac{\sqrt{2}}{10}\)
Solution:
(d) \(\frac{\sqrt{2}}{10}\)
Equation of given line be
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Thus the line passing through the point (2, 3, 4) and having direction ratios < 3, 4, 5 >.
Thus vector equation of line passing through the point whose P.V be \(2 \hat{i}+3 \hat{j}+4 \hat{k}\)
and parallel to \(\vec{b}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) is given by
\(\vec{r}=2 \hat{i}+3 \hat{j}+4 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+5 \hat{k})\)
and given equation of plane be
2x – 2y + z = 5
D’ ratios of normal to plane are < 2, – 2, 1 >
∴ \(\vec{n}=2 \hat{i}-2 \hat{j}+\hat{k}\)
Let θ be the angle between given plane and given line.
Then sin θ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{3^2+4^2+5^2} \sqrt{2^2+(-2)^2+1^2}}\)
= \(\frac{3(2)+4(-2)+5(1)}{\sqrt{9+16+25} \sqrt{4+4+1}}\)
= \(\frac{3}{5 \sqrt{2} \times 3}\)
= \(\frac{1}{5 \sqrt{2}}=\frac{\sqrt{2}}{10}\)
Question 18.
The distance between the planes 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0 is
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{6}\)
(d) none of these
Solution:
(c) \(\frac{1}{6}\)
eqns. of given planes are,
2x + 2y – z + 2 = 0
and 4x + 4y – 2z + 5 = 0
Since both planes are parallel as their d’ratios are proportional.
Let P (x1, y1, z1) be any point on plane (1)
∴ 2x1 + 2y1 – z1 + 2 = 0 …………………….(3)
∴ required distance between || planes = ⊥ distance of point (x1, y1, z1) from plane (2).
= \(\frac{\left|4 x_1+4 y_1-2 z_1+5\right|}{\sqrt{4^2+4^2+(-2)^2}}\)
= \(\frac{\left|2\left(2 x_1+2 y_1-z_1\right)+5\right|}{\sqrt{16+16+4}}\)
= \(\frac{|2(-2)+5|}{6}\)
= \(\frac{1}{6}\)
Question 19.
The angle between the lines \(\vec{r}=(4 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})\) and \(\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-3 \hat{j}+2 \hat{k})\) is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{2}\)
Solution:
(b) \(\frac{\pi}{3}\)
Comparing the given lines with
\(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\)
and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\)
we have, \(\overrightarrow{b_1}=2 \hat{i}+\hat{j}-3 \hat{k}\) ;
\(\overrightarrow{b_2}=\hat{i}-3 \hat{j}+2 \hat{k}\)
Let θ be the acute angle be between given lines
∴ cos θ = \(\frac{\left|\overrightarrow{b_1} \cdot \overrightarrow{b_2}\right|}{\left|\overrightarrow{b_1}\right|\left|\vec{b}_2\right|}\)
= \(\frac{|2 \times 1+1 \times(-3)-3 \times 2|}{\sqrt{4+1+9} \sqrt{1+9+4}}\)
= \(\frac{7}{14}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
Question 20.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{6-z}{5}\) are mutually perpendicular, then k is equal to
(a) – \(\frac{10}{7}\)
(b) \(\frac{10}{7}\)
(c) – 10
(d) 7
Solution:
We know that, the given lines
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\)
and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)
are mutually ⊥ to each other
iff a1a2 + b1b2 + c1c2 = 0
iff – 3 × 3k + 2k × 1 + 2 × (- 5) = 0
⇒ – 7k = 10
⇒ k = – \(\frac{10}{7}\)
Question 21.
If a plane cuts Intercepts of lengths 8, 4 and 4 units on the coordinate axes respectively, then the length of perpendicular from origin to the plane is
(a) \(\frac{3}{8}\) units
(b) \(\frac{8}{3}\) units
(c) 8 units
(d) \(\frac{4}{3}\) units
Solution:
(b) \(\frac{8}{3}\) units
The eqn. of plane using intercept from be given by
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ……………………..(1)
Here plane (1) cut intercepts of lengths 8, 4 and 4 units on coordinates axes.
∴ a = 8 ; b = c = 4
∴ from (1) ;
\(\frac{x}{8}+\frac{y}{4}+\frac{z}{4}\) = 1
⇒ x + 2y + 2z – 8 = 0 ……………….(2)
Thus, length of ⊥ from (0, 0, 0) to plane (2) = \(\frac{|0+0+0-8|}{\sqrt{1+4+4}}\)
= \(\frac{8}{3}\) units
Question 22.
The angle between the lines \(\frac{x+4}{1}=\frac{y-3}{2}=\frac{z+2}{3}\) and \(\frac{x}{3}=\frac{y}{-2}=\frac{z}{1}\) is
(a) sin-1 \(\left(\frac{1}{7}\right)\)
(b) cos-1 \(\left(\frac{2}{7}\right)\)
(c) cos-1 \(\left(\frac{1}{7}\right)\)
(d) none of these
Solution:
Here a1 = 1 ;
a1 = 2 ;
c1 = 3
and a2 = 3 ;
b2 = – 2
c2 = 1
Thus, cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{|1 \times 3+2 \times(-2)+3 \times 1|}{\sqrt{1+4+9} \sqrt{9+4+1}}\)
= \(\frac{2}{14}=\frac{1}{7}\)
⇒ θ = cos \(\left(\frac{1}{7}\right)\)
Question 23.
If a line makes an angle of \(\frac{\pi}{4}\) with the positive directions of each of x-axis and y-axis, then the angle that the line makes
with the positive direction of z-axis is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Solution:
(d) \(\frac{\pi}{2}\)
Let the line makes an angle θ with z-axis. then direction cosines of line are ;
< cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{4}\), cos θ >
i.e. < \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\), cos θ >
Since \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2\) + cos2 θ = 1
⇒ \(\frac{1}{2}\) + \(\frac{1}{2}\) + cos2 θ = 1
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Question 24.
If the planes x + 2y + kz = 5 and 2x + y – 2z = 0 are at right angles, then the value of k is
(a) 2
(b) – 2
(c) \(\frac{1}{2}\)
(d) – \(\frac{1}{2}\)
Solution:
(a) 2
eqns. of given planes are ;
x + 2y + kz = 5 ………………….(1)
and 2x + y – 2z = 0 …………….(2)
D’ratios of planes are < 1, 2, k > and < 2, 1, – 2 >
Since planes (1) and (2) are at right angles.
So their normals are ⊥ to each other.
1 × 2 + 2 × 1 + k (- 2) = 0
⇒ 4 – 2k = 0
⇒ k = 2
Question 25.
The ratio in which the line segment joining the points (- 2, 4, 5) and (3, 5, – 4) is divided by the yz-plane is
(a) 3 : 2
(b) 2 : 3
(c) – 2 : 3
(d) 4 : – 3
Solution:
(b) 2 : 3
Thus by Section formula,
The coordinates of point R which divides the line segment PQ internally in the ratio k : 1 be given by
R \(\left(\frac{3 k-2}{k+1^{-}}, \frac{5 k+4}{k+1}, \frac{-4 k+5}{k+1}\right)\)
It is given that the line PQ is divided by yz-plane i.e. x = 0
⇒ \(\frac{3 k-2}{k+1}\) = 0
k = \(\frac{2}{3}\)
∴ required ratio be k : 1
i.e. \(\frac{2}{3}\) : 1
i.e. 2 : 3
Question 26.
The two lines x = ay + b, z = cy + d and x — a’y + b’, z — e/y + d’ are perpendicular to each other if
(a) aa’ + cc’ = 1
(b) \(\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}\) = – 1
(c) \(\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}\) = 1
(d) aa’ + cc’ = – 1
Solution:
(d) aa’ + cc’ = – 1
The given lines can be written as :
\(\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}\) ………………(1)
and \(\frac{x-b^{\prime}}{a^{\prime}}=\frac{y}{1}=\frac{z-d^{\prime}}{c^{\prime}}\) ……………….(2)
Direction ratios of lines (1) and (2) are ; < a, 1, c > and < a, 1, c >
Since lines (1) and (2) are ⊥ to each other.
∴ aa’ + 1 + cc’ = 0
Question 27.
The distance between the parallel planes ax + by + cz + d = 0 and ax + by + cz + d’ = 0 is
(a) \(\frac{\left|d+d^{\prime}\right|}{\sqrt{a^2+b^2+c^2}}\)
(b) \(\frac{\left|d-d^{\prime}\right|}{\sqrt{a^2+b^2+c^2}}\)
(c) \(\frac{d}{\sqrt{a^2+b^2+c^2}}\)
(d) none of these
Solution:
(b) \(\frac{\left|d-d^{\prime}\right|}{\sqrt{a^2+b^2+c^2}}\)
Given eqns. of planes are;
ax + by + cz + d = 0 …………………..(1)
and ax + by + cz + d’ = 0 …………………..(2)
Distance between parallel planes = distance of point on plane (1) to plane (2)
putting y = 0 = z in eqn.(1) ;
Thus the point(- \(\frac{d}{d}\), 0, o) lies on plane (1)
∴ required distance = \(\frac{\left|a\left(-\frac{d}{a}\right)+0+0+d^{\prime}\right|}{\sqrt{a^2+b^2+c^2}}\)
= \(\frac{\left|d-d^{\prime}\right|}{\sqrt{a^2+b^2+c^2}}\)
Question 28.
Equations of the line passing through (1, 1, 1) and perpendicular to the plane 2x + 3y + z + 5 = 0 are
(a) \(\frac{x+1}{1}=\frac{y-1}{3}=\frac{z-1}{2}\)
(b) \(\frac{x-1}{3}=\frac{y-1}{3}=\frac{z-1}{2}\)
(c) \(\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{1}\)
(d) \(\frac{x-1}{3}=\frac{y-1}{1}=\frac{z-1}{1}\)
Solution:
(c) \(\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{1}\)
eqn. of given plane be,
2x + 3y + z + 5 = 0 …………………..(1)
∴ direction ratios of normal to plane (1) are < 2, 3, 1 >
∴ eqn. of line through the point (1, 1, 1)
having direction ratios < 2, 3, 1 > be given by \(\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{1}\)
Question 29.
The equation of the plane, which makes with coordinate axes, a triangle with its centroid (α, β, γ) is
(a) αx + βy + γz = 3
(b) αx + βy + γz = 1
(c) \(\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}\) = 1
(d) \(\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}\) = 3
Solution:
(d) \(\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}\) = 3
Let the eqn. of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1
Now plane (1) cut coordinate axes at
A (a, 0, 0), B (0, b, 0) and C (0, 0, c).
∴ Centroid of ∆ABC be \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)
Also given centroid of ∆ABC be (α, β, γ)
∴ \(\frac{a}{3}\) = α
⇒ a = 3α;
\(\frac{b}{3}\) = β
⇒ b = 3β
and \(\frac{c}{3}\) = γ
⇒ c = 3γ
∴ from (1) ;
\(\frac{x}{3 \alpha}+\frac{y}{3 \beta}+\frac{z}{3 \gamma}\) = 1
⇒ \(\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}\) = 3
Question 30.
If a variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axes is \(\frac{1}{2}\), then the plane passes through the point
(a) \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\)
(b) (2, 2, 2)
(c) (1, 1, 1)
(d) (0, 0, 0)
Solution:
(b) (2, 2, 2)
Let the eqn. of plane in intercept form be given by
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ……………….(1)
It is given that
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}\)
\(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\) = 1 …………….(2)
From (1) and (2) ; we have
The variable plane pass through the point (2, 2, 2).
Question 31.
If the angle θ between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane 2x – y + \(\sqrt{\lambda}\)z + 4 = 0 is such that sin θ = \(\frac{1}{3}\), then the value of λ is
(a) – \(\frac{4}{3}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{5}{3}\)
(d) – \(\frac{3}{5}\)
Solution:
(c) \(\frac{5}{3}\)
Given direction ratios of line are < 1, 2, 2 > direction ratios of normal to plane are < 2, – 1, \(\sqrt{\lambda}\) >
It is given that θ be the angle between given line and plane s.t
sin θ = \(\frac{1}{3}\)
⇒ \(\frac{1}{3}\) = sin θ
= \(\frac{|1 \times 2+2(-1)+2 \sqrt{\lambda}|}{\sqrt{1+4+4} \sqrt{4+1+\lambda}}\)
= \(\frac{1}{3}=\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}\)
⇒ \(\sqrt{5+\lambda}=2 \sqrt{\lambda}\)
⇒ 5 + λ = 4λ
⇒ λ = \(\frac{5}{3}\)
Question 32.
The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is
(a) 90°
(b) 45°
(c) 30°
(d) 0
Solution:
(a) 90°
eqns. of given lines can be written as 2x = 3y = – z
⇒ \(\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}\) ………………….(1)
and 6x = – y = – 4z
⇒ \(\frac{x}{2}=\frac{-y}{-12}=\frac{z}{-3}\) ……………….(2)
Here a1 = 3 ;
b1 = 2 ;
c1 = – 6 ;
a2 = 2
b2 = – 12 ;
c2 = – 3
Let θ be the angle between two given lines.
Then cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{|3 \times 2+2 \times(-12)-6 \times(-3)|}{\sqrt{9+4+36} \sqrt{4+144+9}}\)
= 0
⇒ θ = \(\frac{\pi}{2}\)
Question 33.
A vector parallel to the line of intersection of the planes \(\vec{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})\) = 5 and \(\vec{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})\) = 3 is
(a) \(2 \hat{i}+7 \hat{j}-13 \hat{k}\)
(b) \(2 \hat{i}-7 \hat{j}+13 \hat{k}\)
(c) \(-2 \hat{i}+7 \hat{j}+13 \hat{k}\)
(d) \(2 \hat{i}+7 \hat{j}+13 \hat{k}\)
Solution:
(c) \(-2 \hat{i}+7 \hat{j}+13 \hat{k}\)
Eqns. of given planes are
\(\vec{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})\) = 5 ……………….(1)
and \(\vec{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})\) = 3 …………………..(2)
∴ vector normal to plane (1) = \(\overrightarrow{n_1}\)
= \(3 \hat{i}-\hat{j}+\hat{k}\)
vector normal to plane (2) = \(\overrightarrow{n_2}\)
= \(\hat{i}+4 \hat{j}-2 \hat{k}\)
also \(\overrightarrow{n_1} \times \overrightarrow{n_2}\) is ⊥ to plane of \(\overrightarrow{n_1} \text { and } \overrightarrow{n_2}\)
∴ \(\overrightarrow{n_1} \text { and } \overrightarrow{n_2}\) is a vector || to the plane of intersection of (1) and (2)
Thus \(\overrightarrow{n_1} \times \overrightarrow{n_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 1 \\
1 & 4 & -2
\end{array}\right|\)
= \(-2 \hat{i}+7 \hat{j}+43 \hat{k}\)
Question 34.
The locus represented by xy +yz = 0 is
(a) A pair of perpendicular lines
(b) A pair of parallel lines
(c) A pair of parallel planes
(d) A pair of perpendicular planes
Solution:
(d) A pair of perpendicular planes
Given equation of locus be
xy + yz = 0
⇒ y (x + z) = 0
⇒ y = 0 or x + z = 0
Now y = 0 represents the XOZ plane.
and x + z = 0 also represents the plane.
Thus the given planes represents the pair of planes
Further D’ ratios of normal to plane y = 0 are < 0, 1, 0 >
and D’ ratio of normal to plane x + z = 0 are < 1, 0, 1 >
Further 0 × 1 + 1 × 0 + 0 × 1 = 0
∴ both planes are perpendicular.
Question 35.
If the planes \(\vec{r} \cdot(2 \hat{j}-\lambda \hat{j}+3 \hat{k})\) = 0 and \(\vec{r} \cdot(\lambda \hat{i}+5 \hat{j}-\hat{k})\) = 5 are perpendicular to each other, then the value of λ2 + λ is
(a) 0
(b) – 2
(c) -1
(d) 2
Solution:
(a) 0
eqn. of given planes are ;
\(\vec{r} \cdot(2 \hat{j}-\lambda \hat{j}+3 \hat{k})\) = 0
and \(\vec{r} \cdot(\lambda \hat{i}+5 \hat{j}-\hat{k})\) = 5
Here \(\overrightarrow{n_1}=2 \hat{i}-\lambda \hat{j}+3 \hat{k}\) ;
\(\overrightarrow{n_2}=\lambda \hat{i}+5 \hat{j}-\hat{k}\)
Given planes (1) and (2) are ⊥ to each other.
∴ \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}\) = 0
⇒ \((2 \hat{j}-\lambda \hat{j}+3 \hat{k}) \cdot(\lambda \hat{i}+5 \hat{j}-\hat{k})\) = 0
⇒ 2λ – 5λ – 3 = 0
⇒ – 3λ = 3
⇒ λ = – 1
Thus λ2 + λ = (- 1)2 – 1
= 1 – 1 = 0
Question 36.
The equation of the plane passing through the point (1, 1, 0) and perpendicular to the line \(\vec{r}=2 \hat{i}+3 \hat{j}+4 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+5 \hat{k})\) is
(a) 3x – 4y + 5z = 7
(b) 3x + 4y + 5z = 7
(c) x + y – 4z = 9
(d) 3x + 4y – 12 = 0
Solution:
(b) 3x + 4y + 5z = 7
eqn. of line be,
\(\vec{r}=2 \hat{i}+3 \hat{j}+4 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+5 \hat{k})\) ……………………..(1)
∴ Direction ratios of line (1) are < 3, 4, 5 >
Thus required eqn. of plane through the point (1, 1, 0) and ⊥ to given line (1) be given by
3 (x – 1) + 4 (y – 1) + 5(z – 0) = 0
⇒ 3x + 4y + 5z = 7
Question 37.
The distance of the point (2, 1, – 1) from the plane x – 2y + 4z = 9 is
(a) \(\frac{\sqrt{13}}{21}\)
(b) \(\frac{13}{21}\)
(c) \(\frac{13}{\sqrt{21}}\)
(d) \(\sqrt{\frac{13}{21}}\)
Solution:
(c) \(\frac{13}{\sqrt{21}}\)
eqn. of given plane be x – 2y + 4z – 9 = 0
∴ required distance of point (2, 1, – 1) from plane x – 2y + 4z – 9 = 0
= \(\frac{|2-2 \times 1+4 \times(-1)-9|}{\sqrt{1^2+(-2)^2+4^2}}\)
= \(\frac{13}{\sqrt{2}}\) units
Question 38.
The angle between a line with direction ratios < 2, 2, 1 > and a line joining the points (3, 1, 4)and (7, 2, 12) is
(a) cos-1 \(\left(\frac{2}{3}\right)\)
(b) cos-1 \(\left(\frac{3}{2}\right)\)
(c) tan-1 \(\left(-\frac{2}{3}\right)\)
(d) none of these
Solution:
(a) cos-1 \(\left(\frac{2}{3}\right)\)
Given direction ratio of line (1) are < 2, 2, 1 >
and direction ratios of line (2) are
< 7 – 3, 2 – 1, 12 – 4 > i.e. < 4, 1, 8 >
Let θ be the angle between gives lines
Then cos θ = \(\frac{|2 \times 4+2 \times 1+1 \times 8|}{\sqrt{4+4+1} \sqrt{16+1+64}}\)
= \(\frac{18}{3 \times 9}=\frac{2}{3}\)
∴ θ = cos-1 \(\left(\frac{2}{3}\right)\)
Question 39.
The lines \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\) and \(\frac{x-x_1}{a^{\prime}}=\frac{y-y_1}{b^{\prime}}=\frac{z-z_1}{c^{\prime}}\) are
(a) parallel
(b) intersecting
(C) skew
(d) coincident
Solution:
(b) intersecting
Clearly both lines passes through the point (x1, y1, z1)
∴ given lines are intersecting.
Question 40.
The vector equation of the line passing through the points A (3, 4, – 7) and B (1, – 1,6) is
(a) \(\vec{r}=3 \hat{i}+4 \hat{j}-7 \hat{k}+\lambda(\hat{i}-\hat{j}+6 \hat{k})\)
(b) \(\vec{r}=3 \hat{i}+4 \hat{j}-8 \hat{k}+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})\)
(c) \(\vec{r}=\hat{i}-\hat{j}+6 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}-7 \hat{k})\)
(d) \(\vec{r}=\hat{i}-\hat{j}+6 \hat{k}+\lambda(4 \hat{i}+3 \hat{j}-\hat{k})\)
Solution:
(b) \(\vec{r}=3 \hat{i}+4 \hat{j}-8 \hat{k}+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})\)
The vector eqn. of line through the point whose position vector \(\vec{a}=3 \hat{i}+4 \hat{j}-7 \hat{k}\) and || to vector whose P.V
\(\vec{b}=(\hat{i}-\hat{j}+6 \hat{k})-(3 \hat{i}+4 \hat{j}-7 \hat{k})\)
i.e. \(\vec{b}=-2 \hat{i}-5 \hat{j}+13 \hat{k}\)
be given by \(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e. \(\vec{r}=3 \hat{i}+4 \hat{j}-7 \hat{k}+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})\)
Question 41.
The vector equation of the line passing through the point (- 1, 5, 4) and perpendicular to the plane z = 0 is
(a) \(\vec{r}=-\hat{i}+5 \hat{j}+4 \hat{k}+\lambda(\hat{i}+\hat{j})\)
(b) \(\vec{r}=-\hat{i}+5 \hat{j}+(4+\lambda) \hat{k}\)
(c) \(\vec{r}=\hat{i}-5 \hat{j}-4 \hat{k}+\lambda \hat{k}\)
(d) \(\overrightarrow{r}=\lambda \hat{k}\) [CBSE 2020]
Solution:
eqn. of given plane be z = 0
and direction ratios of normal to plane are < 0, 0, 1 >
Thus vector eqn. of line through the point (- 1, 5, 4) having P.V \(\vec{a}=-\hat{i}+5 \hat{j}+4 \hat{k}\)
and parallel to vector \(\vec{b}=\hat{k}\) be given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\vec{r}=(-\hat{i}+5 \hat{j}+4 \hat{k})+\lambda \hat{k}\)
⇒ \(\vec{r}=(-\hat{i}+5 \hat{j}+4 \hat{k})+\lambda \hat{k}\),
where λ be the parameter.
Question 42.
The line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is parallel to the plane
(a) 2x + 3y + 4z = 0
(b) 3x + 4y – 5z + 2 = 0
(c) 2x + y – 2z = 0
(d) x – y + z = 2
Solution:
(c) 2x + y – 2z = 0
eqn. of given line be,
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Now line (1) is parallel to plane if normal to plane is ⊥ to given line.
In option (c) ;
D’ratios of normal to plane are < 2, 1, – 2 >
Here 3 × 2 + 4 × 1 + 5 × (- 2) = 0.