Three Dimensional Geometry jee mains questions with solutions

Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

$\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 6$
$\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$
Both (a) and (b)
None of the above

1

2

3

4

(2)

Ans. $\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Explanation:

Let $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} .$ Then,

$\hat{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$

Hence, the required equation of the plane is

$\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

$\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$
$α x + β y + γ y = 3 α β γ$
$\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = \frac{1}{2}$
None of these

1

2

3

4

(1)

Ans. $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

Explanation:

Let the equation of the required plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ...(i)

It meets coordinate axes in points

$A (a, 0, 0), B (0, b, 0), C (0, 0, c)$

The centroid of $∆ A B C$ is $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$

$\Rightarrow$ $\frac{a}{3} = α, \frac{b}{3} = β, \frac{c}{3} = γ$

$\Rightarrow$ $a = 3 α, b = 3 β, c = 3 γ$

Hence, the required plane is

$\frac{x}{3 α} + \frac{y}{3 β} + \frac{z}{3 γ} = 1$

i.e., $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

0
1
2
3

1

2

3

4

(1)

Ans. 0

Explanation:

The given line is

$\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$

$\Rightarrow$ $\frac{x - 2}{2} = \frac{y - \frac{5}{2}}{- \frac{3}{2}} = \frac{z + 1}{0}$

This shows that the given line passes through the point $(2, \frac{5}{2}, - 1)$ and has direction ratios $(2, - \frac{3}{2}, 0) .$ Thus, given line passes through the point having position vector $\vec{a} = 2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}$ and is parallel to the vector $\vec{b} = (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k}) .$

So, its vector equation is

$\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k})$

Hence, $p = 0$

Q 14 :

The equation of the plane through (1, 2, 3) and parallel to the plane $2 x + 3 y - 4 z = 0$ is:

$2 x + 3 y + 4 z = 4$
$2 x + 3 y + 4 z + 4 = 0$
$2 x - 3 y + 4 z + 4 = 0$
$2 x + 3 y - 4 z + 4 = 0$

1

2

3

4

(4)

Ans. $2 x + 3 y - 4 z + 4 = 0$

Explanation:

Plane parallel to the plane $2 x + 3 y - 4 z = 0$ is

$2 x + 3 y - 4 z + k = 0$ ...(i)

Also plane (i) is passing through (1, 2, 3).

$∴$ $(2) (1) + (3) (2) - (4) (3) + k = 0$

$\Rightarrow$ $k = 4$

$∴$ Required plane is $2 x + 3 y - 4 z + 4 = 0 .$

Q 15 :

The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2 x + 3 y + 4 z + 5 = 0$ , the point (1, 1, 1) is:

$20 x + 23 y + 26 z - 69 = 0$
$20 x + 23 y + 26 z + 69 = 0$
$23 x + 20 y + 26 z - 69 = 0$
None of the above

1

2

3

4

(1)

Ans. $20 x + 23 y + 26 z - 69 = 0$

Explanation:

$(x + y + z - 6) + λ (2 x + 3 y + 4 z + 5) = 0$

$\Rightarrow$ $λ = \frac{3}{14}$

$\Rightarrow$ $20 x + 23 y + 26 z - 69 = 0$

Q 16 :

If the given planes $a x + b y + c z + d = 0$ and $a' x + b' y + c' z + d' = 0$ be mutually perpendicular, then:

$\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}$
$\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} = 0$
$a a' + b b' + c c' + d d' = 0$
$a a' + b b' + c c' = 0$

1

2

3

4

(4)

Ans. $a a' + b b' + c c' = 0$

Explanation:

It is a fundamental concept.

Q 17 :

If the planes $x + 2 y + k z = 0$ and $2 x + y - 2 z = 0$ are at right angles, then the value of $k$ is:

$- \frac{1}{2}$
$\frac{1}{2}$
$- 2$
$2$

1

2

3

4

(4)

Ans. 2

Explanation:

Obviously, $2 (1) + 1 (2) - 2 (k) = 0$

$\Rightarrow$ $k = 2$

Q 18 :

The intercepts of the plane $5 x - 3 y + 6 z = 60$ on the coordinate axes are:

$(10, 20, - 10)$
$(10, - 20, 12)$
$(12, - 20, 10)$
$(12, 20, - 10)$

1

2

3

4

(3)

Ans. $(12, - 20, 10)$

Explanation:

$\frac{5 x}{60} - \frac{3 y}{60} + \frac{6 z}{60} = 1$

$\Rightarrow$ $\frac{x}{12} - \frac{y}{20} + \frac{z}{10} = 1$

$Hence, the intercepts are (12, - 20, 10)$

Q 19 :

The two planes $x - 2 y + 4 z = 10$ and $18 x + 17 y + k z = 50$ are perpendicular, if $k$ is equal to:

$- 4$
$4$
$2$
$- 2$

1

2

3

4

(2)

Ans. 4

Explanation:

$\Rightarrow$ $18 \times 1 + 17 \times (- 2) + 4 \times k = 0$

$\Rightarrow$ $18 - 34 + 4 k = 0$ $[using a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0]$

$\Rightarrow$ $4 k = 16 \Rightarrow k = 4$

Q 20 :

Let the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,$ then the value of $p$ is:

$\frac{3}{7}$
$\frac{7}{3}$
$\frac{4}{3}$
$\frac{3}{4}$

1

2

3

4

(2)

Ans. $\frac{7}{3}$

Explanation:

The distance of point $(1, 1, p)$ from the plane

$\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$

or (in Cartesian form) $3 x + 4 y - 12 z + 13 = 0$ is

$d_{1} =$ $| \frac{3 \times 1 + 4 \times 1 - 12 \times p + 13}{\sqrt{3^{2} + 4^{2} + {(- 12)}^{2}}} |$

$=$ $| \frac{3 + 4 - 12 p + 13}{\sqrt{169}} |$ $=$ $| \frac{20 - 12 p}{13} |$ $\dots (i)$

The distance of the point $(- 3, 0, 1)$ from the plane $3 x + 4 y - 12 z + 13 = 0$ is

$d_{2} =$ $| \frac{3 \times (- 3) + 4 \times 0 - 12 \times 1 + 13}{\sqrt{3^{2} + 4^{2} + {(- 12)}^{2}}} |$

$= \frac{8}{13}$

According to the given condition,

$d_{1} = d_{2}$

$\Rightarrow$ $| \frac{20 - 12 p}{13} |$ $= \frac{8}{13}$

$\Rightarrow$ $\frac{20 - 12 p}{13} = \pm \frac{8}{13}$

Taking +ve sign, we get

$\frac{20 - 12 p}{13} = \frac{8}{13}$

$\Rightarrow$ $p = 1$

Taking $-$ ve sign, we get

$\frac{20 - 12 p}{13} = - \frac{8}{13}$

$\Rightarrow$ $p = \frac{28}{12} = \frac{7}{3}$

Q 21 :

The distance of the plane $\vec{r} \cdot (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 1$ from the origin is:

1
7
$\frac{1}{7}$
None of these

1

2

3

4

(1)

Ans. 1

Explanation:

The distance of the plane $\vec{r} \cdot (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 1$ from the origin is equal to 1.

Q 22 :

The length of the perpendicular from the origin to the plane $3 x + 4 y + 12 z = 52$ is:

$3$
$- 4$
$5$
None of these

1

2

3

4

(4)

Ans. None of these

Explanation:

$p =$ $| \frac{- 52}{\sqrt{9 + 16 + 144}} |$ $=$ $| \frac{- 52}{13} |$ $=$ $| - 4 |$ $= 4$

Q 23 :

The equation of the plane which is parallel to the plane $x - 2 y + 2 z = 5$ and whose distance from the point (1, 2, 3) is 1, is:

$x - 2 y + 2 z = 3$
$x - 2 y + 2 z + 3 = 0$
$x - 2 y + 2 z = 6$
$x - 2 y + 2 z + 6 = 0$

1

2

3

4

(3)

Ans. $x - 2 y + 2 z = 6$

Explanation:

$Equation of plane parallel to x - 2 y + 2 z = 5 is$

$x - 2 y + 2 z + k = 0$ ...(i)

$Now, according to question,$

$\frac{1 - 4 + 6 + k}{\sqrt{9}} = \pm 1$

$or$ $k + 3 = \pm 3$

$\Rightarrow$ $k = 0 or - 6$

$∴$ $x - 2 y + 2 z - 6 = 0$

$or$ $x - 2 y + 2 z = 6$

Q 24 :

The distance of the plane $6 x - 3 y + 2 z - 14 = 0$ from the origin is:

2
1
14
8

1

2

3

4

(1)

Ans. 2

Explanation:

Distance of plane from origin

$= \frac{d}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{14}{\sqrt{6^{2} + 3^{2} + 2^{2}}}$

$= \frac{14}{\sqrt{49}} = \frac{14}{7} = 2$

Q 25 :

The distance of the origin (0, 0, 0) from the plane $- 2 x + 6 y - 3 z = - 7$ is:

$1 unit$
$\sqrt{2} units$
$2 \sqrt{2} units$
$3 units$

1

2

3

4

(1)

Ans. 1 unit

Explanation:

$Given, equation of plane is$

$- 2 x + 6 y - 3 z = - 7$

$\Rightarrow$ $- 2 x + 6 y - 3 z + 7 = 0$

$We know distance of point (x_{1}, y_{1}, z_{1}) from a plane A x + B y + C z + D = 0 is given as$

$Distance =$ $| \frac{A x_{1} + B y_{1} + C z_{1} + D}{\sqrt{A^{2} + B^{2} + C^{2}}} |$

$∴$ $Distance =$ $| \frac{- 2 (0) + 6 (0) + (- 3) (0) + 7}{\sqrt{4 + 36 + 9}} |$

$=$ $| \frac{7}{\sqrt{49}} |$ $=$ $| \frac{7}{7} |$ $= 1 unit$

Q 26 :

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lie in the plane $x + 3 y - α z + β = 0,$ then $(α, β)$ equals:

$(- 6, 7)$
$(- 5, 5)$
$(5, - 15)$
$(6, - 17)$

1

2

3

4

(1)

Ans. $(- 6, 7)$

Explanation:

Since the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane

$x + 3 y - α z + β = 0$

Therefore, the point $(2, 1, - 2)$ lies on the plane.

i.e., $2 + 3 + 2 α + β = 0$

or $2 α + β + 5 = 0$ $\dots (i)$

Also, the normal to the plane will be perpendicular to the line.

Therefore,

$3 \times 1 - 5 \times 3 + 2 \times (- α) = 0$

$\Rightarrow$ $α = - 6$

From equation (i), we have

$β = 7$

Therefore, $(α, β) = (- 6, 7)$

Q 27 :

The line $\frac{x - 1}{1} = \frac{y - 2}{- 2} = \frac{z - 1}{3}$ and the plane $x + 2 y + z = 6$ meet at:

no point
only one point
infinitely many points
none of these

1

2

3

4

(3)

Ans. infinitely many points

Explanation:

Direction ratios of given line are 1, -2, 3 and the d.r. of normal to the given plane are 1, 2, 1.

Since 1 × 1 + (-2) × 2 + 3 × 1 = 0, therefore, the line is parallel to the plane.

Also, the base point of the line (1, 2, 1) lies in the given plane.

(1 + 2 × 2 + 1 = 6 is true)

Hence, the given line lies in the given plane.

Q 28 :

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if:

$k = 3$ or $- 2$
$k = 0$
$k = 1$ or $- 1$
$k = 0$ or $- 3$

1

2

3

4

(4)

Ans. $k = 0$ or $- 3$

Explanation:

We know, if two lines are coplanar then

$| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |$ $= 0$

Therefore, $| \begin{matrix} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix} |$ $= 0$

$\Rightarrow$ $| \begin{matrix} - 1 & 1 & 1 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix} |$ $= 0$

Applying $C_{1} \to C_{1} + C_{2}$ and $C_{2} \to C_{2} - C_{3}$

$\Rightarrow$ $| \begin{matrix} 0 & 0 & 1 \\ 2 & 1 + k & - k \\ k + 2 & 1 & 1 \end{matrix} |$ $= 0$

Expanding along $C_{1}$ ,

$1 {2 - (1 + k) (2 + k)} = 0$

$2 - (2 + 3 k + k^{2}) = 0$

$k^{2} + 3 k = 0$

$k (k + 3) = 0$

$∴$ $k = 0, - 3$

Q 29 :

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

$2 x + 3 y + 4 z = 0$
$3 x + 4 y - 5 z = 7$
$2 x + y - 2 z = 0$
$x - y + z = 2$

1

2

3

4

(3)

Ans. $2 x + y - 2 z = 0$

Explanation:

$D.R.:$ $3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{a}$ $(say)$

$\vec{a} \cdot \vec{n} = 0$

$∴$ $(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \cdot \vec{n} = 0$

$From option (3)$

$(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + \hat{j} - 2 \hat{k})$ $= 3 (2) + 4 (1) + 5 (- 2)$

$= 0$

Q 30 :

The point of intersection of the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z,$ is:

$(1, 1, 1)$
$(1, - 1, - 1)$
$(- 1, 1, - 1)$
$(- 1, - 1, - 1)$

1

2

3

4

(4)

Ans. $(- 1, - 1, - 1)$

Explanation:

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

$\Rightarrow$ $x = 2 λ + 1, y = 3 λ + 2, z = 4 λ + 3$

Also, $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = μ$

$\Rightarrow$ $x = 5 μ + 4, y = 2 μ + 1, z = μ$

Now, $2 λ + 1 = 5 μ + 4$

$3 λ + 2 = 2 μ + 1$

$4 λ + 3 = μ$

On solving, $λ = - 1, μ = - 1$

$x = - 1, y = - 1, z = - 1$

$∴$ $Point = (- 1, - 1, - 1)$

Q 31 :

The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is:

3
2
1
0

1

2

3

4

(1)

Ans. 3

Explanation:

$\frac{x_{0} - 3}{1} = \frac{y_{0} - 4}{2} = \frac{z_{0} - 5}{2} = k$

$x_{0} = k + 3, y_{0} = 2 k + 4, z_{0} = 2 k + 5$

$We have, x_{0} + y_{0} + z_{0} = 17$

$5 k + 12 = 17$

$k = 1$

$x_{0} = 4, y_{0} = 6, z_{0} = 7$

$point = (4, 6, 7)$

$Distance between (3, 4, 5) and (4, 6, 7)$

$= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}}$

$= \sqrt{1 + 4 + 4}$

$= 3$

Q 32 :

The equation of plane passing through the point $(0, 7, - 7)$ and containing the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is:

$x - y - z = 0$
$4 x + y + z = 0$
$x + y + z = 0$
$3 x + 2 y + 2 z = 0$

1

2

3

4

(3)

Ans. $x + y + z = 0$

Explanation:

Equation of plane passing through $(0, 7, - 7)$ is

$a (x - 0) + b (y - 7) + c (z + 7) = 0$ ...(1)

The line passes through $(- 1, 3, - 2)$

Therefore, $a + 4 b - 5 c = 0$ ...(2)

and $- 3 a + 2 b + c = 0$ ...(3)

On solving both equations,

$\frac{a}{14} = \frac{b}{14} = \frac{c}{14} = k$

Therefore, $x + y + z = 0$

Q 33 :

A line $L : \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point $' S'$ on the line $(L)$ in the given plane $(P)$ is (2, 3, 1), then co-ordinates of point S, is:

$(1, 0, 3)$
$(0, - 1, 3)$
$(- 2, - 3, - 1)$
$(4, 7, - 1)$

1

2

3

4

(4)

Ans. $(4, 7, - 1)$

Explanation:

$Line L$

$\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1} = λ (let)$

$x = λ + 2, y = 2 λ + 3, z = - λ + 1$

$direction ratio of plane (2, 0, 8)$

$Than$

$2 (λ + 2) + 0 (2 λ + 3) + 8 (- λ + 1) = 0$

$2 λ + 4 + 0 - 8 λ + 8 = 0$

$- 6 λ = - 12$

$λ = 2$

$Co-ordinate of point S = (2 + 2, 2 \times 2 + 3, - 2 + 1)$

$S (4, 7, - 1)$

Q 34 :

The angle between the two planes $x + y - z = 3$ and $3 x + 2 y + z = 5$ is:

$\cos^{- 1} 4$
$\cos^{- 1} \frac{2 \sqrt{42}}{21}$
$\cos^{- 1} \frac{1}{4}$
$\cos^{- 1} \frac{1}{\sqrt{42}}$

1

2

3

4

(2)

Ans. $\cos^{- 1} \frac{2 \sqrt{42}}{21}$

Explanation:

$P_{1} = x + y - z = 3$

$P_{2} = 3 x + 2 y + z = 5$

${\vec{x}}_{1} = \hat{i} + \hat{j} - \hat{k}$

${\vec{x}}_{2} = 3 \hat{i} + 2 \hat{j} + \hat{k}$

$\cos θ =$ $\frac{| {\vec{x}}_{1} \cdot {\vec{x}}_{2} |}{| {\vec{x}}_{1} | | {\vec{x}}_{2} |}$ $= \frac{| 3 + 2 - 1 |}{\sqrt{3} \sqrt{14}}$

$= \frac{4}{\sqrt{42}} = \frac{4 \sqrt{42}}{42}$

$\cos θ = \frac{2 \sqrt{42}}{21}$

$∴$ $θ = \cos^{- 1} (\frac{2 \sqrt{42}}{21})$

Q 35 :

The vector equation of the line joining the points $(- 2, - 3, - 4)$ and $(1, - 2, 4)$ is:

$\vec{r} = (- 2 \hat{i} - 3 \hat{j} - 4 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 4 \hat{k})$
$\vec{r} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + λ (3 \hat{i} - \hat{j} + 8 \hat{k})$
$\vec{r} = (- 2 \hat{i} - 3 \hat{j} - 4 \hat{k}) + λ (3 \hat{i} + \hat{j} + 8 \hat{k})$
$\vec{r} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + λ (3 \hat{i} + \hat{j} + 8 \hat{k})$

1

2

3

4

(3)

Ans. $\vec{r} = (- 2 \hat{i} - 3 \hat{j} - 4 \hat{k}) + λ (3 \hat{i} + \hat{j} + 8 \hat{k})$

Explanation:

$A (- 2, - 3, - 4) and B (1, - 2, 4)$

$\vec{a} = - 2 \hat{i} - 3 \hat{j} - 4 \hat{k}$

$\vec{b} = \hat{i} - 2 \hat{j} + 4 \hat{k}$

$\vec{r} = (- 2 \hat{i} - 3 \hat{j} - 4 \hat{k}) + λ (3 \hat{i} + \hat{j} + 8 \hat{k})$

Q 36 :

If the equation of a floor of a room is given by $x + y - z + 4 = 0$ and the equation of roof is given by $x + y - z + 5 = 0,$ then the height of the room is:

$\frac{1}{6}$ units
$\frac{1}{3}$ units
$\frac{1}{\sqrt{3}}$ units
$\frac{1}{\sqrt{6}}$ units

1

2

3

4

(2)

Ans. $\frac{1}{3}$ units

Explanation:

$\frac{| 5 - 4 |}{\sqrt{5 + 4}}$ $= \frac{1}{3} unit$

Q 37 :

Find the locus of a point whose distance from $x$ -axis is twice the distance from the point $(1, - 1, 2)$ :

$y^{2} + 2 x - 2 y - 4 z + 6 = 0$
$x^{2} + 2 x - 2 y - 4 z + 6 = 0$
$x^{2} - 2 x + 2 y - 4 z + 6 = 0$
$z^{2} - 2 x + 2 y - 4 z + 6 = 0$

1

2

3

4

(3)

$Let the point P (x_{p}, y_{p}, z_{p}) be the required point,$

$the distance of the point from x-axis is \sqrt{y_{p}^{2} + z_{p}^{2}}, The distance from the point (1, - 1, 2) is$

$\Rightarrow$ $y_{p}^{2} + z_{p}^{2} = {(x_{p} - 1)}^{2} + {(y_{p} + 1)}^{2} + {(z_{p} - 2)}^{2} \Rightarrow x_{p}^{2} - 2 x_{p} + 2 y_{p} - 4 z_{p} + 6 = 0$

$Therefore, the locus of point P is$ $x^{2} - 2 x + 2 y - 4 z + 6 = 0$

Q 38 :

Let OA, OB, OC be coterminous edges of a cuboid. If $l, m, n$ be the shortest distance between the sides OA, OB, OC and their respective skew body diagonals to them, respectively, then find $\frac{(\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}})}{(\frac{1}{O A^{2}} + \frac{1}{O B^{2}} + \frac{1}{O C^{2}})}$

(2)

$Let l = m = n = \frac{1}{\sqrt{2}}$

Q 39 :

The perpendicular distance of point $(2, 0, - 3)$ from line which passes through the point $(0, 2, - 4)$ and perpendicular to the lines $\vec{r} = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k}),$ $\vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k})$ , where $λ$ , $μ \in R$ is

$\frac{\sqrt{219}}{3}$
$\frac{\sqrt{78}}{3}$
$\frac{\sqrt{52}}{3}$
$\frac{\sqrt{126}}{3}$

1

2

3

4

(2)

Q 40 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α, β$ respectively. If the variance of random variable $α - β$ is $\frac{k_{1}}{k_{2}}$ , where $k_{1}$ and $k_{2}$ are co-prime, then $k_{1} - k_{2} =$ _______

21
29
13
19

1

2

3

4

(2)

$α - β$	$5$	$4$	$3$	$2$	$1$	$0$	$- 1$	$- 2$	$- 3$	$- 4$	$- 5$
$Prob$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

Topic Question Set

Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

Q 14 :

The equation of the plane through (1, 2, 3) and parallel to the plane $2 x + 3 y - 4 z = 0$ is:

Q 15 :

The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2 x + 3 y + 4 z + 5 = 0$ , the point (1, 1, 1) is:

Q 16 :

If the given planes $a x + b y + c z + d = 0$ and $a' x + b' y + c' z + d' = 0$ be mutually perpendicular, then:

Q 17 :

If the planes $x + 2 y + k z = 0$ and $2 x + y - 2 z = 0$ are at right angles, then the value of $k$ is:

Q 18 :

The intercepts of the plane $5 x - 3 y + 6 z = 60$ on the coordinate axes are:

Q 19 :

The two planes $x - 2 y + 4 z = 10$ and $18 x + 17 y + k z = 50$ are perpendicular, if $k$ is equal to:

Q 20 :

Let the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,$ then the value of $p$ is:

Q 21 :

The distance of the plane $\vec{r} \cdot (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 1$ from the origin is:

Q 22 :

The length of the perpendicular from the origin to the plane $3 x + 4 y + 12 z = 52$ is:

Q 23 :

The equation of the plane which is parallel to the plane $x - 2 y + 2 z = 5$ and whose distance from the point (1, 2, 3) is 1, is:

Q 24 :

The distance of the plane $6 x - 3 y + 2 z - 14 = 0$ from the origin is:

Q 25 :

The distance of the origin (0, 0, 0) from the plane $- 2 x + 6 y - 3 z = - 7$ is:

Q 26 :

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lie in the plane $x + 3 y - α z + β = 0,$ then $(α, β)$ equals:

Q 27 :

The line $\frac{x - 1}{1} = \frac{y - 2}{- 2} = \frac{z - 1}{3}$ and the plane $x + 2 y + z = 6$ meet at:

Q 28 :

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if:

Q 29 :

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

Q 30 :

The point of intersection of the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z,$ is:

Q 31 :

The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is:

Q 32 :

The equation of plane passing through the point $(0, 7, - 7)$ and containing the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is:

Q 33 :

A line $L : \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point $' S'$ on the line $(L)$ in the given plane $(P)$ is (2, 3, 1), then co-ordinates of point S, is:

Q 34 :

The angle between the two planes $x + y - z = 3$ and $3 x + 2 y + z = 5$ is:

Q 35 :

The vector equation of the line joining the points $(- 2, - 3, - 4)$ and $(1, - 2, 4)$ is:

Q 36 :

If the equation of a floor of a room is given by $x + y - z + 4 = 0$ and the equation of roof is given by $x + y - z + 5 = 0,$ then the height of the room is:

Q 37 :

Find the locus of a point whose distance from $x$ -axis is twice the distance from the point $(1, - 1, 2)$ :

Q 40 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α, β$ respectively. If the variance of random variable $α - β$ is $\frac{k_{1}}{k_{2}}$ , where $k_{1}$ and $k_{2}$ are co-prime, then $k_{1} - k_{2} =$ _______

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : The vector equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2i^-3j^+4k^ is:

Q 12 : A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (α,β,γ). The equation of the plane is:

Q 13 : If vector equation of the plane x-22=2y-5-3=z+1, is r→=(2i^+52j^-k^)+λ(2i^-32j^+pk^), then p is equal to:

Q 14 : The equation of the plane through (1, 2, 3) and parallel to the plane 2x+3y-4z=0 is:

Q 15 : The equation of the plane passing through the intersection of the planes x+y+z=6 and 2x+3y+4z+5=0, the point (1, 1, 1) is:

Q 16 : If the given planes ax+by+cz+d=0 and a'x+b'y+c'z+d'=0 be mutually perpendicular, then:

Q 17 : If the planes x+2y+kz=0 and 2x+y-2z=0 are at right angles, then the value of k is:

Q 18 : The intercepts of the plane 5x-3y+6z=60 on the coordinate axes are:

Q 19 : The two planes x-2y+4z=10 and 18x+17y+kz=50 are perpendicular, if k is equal to:

Q 20 : Let the points (1,1,p) and (-3,0,1) be equidistant from the plane r→·(3i^+4j^-12k^)+13=0, then the value of p is:

Q 21 : The distance of the plane r→·(27i^+37j^-67k^)=1 from the origin is:

Q 22 : The length of the perpendicular from the origin to the plane 3x+4y+12z=52 is:

Q 23 : The equation of the plane which is parallel to the plane x-2y+2z=5 and whose distance from the point (1, 2, 3) is 1, is:

Q 24 : The distance of the plane 6x-3y+2z-14=0 from the origin is:

Q 25 : The distance of the origin (0, 0, 0) from the plane -2x+6y-3z=-7 is:

Q 26 : Let the line x-23=y-1-5=z+22 lie in the plane x+3y-αz+β=0, then (α,β) equals:

Q 27 : The line x-11=y-2-2=z-13 and the plane x+2y+z=6 meet at:

Q 28 : The lines x-21=y-31=z-4-k and x-1k=y-42=z-51 are coplanar if:

Q 29 : The line x-23=y-34=z-45 is parallel to the plane:

Q 30 : The point of intersection of the lines x-12=y-23=z-34 and x-45=y-12=z, is:

Q 31 : The distance between the point (3, 4, 5) and the point where the line x-31=y-42=z-52 meets the plane x+y+z=17, is:

Q 32 : The equation of plane passing through the point (0,7,-7) and containing the line x+1-3=y-32=z+21 is:

Q 33 : A line L:x-21=y-32=z-1-1 is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point 'S' on the line (L) in the given plane (P) is (2, 3, 1), then co-ordinates of point S, is:

Q 34 : The angle between the two planes x+y-z=3 and 3x+2y+z=5 is:

Q 35 : The vector equation of the line joining the points (-2,-3,-4) and (1,-2,4) is:

Q 36 : If the equation of a floor of a room is given by x+y-z+4=0 and the equation of roof is given by x+y-z+5=0, then the height of the room is:

Q 37 : Find the locus of a point whose distance from x-axis is twice the distance from the point (1,-1,2):

Q 38 : Let OA, OB, OC be coterminous edges of a cuboid. If l, m, n be the shortest distance between the sides OA, OB, OC and their respective skew body diagonals to them, respectively, then find (1l2+1m2+1n2)(1OA2+1OB2+1OC2)

Q 39 : The perpendicular distance of point (2,0,-3) from line which passes through the point (0,2,-4) and perpendicular to the lines r→=(-3i^+2k^)+λ(2i^+3j^+5k^), r→=(i^-2j^+k^)+μ(-i^+3j^+2k^), where λ, μ∈R is

Q 40 : Two dice A and B are rolled. Let the numbers obtained on A and B be α,β respectively. If the variance of random variable α-β is k1k2, where k1 and k2 are co-prime, then k1-k2= _______

Want Us to Email you About Special Offers & Updates?

Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

Q 14 :

The equation of the plane through (1, 2, 3) and parallel to the plane $2 x + 3 y - 4 z = 0$ is:

Q 15 :

The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2 x + 3 y + 4 z + 5 = 0$ , the point (1, 1, 1) is:

Q 16 :

If the given planes $a x + b y + c z + d = 0$ and $a' x + b' y + c' z + d' = 0$ be mutually perpendicular, then:

Q 17 :

If the planes $x + 2 y + k z = 0$ and $2 x + y - 2 z = 0$ are at right angles, then the value of $k$ is:

Q 18 :

The intercepts of the plane $5 x - 3 y + 6 z = 60$ on the coordinate axes are:

Q 19 :

The two planes $x - 2 y + 4 z = 10$ and $18 x + 17 y + k z = 50$ are perpendicular, if $k$ is equal to:

Q 20 :

Let the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,$ then the value of $p$ is:

Q 21 :

The distance of the plane $\vec{r} \cdot (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 1$ from the origin is:

Q 22 :

The length of the perpendicular from the origin to the plane $3 x + 4 y + 12 z = 52$ is:

Q 23 :

The equation of the plane which is parallel to the plane $x - 2 y + 2 z = 5$ and whose distance from the point (1, 2, 3) is 1, is:

Q 24 :

The distance of the plane $6 x - 3 y + 2 z - 14 = 0$ from the origin is:

Q 25 :

The distance of the origin (0, 0, 0) from the plane $- 2 x + 6 y - 3 z = - 7$ is:

Q 26 :

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lie in the plane $x + 3 y - α z + β = 0,$ then $(α, β)$ equals:

Q 27 :

The line $\frac{x - 1}{1} = \frac{y - 2}{- 2} = \frac{z - 1}{3}$ and the plane $x + 2 y + z = 6$ meet at:

Q 28 :

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if:

Q 29 :

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

Q 30 :

The point of intersection of the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z,$ is:

Q 31 :

The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is:

Q 32 :

The equation of plane passing through the point $(0, 7, - 7)$ and containing the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is:

Q 33 :

A line $L : \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point $' S'$ on the line $(L)$ in the given plane $(P)$ is (2, 3, 1), then co-ordinates of point S, is:

Q 34 :

The angle between the two planes $x + y - z = 3$ and $3 x + 2 y + z = 5$ is:

Q 35 :

The vector equation of the line joining the points $(- 2, - 3, - 4)$ and $(1, - 2, 4)$ is:

Q 36 :

If the equation of a floor of a room is given by $x + y - z + 4 = 0$ and the equation of roof is given by $x + y - z + 5 = 0,$ then the height of the room is:

Q 37 :

Find the locus of a point whose distance from $x$ -axis is twice the distance from the point $(1, - 1, 2)$ :

Q 40 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α, β$ respectively. If the variance of random variable $α - β$ is $\frac{k_{1}}{k_{2}}$ , where $k_{1}$ and $k_{2}$ are co-prime, then $k_{1} - k_{2} =$ _______