Three Dimensional Geometry jee mains questions with solutions

Q 11 :

Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]

(11)

We have, $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t$ (say)

$\vec{Q R} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0$

$\Rightarrow (t - 1) + (2 t - 5) \times 2 + (3 t - 2) \times 3 = 0 \Rightarrow t = \frac{17}{14}$

$\Rightarrow R \equiv (\frac{17}{14}, \frac{48}{14}, \frac{79}{14})$

Since, R is the mid point of PQ

$∴ \frac{α + 1}{2} = \frac{17}{14}, \frac{β + 6}{2} = \frac{48}{14}, \frac{γ + 4}{2} = \frac{79}{14}$

$\Rightarrow α = \frac{10}{7}, β = \frac{6}{7}, γ = \frac{51}{7}$

Hence, $2 α + β + γ = 2 \times \frac{10}{7} + \frac{6}{7} + \frac{51}{7} = \frac{20 + 6 + 51}{7} = \frac{77}{7} = 11$ .

Q 12 :

The square of the distance of the image of the point (6, 1, 5) in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is _________. [2024]

(62)

Let $L : \frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4} = λ$ be the given line.

Any point on L is given by $P (3 λ + 1, 2 λ, 4 λ + 2)$

Direction ratios of L is given by $3 \hat{i} + 2 \hat{j} + 4 \hat{k}$

Now, AP $⊥$ L so we have

$3 (3 λ + 1 - 6) + 2 (2 λ - 1) + 4 (4 λ + 2 - 5) = 0$

$\Rightarrow 9 λ - 15 + 4 λ - 2 + 16 λ - 12 = 0$

$\Rightarrow 29 λ = 29 \Rightarrow λ = 1$

So, P(4, 2, 6) is the mid point of AB

$\Rightarrow 4 = \frac{x + 6}{2}, 2 = \frac{y + 1}{2}, 6 = \frac{z + 5}{2}$

$\Rightarrow$ B(x, y, z) = (2, 3, 7) is the image of A

Required distance = $\sqrt{4 + 9 + 49} = \sqrt{62}$ .

Q 13 :

Let the line of the shortest distance between the lines

$L_{1} : \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $L_{2} : \vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$

intersect $L_{1}$ and $L_{2}$ at P and Q respectively. If $(α, β, γ)$ is the mid point of the line segment PQ, then $2 (α + β + γ)$ is equal to __________. [2024]

(21)

We have, $L_{1} : \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

$L_{2} : \vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$

Here, ${\vec{b}}_{1} = \hat{i} - \hat{j} + \hat{k}, {\vec{b}}_{2} = \hat{i} + \hat{j} - \hat{k}$

${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix} |$ $= 0 \hat{i} + 2 \hat{j} + 2 \hat{k}$

Direction ratios of line perpendicular to $L_{1}$ and $L_{2}$

$= ((4 + μ) - (1 + λ), (5 + μ) - (2 - λ), (6 - μ) - (3 + λ))$

$= 3 + μ - λ, 3 + μ + λ, 3 - μ - λ$

$∴ \frac{3 + μ - λ}{0} = \frac{3 + μ + λ}{2} = \frac{3 - μ - λ}{2}$

On solving, we get

$λ = \frac{3}{2}, μ = \frac{- 3}{2}$

Point $P \equiv (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$ and $Q \equiv (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$

$∴$ Mid point of AB = $(\frac{5}{2}, 2, 6) = (α, β, γ)$

$∴ 2 (α + β + γ)$ = 5 + 4 + 12 = 21.

Q 14 :

The lines $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16}$ and $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1}$ intersect at the point P. If the distance of P from the line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is $l$ , then $14 l^{2}$ is equal to _________. [2024]

(108)

We have, $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16} = λ$ (say)

$\Rightarrow x = 2 λ + 2, y = - 2 λ, z = 16 λ + 7$

Also, $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1} = μ$ (say)

$\Rightarrow x = 4 μ - 3, y = 3 μ - 2, z = μ - 2$

Lines are intersecting at point P.

$∴ 2 λ + 2 = 4 μ - 3 \Rightarrow 2 λ - 4 μ = - 5$ ... (i)

$- 2 λ = 3 μ - 2 \Rightarrow - 2 λ - 3 μ = - 2$ ... (ii)

On solving (i) and (ii), we get

$λ = - \frac{1}{2}$ and $μ = 1$

$∴$ Point P is (1, 1, –1)

Now, $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1} = k$ (say)

$\Rightarrow$ x = 2k – 1, y = 3k + 1, z = k + 1

D.r.'s of PQ : 2k – 2, 3k, k + 2

D.r.'s of line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is 2, 3, 1

As both line are perpendicular to each other.

2(2k –2) + 3(3k) + 1(k +2) = 0

$\Rightarrow 14 k = 2 \Rightarrow k = \frac{1}{7}$

Thus, point Q is $(- \frac{5}{7}, \frac{10}{7}, \frac{8}{7})$

$∴ P Q = \sqrt{{(1 + \frac{5}{7})}^{2} + {(1 - \frac{10}{7})}^{2} + {(- 1 - \frac{8}{7})}^{2}} = \sqrt{\frac{378}{49}}$

Also, $P Q^{2} = l^{2} = \frac{378}{49} \Rightarrow 14 l^{2} = \frac{378}{49} \times 14 = 108$ .

Q 15 :

A line with direction ratios 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively at the points P and Q. If the length of the perpendicular from the point (1, 2, 12) to the line PQ is $l$ , then $l^{2}$ is __________. [2024]

(65)

We have, $l_{1}$ : $\frac{x}{1} = \frac{y + 2}{1} = \frac{z}{1} = λ \in R$

Coordinates of point P are $(λ, λ - 2, λ)$

$l_{2}$ : $\frac{x + 2}{2} = \frac{y}{1} = \frac{z}{1} = μ \in R$

Coordinates of point Q are $(2 μ - 2, μ, μ)$

D.r.'s of PQ are $(2 μ - 2 - λ, μ - λ + 2, μ - λ)$

Also, D.r.'s of line PQ is (2, 1, 2)

$∴ \frac{2 μ - 2 - λ}{2} = \frac{μ - λ + 2}{1} = \frac{μ - λ}{2}$

$\Rightarrow λ = 6, μ = 2$

$∴$ Coordinates of point P are (6, 4, 6) and coordinates of point Q are (2, 2, 2).

Equations of line PQ is $\frac{x - 2}{2} = \frac{y - 2}{1} = \frac{z - 2}{2} = k \in R$

Now, from condition for perpendicularity,

Since, AB $⊥$ PQ, then,

2(2k + 1) + (1)k + 2(2k – 10) = 0

$\Rightarrow 9 k = 18 \Rightarrow k = 2$

Therefore, point A is (6, 4, 6)

Now, perpendicular distance from B(1, 2, 12) to line PQ is given by

$l$ $= \sqrt{{(6 - 1)}^{2} + {{(4 - 2)}^{2} + (6 - 12)}^{2}}$ $\Rightarrow l^{2}$ = 25 + 4 + 36 = 65.

Q 16 :

Let O be the orgin, and M and N be the points on the lines $\frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3}$ and $\frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9}$ respectively such that MN is the shortest distance between the given lines. Then $\vec{O M} \cdot \vec{O N}$ is equal to __________. [2024]

(9)

Let, $L_{1} : \frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3} = λ \in R$

$L_{2} : \frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9} = μ \in R$

M is a point on $L_{1}$ .

So, $M = (4 λ + 5, λ + 4, 3 λ + 5)$

and N is a point on $L_{2}$ .

So, $N = (12 μ - 8, 5 μ - 2, 9 μ - 11)$

The direction ratios of MN is

$(12 μ - 4 λ - 13, 5 μ - λ - 6, 9 μ - 3 λ - 16)$

$∵ M N ⊥ L_{1} and M N ⊥ L_{2}$

$∴ 4 (12 μ - 4 λ - 13) + 5 μ - λ - 6 + 3 (9 μ - 3 λ - 16) = 0$

$\Rightarrow 48 μ - 16 λ - 52 + 5 μ - λ - 6 + 27 μ - 9 λ - 48 = 0$

$\Rightarrow 80 μ - 26 λ - 106 = 0 \Rightarrow 40 μ - 13 λ - 53 = 0$ ... (i)

Also,

$12 (12 μ - 4 λ - 13) + 5 (5 μ - λ - 6) + 9 (9 μ - 3 λ - 16) = 0$

$144 μ - 48 λ - 156 + 25 μ - 5 λ - 30 + 81 μ - 27 λ - 144 = 0$

$\Rightarrow 250 μ - 80 λ - 330 = 0$ $\Rightarrow 25 μ - 8 λ - 33 = 0$ ... (ii)

On solving equation (i) and (ii), we get $μ = 1$ and $λ = - 1$

M = (1, 3, 2) and N = (4, 3, –2)

$∴ \vec{O M} = \hat{i} + 3 \hat{j} + 2 \hat{k} and \vec{O N} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$

$\vec{O M} \cdot \vec{O N}$ = 4 + 9 – 4 = 9.

Q 17 :

If $d_{1}$ is the shortest distance between the lines x + 1 = 2y = –12z, x = y + 2 = 6z – 6 and $d_{2}$ is the shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}, \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ , then the value of $\frac{32 \sqrt{3} d_{1}}{d_{2}}$ is ___________. [2024]

(16)

We have, $l_{1}$ : x + 1 = 2y = –12z and $l_{2}$ : x = y + 2 = 6z – 6

So, $l_{1}$ : $\frac{x - (- 1)}{1} = \frac{y - 0}{\frac{1}{2}} = \frac{z - 0}{\frac{- 1}{12}}$ and $l_{2}$ : $\frac{x - 0}{1} = \frac{y - (- 2)}{1} = \frac{z - 1}{\frac{1}{6}}$

These lines can be written in vector form as

${\vec{r}}_{1} = (- \hat{i} + 0 \hat{j} + 0 \hat{k}) + λ (\hat{i} + \frac{\hat{j}}{2} - \frac{\hat{k}}{12})$ and ${\vec{r}}_{2} = (0 \hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} + \frac{1}{6} \hat{k})$

For, $\vec{r_{1}} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r_{2}} = \vec{a_{2}} + λ \vec{b_{2}}$

Shortest distance between $l_{1}$ and $l_{2}$ is given by

$d_{1} =$ $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$ $=$ $| \frac{(\hat{i} + 2 \hat{j} + \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 / 2 & - 1 / 12 \\ 1 & 1 & 1 / 6 \end{matrix} |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$

$=$ $| \frac{(\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\frac{1}{6} \hat{i} + \frac{1}{4} \hat{j} + \frac{1}{2} \hat{k})}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{4}}} |$ $=$ $| \frac{\frac{1}{6} + \frac{1}{2} + \frac{1}{2}}{\sqrt{\frac{49}{144}}} |$ $= \frac{7}{6} \times \frac{12}{7} = 2$

Similarly, $l_{3}$ : $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $l_{4}$ : $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$

These lines can be written in vector form as

${\vec{r}}_{1} = (\hat{i} - 8 \hat{j} + 4 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 5 \hat{k})$

${\vec{r}}_{2} = (\hat{i} + 2 \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} - 3 \hat{k})$

Shortest distance between $l_{3}$ and $l_{4}$ is given as

$d_{2} =$ $| \frac{(10 \hat{j} + 2 \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} |}{| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} | |} |$

$=$ $| \frac{(10 \hat{j} + 2 \hat{k}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k})}{\sqrt{16^{2} + 16^{2} + 16^{2}}} |$ $= \frac{160 + 32}{16 \sqrt{3}} = \frac{12}{\sqrt{3}} = 4 \sqrt{3}$

Now, $\frac{32 \sqrt{3} d_{1}}{d_{2}} = \frac{32 \sqrt{3} \times 2}{4 \sqrt{3}} = 16$ .

Q 18 :

Let a line passing through the point (–1, 2, 3) intersect the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{- 2}$ at $M (α, β, γ)$ and $L_{2} : \frac{x + 2}{- 3} = \frac{y - 2}{- 2} = \frac{z - 1}{4}$ at N(a, b, c). Then the value of $\frac{{(α + β + γ)}^{2}}{{(a + b + c)}^{2}}$ equals __________. [2024]

(196)

We have, $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{- 2} = λ \in R$

$∴ M \equiv (3 λ + 1, 2 λ + 2, - 2 λ - 1)$

$L_{2} : \frac{x + 2}{- 3} = \frac{y - 2}{- 2} = \frac{z - 1}{4} = μ \in R$

$∴ N \equiv (- 3 μ — 2, - 2 μ + 2, 4 μ + 1)$

$α + β + γ = 3 λ + 2 \Rightarrow a + b + c = - μ + 1$

Direction ratio's of line AM = $< 3 λ + 2, 2 λ, - 2 λ - 4 >$

Direction ratio's of line AN = $< - 3 μ — 1, - 2 μ, 4 μ - 2 >$

$\Rightarrow \frac{3 λ + 2}{- 3 μ - 1} = \frac{2 λ}{- 2 μ} = \frac{- 2 λ - 4}{4 μ - 2}$

$\Rightarrow \frac{3 λ + 2}{- 3 μ - 1} = \frac{λ}{- μ} and \frac{λ}{- μ} = \frac{- 2 λ - 4}{4 μ - 2}$

$\Rightarrow - 3 λ μ - 2 μ = - 3 μ λ - λ and 4 μ λ - 2 λ = 2 μ λ + 4 μ$

$\Rightarrow 2 μ = λ and 2 μ λ = 2 λ + 4 μ$

$\Rightarrow λ^{2} = 2 λ + 2 λ \Rightarrow λ^{2} = 4 λ$

$\Rightarrow λ (λ - 4) = 0 \Rightarrow λ = 0, λ = 4$

$∴ \frac{{(α + β + γ)}^{2}}{{(a + b + c)}^{2}} = \frac{{(3 \times 4 + 2)}^{2}}{{(- 2 + 1)}^{2}} = \frac{{(14)}^{2}}{1} = 196$ .

Q 19 :

Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If $∠ Q P R$ is a right angle, then $12 a^{2}$ is equal to __________. [2024]

(12)

Line $L_{1}$ is given by y = x, z = 1 can be expressed as

$L_{1} : \frac{x}{1} = \frac{y}{1} = \frac{z - 1}{0} = α$

$\Rightarrow x = α, y = α, z = 1$

Let the coordinate of Q on $L_{1}$ be $(α, α, 1)$

Line $L_{2}$ given by y = –x, z = –1 can be expressed as

$L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z + 1}{0} = β$ (say)

$\Rightarrow x = β, y = - β, z = - 1$

Let the coordinates of R on $L_{2}$ be $(β, - β, - 1)$

Direction ratios of PQ are $(a - α, a - α, a - 1)$ .

Now $P Q ⊥ L_{1}$

$∴ 1 (a - α) + 1 (a - α) + 0 (a - 1) = 0 \Rightarrow a = α$

Hence Q(a, a, 1)

Direction ratios of PR are $a - β, a + β, a + 1$

Now $P R ⊥ L_{2}$

$∴ 1 (a - β) + (- 1) (a + β) + 0 (a + 1) = 0 \Rightarrow β = 0$

Hence R(0, 0, —1)

Now, as $∠ Q P R = 90 °$

$\Rightarrow$ (a – a)(a – 0) + (a – a)(a – 0) + (a – 1)(a + 1) = 0

$\Rightarrow$ (a –1)(a + 1) = 0 $\Rightarrow$ a = 1 or a = –1

a = 1, rejected as P and Q are different points

$\Rightarrow$ a = –1, then $12 a^{2} = 12 \times {(- 1)}^{2} = 12$ .

Q 20 :

A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __________. [2024]

(22)

Equation of the line through A(4, –6, –2) and B(16, –2, 4) is

$\frac{x - 4}{16 - 4} = \frac{y + 6}{- 2 + 6} = \frac{z + 2}{4 + 2} = λ, λ \in R$

$\Rightarrow \frac{x - 4}{12} = \frac{y + 6}{4} = \frac{z + 2}{6} = λ$

$∴ x = 12 λ + 4, y = 4 λ - 6, z = 6 λ - 2$

Let $a = 12 λ + 4, b = 4 λ - 6, c = 6 λ - 2$ .

Now ${(12 λ + 4 - 4)}^{2} + {(4 λ - 6 + 6)}^{2} + {(6 λ - 2 + 2)}^{2} = {(21)}^{2}$

$144 λ^{2} + 16 λ^{2} + 36 λ^{2} = 441$

$\Rightarrow 196 λ^{2} = 441$

$λ^{2} = \frac{441}{196} \Rightarrow λ = \pm \frac{21}{14} = \pm \frac{3}{2}$

When $λ = \frac{3}{2}$

$a = 12 \times \frac{3}{2} + 4 = 22; b = 4 \times \frac{3}{2} - 6 = 0; c = 6 \times \frac{3}{2} - 2 = 7$

When $λ = \frac{- 3}{2}$

$a = 12 \times \frac{- 3}{2} + 4 = - 14; b = 4 \times \frac{- 3}{2} - 6 = - 12; c = 6 \times \frac{- 3}{2} - 2 = - 11$

Distance between (22, 0, 7) and (4, –12, 3)

$= \sqrt{18^{2} + 12^{2} + 4^{2}} = \sqrt{484} = 22$ .

Q 21 :

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is $\frac{m}{n}$ then: [2025]

$m - 5 \sqrt{21} n = 0$
$5 m - 2 \sqrt{21} n = 0$
$2 m - 5 \sqrt{21} n = 0$
$5 m - 21 \sqrt{2} n = 0$

1

2

3

4

(3)

We have, equation of line L is

$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = λ (say)$

$∴$ Coordinates of $M (5 λ - 3, 2 λ + 1, 3 λ - 4)$

Dr's of PM are $< 5 λ - 3, 2 λ - 1, 3 λ - 7 >$

Dr's of line L are < 5, 2, 3 >

As PM $⊥$ L, so, $(5 λ - 3) 5 + (2 λ - 1) 2 + (3 λ - 7) 3 = 0$

$\Rightarrow λ = 1$

$∴$ Coordinates of M is (2, 3, –1)

Now, $P M = \sqrt{{(2 - 0)}^{2} + {(3 - 2)}^{2} + {(- 1 - 3)}^{2}} = \sqrt{4 + 1 + 16} = \sqrt{21}$

$∴$ Area of $△ P Q R = \frac{1}{2} \times Q R \times P M = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{m}{n}$

$\Rightarrow 2 m - 5 \sqrt{21} n = 0$ .

Q 22 :

The line $L_{1}$ is parallel to the vector $\vec{a} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point (7, 6, 2) and the line $L_{2}$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point (5,3, 4). The shortest distance between the lines $L_{1}$ and $L_{2}$ is : [2025]

$\frac{23}{\sqrt{38}}$
$\frac{21}{\sqrt{57}}$
$\frac{23}{\sqrt{57}}$
$\frac{21}{\sqrt{38}}$

1

2

3

4

(1)

We have, $L_{1} : 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + λ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$

$L_{2} : 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + μ (2 \hat{i} + \hat{j} + 3 \hat{k})$

Here, ${\vec{a}}_{1} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$

${\vec{b}}_{1} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} and {\vec{b}}_{2} = 2 \hat{i} + \hat{j} + 3 \hat{k}$

Now, ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 2 & 4 \\ 2 & 1 & 3 \end{matrix} |$

$= \hat{i} (6 - 4) - \hat{j} (- 9 - 8) + \hat{k} (- 3 - 4)$

$= 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$

Also, ${\vec{a}}_{2} - {\vec{a}}_{1} = - 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$

Shortest distance between $L_{1}$ and $L_{2}$

$= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} = \frac{| - 4 - 51 - 14 |}{\sqrt{4 + 289 + 49}}$

$= \frac{69}{\sqrt{342}} = \frac{69}{3 \sqrt{38}} = \frac{23}{\sqrt{38}}$

Q 23 :

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is $Q (α, β, γ)$ , then $α + β + γ$ is equal to : [2025]

18
$13$
$\frac{47}{3}$
$\frac{46}{3}$

1

2

3

4

(4)

Equation of line AB is given by

$\frac{x - 4}{- 1} = \frac{y - 7}{- 2} = \frac{z - 1}{2} = λ (say)$

Any point on line AB is given by $(- λ + 4, - 2 λ + 7, 2 λ + 1)$

Now, mid-point of PQ lies on AB i.e.,

$(\frac{1 + α}{2}, \frac{β}{2}, \frac{3 + γ}{2})$ is a point on AB

$∴ \frac{1 + α}{2} = - λ + 4, \frac{β}{2} = - 2 λ + 7$ and $\frac{3 + γ}{2} = 2 λ + 1$

$\Rightarrow α = - 2 λ + 7, β = 14 - 4 λ and γ = 4 λ - 1$ ... (i)

Now, $\vec{P Q}$ is perpendicular to $\vec{A B}$

$∴ (α - 1) (- 1) + β (- 2) + (γ - 3) (2) = 0$

$\Rightarrow - α - 2 β + 2 γ = 5$

Now, substituting the values of $α$ , $β$ and $γ$ from equation (i) we get

$- (- 2 λ + 7) - 2 (14 - 4 λ) + 2 (4 λ - 1) = 5$

$\Rightarrow 2 λ - 7 - 28 + 8 λ + 8 λ - 2 = 5$

$\Rightarrow 18 λ = 42 \Rightarrow λ = \frac{7}{3}$

$∴ α = 7 - 2 \times \frac{7}{3} = \frac{7}{3}, β = 14 - 4 \times \frac{7}{3} = \frac{14}{3}$

and $γ = 4 \times \frac{7}{3} - 1 = \frac{25}{3}$

$∴ α + β + γ = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$ .

Q 24 :

Line $L_{1}$ passes through the point (1, 2, 3) and is parallel to z-axis. Line $L_{2}$ passes through the point ( $λ$ , 5, 6) and is parallel to y-axis. Let for $λ = λ_{1}, λ_{2}$ , $λ_{2} < λ_{1}$ , the shortest distance between the two lines be 3. Then the square of the distance of the point $(λ_{1}, λ_{2}, 7)$ from the line $L_{1}$ is [2025]

40
32
25
37

1

2

3

4

(3)

$L_{1} \equiv \frac{x - 1}{0} = \frac{y - 2}{0} = \frac{z - 3}{1}$

[ $∵ L_{1}$ is parallel to z-axis and passing through (1, 2, 3)]

and $L_{2} \equiv \frac{x - λ}{0} = \frac{y - 5}{1} = \frac{z - 6}{0}$

[ $∵ L_{2}$ is parallel to y-axis and passing through $(λ, 5, 6)$ ]

Now, shortest distance between $L_{1}$ and $L_{2}$ is given by

$\frac{| \begin{matrix} λ - 1 & 3 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} |}{\sqrt{{(- 1)}^{2} + 0 + 0}}$ $=$ $| λ - 1 |$

$\Rightarrow | λ - 1 | = 3 \Rightarrow λ = 4, - 2$

$\Rightarrow λ_{1} = 4, λ_{2} = - 2$ [ $∵ λ_{2} < λ_{1}$ ]

Let the foot the perpendicular from point P(4, –2, 7) to the line $L_{1}$ is $Q (1, 2, μ + 3)$ .

Direction ratios of the QP are (3, –4, 4 – $μ$ )

$∵$ QP is perpendicular to $L_{1}$

So, $3 \times 0 - 4 \times 0 + (4 - μ) \times 1 = 0 \Rightarrow μ = 4$

The coordinates of point Q are (1, 2, 7).

$∴ P Q^{2} = 9 + 16 + 0 \Rightarrow P Q^{2} = 25$

Q 25 :

Let a line passing through the point (4, 1, 0) intersect the line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ at the point $A (α, β, γ)$ and the line $L_{2} : x - 6 = y = - z + 4$ at the point B(a, b, c). Then $| \begin{matrix} 1 & 0 & 1 \\ α & β & γ \\ a & b & c \end{matrix} |$ is equal to [2025]

12
6
8
16

1

2

3

4

(3)

$L_{1} = \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = t (say)$

$L_{2} = \frac{x - 6}{1} = \frac{y}{1} = \frac{z - 4}{- 1} = μ (say)$

A(2t + 1, 3t + 2, 4t + 3) is any point on $L_{1}$ .

And $B (μ + 6, μ, 4 - μ)$ is any point on $L_{2}$ .

$∴$ D.R. of PA = 2t – 3, 3t + 1, 4t + 3 and D.R. of $P B = μ + 2, μ - 1, 4 - μ$ , where the coordinate of P is (4, 1, 0).

$\frac{2 t - 3}{μ + 2} = \frac{3 t + 1}{μ - 1} = \frac{4 t + 3}{4 - μ}$

[Direction ratios of PA and PB are proportional]

$2 t μ - 2 t - 3 μ + 3 = 3 t μ + 6 t + μ + 2$

$\Rightarrow t μ + 8 t + 4 μ - 1 = 0$ ... (i)

Also, $12 t - 3 μ t + 4 - μ = 4 μ t + 3 μ - 4 t - 3$

$\Rightarrow 7 μ t - 16 t + 4 μ - 7 = 0$ ... (ii)

And $8 t - 2 μ t - 12 + 3 μ = 4 μ t + 8 t + 3 μ + 6$

$\Rightarrow 6 μ t = - 18 \Rightarrow μ t = - 3$

From equation (i), we get

$8 t + 4 μ = 4 \Rightarrow 2 t + μ = 1$ ... (iii)

From equation (ii), we get

$- 21 - 16 t + 4 μ - 7 = 0$

$16 t - 4 μ = - 28 \Rightarrow 4 t - μ = - 7$ ... (iv)

From equation (iii) and (iv), we get

$∴ t = - 1, μ = 3$

$∴ A (- 1, - 1, - 1) and B (9, 3, 1)$

Now, $| \begin{matrix} 1 & 0 & 1 \\ α & β & γ \\ a & b & c \end{matrix} |$ $=$ $| \begin{matrix} 1 & 0 & 1 \\ - 1 & - 1 & - 1 \\ 9 & 3 & 1 \end{matrix} |$

1(–1 + 3) – 0 + 1(–3 + 9) = 8.

Q 26 :

The distance of the point (7, 10, 11) from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is [2025]

14
18
16
12

1

2

3

4

(1)

Let $P \equiv (7, 10, 11)$

Any point on the line

$L_{1} : \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} = λ$

has coordinates

$x = λ + 4, y = 4, z = 3 λ + 2$

i.e., $Q (λ + 4, 4, 3 λ + 2)$

$∵$ Line PQ is parallel to line

$L_{2} : \frac{x - 9}{2} = \frac{y - 3}{3} = \frac{z - 17}{6}$

$\Rightarrow \frac{λ + 4 - 7}{2} = \frac{4 - 10}{3} = \frac{3 λ + 2 - 11}{6} \Rightarrow λ = - 1$

$∴$ Q(3, 4, –1) is the point on the line $L_{1}$ .

Hence, $P Q = \sqrt{16 + 36 + 144} = 14$ units.

Q 27 :

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $3 \sqrt{30}$ . Then the positive value of $5 α + β$ is [2025]

46
48
42
40

1

2

3

4

(1)

Given line are $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$

Let $A (3, α, 3)$ and $B (- 3, - 7, β)$

$∴ \vec{B A} = 6 \hat{i} + (α + 7) \hat{j} + (3 - β) \hat{k}$

Now, $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix} |$ $= - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

$∴$ Shortest distance between lines $=$ $| \frac{\vec{B A} \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$ $= 3 \sqrt{3}$

$\Rightarrow 36 + 15 α + 105 - 9 + 3 β = 270 \Rightarrow 5 α + β = 46$ .

Q 28 :

Let A and B be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both A and B are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then $\vec{O A} \cdot \vec{O B}$ is equal to [2025]

49
62
21
47

1

2

3

4

(4)

We have, $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2} = λ (say)$

$\Rightarrow$ Let foot of perpendicular from P(1, 2, 3) on L is

$Q = (3 λ + 6, 2 λ + 7, - 2 λ + 7)$

Now, $3 (3 λ + 6 - 1) + 2 (2 λ + 7 - 2) - 2 (- 2 λ + 7 - 3) = 0$ [ $∵ \vec{P Q} ⊥ L$ ]

$\Rightarrow 17 λ = - 17 \Rightarrow λ = - 1$

Now, distance of A from Q(3, 5, 9) is the foot of perpendicular.

Let any point on line L is $A (3 μ + 6, 2 μ + 7, - 2 μ + 7)$

Now, distance of A from Q is $2 \sqrt{17}$

$\Rightarrow {(3 μ + 6 - 3)}^{2} + {(2 μ + 7 - 5)}^{2} + {(- 2 μ + 7 - 9)}^{2} = {(2 \sqrt{17})}^{2}$

$\Rightarrow 9 μ^{2} + 9 + 18 μ + {4 μ^{2} + 4 + 8 μ + 4 μ}^{2} + 4 + 8 μ = 68$

$\Rightarrow 17 μ^{2} + 17 + 34 μ = 68 \Rightarrow μ^{2} + 2 μ + 1 = 4$

$\Rightarrow μ^{2} + 2 μ - 3 = 0 \Rightarrow (μ + 3) (μ - 1) = 0 \Rightarrow μ = - 3 or 1$

$∴ A (- 3, 1, 13) and B (9, 9, 5)$ are the points on the line L.

$∴ \vec{O A} \cdot \vec{O B} = - 27 + 9 + 65 = 47$ .

Q 29 :

Let A be the point of intersection of the lines $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$ . Let B and C be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $A B = A C = \sqrt{15}$ . Then the square of the area of the triangle ABC is [2025]

57
54
63
60

1

2

3

4

(2)

We have, $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$

Angle between $L_{1}$ and $L_{2}$ ,

$\cos θ =$ $| \frac{3 \times 1 + 4 \times 0 + 5 (- 1)}{\sqrt{1 + 0 + 1} \sqrt{9 + 16 + 25}} |$ $= \frac{1}{5}$

$\Rightarrow \sin θ = \frac{\sqrt{24}}{5}$

Now, area of $△ A B C = \frac{1}{2} a b \sin θ = \frac{1}{2} {(\sqrt{15})}^{2} (\frac{\sqrt{24}}{5})$

So, square of area = $\frac{15 \times 15 \times 24}{4 \times 25} = 54$ .

Q 30 :

Let the values of p, for which the shortest distance between the lines $\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and $\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ is $\frac{1}{\sqrt{6}}$ , be a, b, (a < b). Then the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is : [2025]

9
$\frac{2}{3}$
18
$\frac{3}{2}$

1

2

3

4

(2)

We have lines $\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$

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

where, $\vec{a} = - \hat{i} + 0 \hat{j} + 0 \hat{k}$

$\vec{b} = p \hat{i} + 2 \hat{j} + \hat{k} then \vec{a} - \vec{b} = (- 1 - p) \hat{i} - 2 \hat{j} - \hat{k}$

$\vec{p} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} and \vec{q} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Then, $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix} |$

$\Rightarrow \vec{p} \times \vec{q} = \hat{i} (16 - 15) - \hat{j} (12 - 10) + \hat{k} (9 - 8)$

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

Now, shortest distance $=$ $| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$\Rightarrow \frac{1}{\sqrt{6}} = \frac{| (- p — 1) + 4 - 1 |}{\sqrt{1 + 4 + 1}} \Rightarrow$ $| - p + 2 |$ $= 1$

$\Rightarrow$ p = 3 and 1, then a = 1 and b = 3 ( $∵$ a < b)

So, length of latus rectum of ellipse $\frac{x^{2}}{1^{2}} + \frac{y^{2}}{3^{2}} = 1$ is $\frac{2 a^{2}}{b} = \frac{2 \times 1}{3} = \frac{2}{3}$ .

Q 31 :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

–3
$- \frac{3}{2}$
$\frac{3}{2}$
3

1

2

3

4

(1)

The given lines are $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ ,

$\vec{n} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & α & 1 \end{matrix} |$

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

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

Shortest distance = $| \frac{\vec{B A} \cdot \vec{n}}{| \vec{n} |} |$ $=$ $| \frac{(\hat{i} + 2 \hat{j} - 2 \hat{k}) \cdot \vec{n}}{| \vec{n} |} |$ $= \frac{5}{\sqrt{6}}$ [Given]

$=$ $| \frac{13 - 8 α}{\sqrt{{(3 - 4 α)}^{2} + 4 + {(2 α - 3)}^{2}}} |$ $= \frac{5}{\sqrt{6}}$

$\Rightarrow 6 {(13 - 8 α)}^{2} = 25 ({(4 α - 3)}^{2} + {(2 α - 3)}^{2} + 4)$

$\Rightarrow 6 (64 α^{2} - 208 α + 169) = (25 (20 α^{2} - 36 α + 22))$

$\Rightarrow 116 α^{2} + 348 α - 464 = 0$

$∴$ Sum of roots $α_{1}$ and $α_{2} = \frac{- 348}{116} = - 3$ .

Q 32 :

Let the line L pass through (1, 1, 1) and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then which of the following points lies on the line L? [2025]

(10, –29, –50)
(4, 22, 7)
(7, 15, 13)
(5, 4, 3)

1

2

3

4

(3)

Equation of line L passes through (1, 1, 1) is $L : \frac{x - 1}{a} = \frac{y - 1}{b} = \frac{z - 1}{c}$ , where a, b, c are direction ratios.

Let $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ (say)$

Any point on $L_{1}$ be $A (2 λ + 1, 3 λ - 1, 4 λ + 1)$

Let $L_{2} : \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1} = μ (say)$

Any point on $L_{2}$ be $B (μ + 3, 2 μ + 4, μ)$

Direction ratio of L be : $< 2 λ, 3 λ - 2, 4 λ > or < μ + 2, 2 μ + 3, μ - 1 >$

Now, $\frac{2 λ}{μ + 2} = \frac{3 λ - 2}{2 μ + 3} = \frac{4 λ}{μ - 1} \Rightarrow λ = \frac{- 6}{5} and μ = - 5$

$∴ < a, b, c > \equiv < - 3, - 7, - 6 > or < 3, 7, 6 >$

$∴ L : \frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}$

Hence, (7, 15, 13) lies on the line.

Q 33 :

If the equation of the line passing through the point $(0, - \frac{1}{2}, 0)$ and perpendicular to the lines $\vec{r} = λ (\hat{i} + a \hat{j} + b \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - 6 \hat{k}) + μ (- b \hat{i} + a \hat{j} + 5 \hat{k})$ is $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ , then a + b + c + d is equal to : [2025]

12
10
14
13

1

2

3

4

(3)

Since, given line is perpendicular to both the line, i.e., $(\hat{i} + a \hat{j} + b \hat{k}) \times (- b \hat{i} + a \hat{j} + 5 \hat{k})$

Also, parallel vector along the required line is

$(\hat{i} + a \hat{j} + b \hat{k}) \times (- b \hat{i} + a \hat{j} + 5 \hat{k})$

$= (5 a - a b) \hat{i} - (b^{2} + 5) \hat{j} + (a + a b) \hat{k}$

$∴$ Dr's of required lines are $(5 a - a b), - (b^{2} + 5), (a + a b)$

But given Dr's of required line are –2, d, –4

$∴ \frac{5 a - a b}{- 2} = \frac{- (b^{2} + 5)}{d} = \frac{a + a b}{- 4}$ ... (i)

Since, point $(0, \frac{- 1}{2}, 0)$ lies on $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ ,

$\Rightarrow \frac{0 - 1}{- 2} = \frac{\frac{- 1}{2} + 4}{d} = \frac{0 - c}{- 4} \Rightarrow d = 7, c = 2$

From (i), $\frac{5 a - a b}{- 2} = \frac{- b^{2} - 5}{7} = \frac{a + a b}{- 4}$

$\begin{matrix} \frac{5 a - a b}{- 2} = \frac{a + a b}{- 4} \\ \Rightarrow - 20 a + 4 a b = - 2 a - 2 a b \\ \Rightarrow 18 a = 6 a b \\ \Rightarrow b = 3 \end{matrix} |$ $\begin{matrix} \frac{- b^{2} - 5}{7} = \frac{a + a b}{- 4} \\ \Rightarrow 4 b^{2} + 20 = 7 a + 7 a b \\ \Rightarrow 36 + 20 = 7 a + 21 a \\ \Rightarrow 56 = 28 a \Rightarrow a = 2 \end{matrix}$

$∴$ a + b + c + d = 2 + 3 + 2 + 7 = 14

Q 34 :

Consider the lines $L_{1}$ : x – 1 = y – 2 = z and $L_{2}$ : x – 2 = y = z – 1. Let the feet of the perpendiculars from the point P(5, 1, –3) on the lines $L_{1}$ and $L_{2}$ be Q and R respectively. If the area of the triangle PQR is A, then $4 A^{2}$ is equal to : [2025]

147
143
139
151

1

2

3

4

(1)

We have, the point P(5, 1, –3)

$L_{1}$ : x – 1 = y – 2 = z = $λ$ (say)

$L_{2}$ : x – 2 = y = z – 1 = $μ$ (say)

Any point on $L_{1}$ and $L_{2}$ are given by $Q (λ + 1, λ + 2, λ)$ and $R (μ + 2, μ, μ + 1)$ respectively.

$\vec{P Q} = (λ - 4) \hat{i} + (λ + 1) \hat{j} + (λ + 3) \hat{k}$

Since, $\vec{P Q} ⊥ L_{1}$ , whose direction ratios are < 1, 1, 1 >, So we have

$1 (λ - 4) + 1 (λ + 1) + 1 (λ + 3) = 0$

$\Rightarrow 3 λ = 0 \Rightarrow λ = 0$

$∴$ Q(1, 2, 0) is the foot of perpendicular on $L_{1}$ . Similarly,

$\vec{P R} = (μ - 3) \hat{i} + (μ - 1) \hat{j} + (μ + 4) \hat{k}$

Now, $\vec{P Q} ⊥ L_{2}$ , whose direction ratios are < 1, 1, 1 >, so we have

$1 (μ - 3) + 1 (μ - 1) + 1 (μ + 4) = 0$

$\Rightarrow 3 μ - 4 + 4 = 0 \Rightarrow μ = 0$

$∴$ R(2, 0, 1) is foot of perpendicular on $L_{2}$ .

Now, area $△ P Q R = \frac{1}{2}$ $| \vec{P Q} \times \vec{P R} |$

$\Rightarrow A = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 4 & 1 & 3 \\ - 3 & - 1 & 4 \end{matrix} | |$ $= \frac{1}{2}$ $| 7 \hat{i} + 7 \hat{j} + 7 \hat{k} |$ $= \frac{7 \sqrt{3}}{2}$

$∴ 4 A^{2} = 4 \times {(\frac{1}{2} \times 7 \sqrt{3})}^{2} = 147$

Q 35 :

Let the values of $λ$ for which the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - λ}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{\sqrt{6}}$ be $λ_{1}$ and $λ_{2}$ . Then the radius of the circle passing through the points (0, 0), $(λ_{1}, λ_{2})$ and $(λ_{2}, λ_{1})$ is [2025]

3
4
$\frac{\sqrt{2}}{3}$
$\frac{5 \sqrt{2}}{3}$

1

2

3

4

(4)

$\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{b} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$

represents direction vector of the given lines.

$∴ \vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$

$= \hat{i} (15 - 16) - \hat{j} (10 - 12) + \hat{k} (8 - 9) = - \hat{i} + 2 \hat{j} - \hat{k}$

The given lines passes through ${\vec{p}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{p}}_{2} = λ \hat{i} + 4 \hat{j} + 5 \hat{k}$

$∴ {\vec{p}}_{2} - {\vec{p}}_{1} = (λ - 1) \hat{i} + 2 \hat{j} + 2 \hat{k}$

Shortest distance between given lines $=$ $| \frac{({\vec{p}}_{2} - {\vec{p}}_{1}) \cdot (\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |} |$

$\Rightarrow \frac{1}{\sqrt{6}} =$ $| \frac{- λ + 1 + 4 - 2}{\sqrt{1 + 4 + 1}} |$

$\Rightarrow$ $| - λ + 3 |$ $= 1 \Rightarrow λ = 3 \pm 1 \Rightarrow λ = 4, 2$

Radius of circle passing through points (0, 0), (4, 2), (2, 4) is given by $\frac{a b c}{4 △}$ .

$= \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2} | \begin{matrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{matrix} |} = \frac{20 \times 2 \sqrt{2}}{2 \times 12} = \frac{5 \sqrt{2}}{3}$

Q 36 :

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_{1}$ and $L_{2}$ ? [2025]

$(- \frac{5}{3}, - 7, 1)$
$(2, 3, \frac{1}{3})$
$(\frac{14}{3}, - 3, \frac{22}{3})$
$(\frac{8}{3}, - 1, \frac{1}{3})$

1

2

3

4

(3)

We have, $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$

Let $A (2 λ + 1, 3 λ + 2, 4 λ + 3)$ and $B (3 μ + 2, 4 μ + 4, 5 μ + 5)$ lies on the line $L_{1}$ and $L_{2}$ respectively.

Direction ratios of AB is

$< 3 μ - 2 λ + 1, 4 μ - 3 λ + 2, 5 μ - 4 λ + 2 >$

As $A B ⊥ L_{1}$ , we have

$2 (3 μ - 2 λ + 1) + 3 (4 μ - 3 λ + 2) + 4 (5 μ - 4 λ + 2) = 0$

$\Rightarrow 6 μ - 4 λ + 2 + 12 μ - 9 λ + 6 + 20 μ - 16 λ + 8 = 0$

$\Rightarrow 38 μ - 29 λ + 16 = 0$ ... (i)

Also, $A B ⊥ L_{2}$ , we have

$3 (3 μ - 2 λ + 1) + 4 (4 μ - 3 λ + 2) + 5 (5 μ - 4 λ + 2) = 0$

$\Rightarrow 50 μ - 38 λ + 21 = 0$ ... (ii)

Solving (i) and (ii), we get

$λ = \frac{1}{3} and μ = \frac{- 1}{6}$

$∴$ Coordinate of A is $(\frac{5}{3}, 3, \frac{13}{3})$ and $B (\frac{3}{2}, \frac{10}{3}, \frac{25}{6})$

Equation of AB is given by

$\frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{- \frac{1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}} \Rightarrow x - \frac{5}{3} = \frac{y - 3}{- 2} = z - \frac{13}{3}$

$∴$ Point $(\frac{14}{3}, - 3, \frac{22}{3})$ lies on line AB.

Q 37 :

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point P(2, –10, 1) is: [2025]

6
$3 \sqrt{5}$
$5 \sqrt{2}$
$4 \sqrt{3}$

1

2

3

4

(2)

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

Any point on the line is given by

$M (x, y, z) \equiv (2 λ + 1, - λ - 2, 2 λ - 3)$

Also, $\vec{P M} = (2 λ + 1 - 2) \hat{i} + (- λ - 2 + 10) \hat{j} + (2 λ - 3 - 1) \hat{k}$

$= (2 λ - 1) \hat{i} + (- λ + 8) \hat{j} + (2 λ - 4) \hat{k}$ ... (i)

$∵ \vec{P M} \cdot \vec{n} = 0$

$\Rightarrow (2 λ - 1) 2 + (- λ + 8) (- 1) + (2 λ - 4) 2 = 0$

$\Rightarrow 4 λ - 2 + λ - 8 + 4 λ - 8 = 0$

$\Rightarrow 9 λ - 18 = 0 \Rightarrow λ = 2$

$∴ | \vec{P M} | = \sqrt{3^{2} + 6^{2} + 0^{2}} = \sqrt{45} = 3 \sqrt{5}$ [Using (i)]

Q 38 :

Let a line pass through two distinct pointsP(–2, –1, 3) and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of $△$ PQR is equal to : [2025]

136
144
148
140

1

2

3

4

(1)

Equation of line passing through P(–2, –1, 3) and parallel to vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ is $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ .

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

$\Rightarrow Q = (3 λ - 2, 2 λ - 1, 2 λ + 3), λ \in ℝ - {0}$

Also, $| \vec{Q R} |$ $= 5 = \sqrt{{(3 λ - 3)}^{2} + {(2 λ - 4)}^{2} + {(2 λ)}^{2}}$

$∴ 17 λ^{2} - 34 λ + 25 = 25 \Rightarrow λ = 2 (∵ λ \neq 0)$

Q(4, 3, 7), P(–2, –1, 3), R(1, 3, 3)

$∴ Area of △ P Q R = \frac{1}{2}$ $| \vec{P Q} \times \vec{P R} |$

$△ = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \end{matrix} | |$

$△ =$ $| - 8 \hat{i} + 6 \hat{j} + 6 \hat{k} |$ $= \sqrt{136} ∴ △^{2} = 136$

Q 39 :

Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ . Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

$8 \sqrt{15}$
$\sqrt{30}$
$9 \sqrt{15}$
$3 \sqrt{30}$

1

2

3

4

(4)

Let $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2} = λ$ then

Point $P = (7 λ + 3, - λ + 2, - 2 λ - 1)$

Now, Dr's of $Q P = (7 λ - 7, - λ + 5, - 2 λ)$

$∴ (7 λ - 7) (7) + (- λ + 5) (- 1) + (- 2 λ) (- 2) = 0$

$\Rightarrow 49 λ - 49 + λ - 5 + 4 λ = 0 \Rightarrow λ = 1$

$∴ P = (10, 1, - 3)$

$∴ \vec{P Q} = - 4 \hat{j} + 2 \hat{k}, \vec{P R} = - 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ [ $∵$ R = (3, –2, 1)]

$∴ Area of △ P Q R = \frac{1}{2}$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & - 4 & 2 \\ - 7 & - 3 & 4 \end{matrix} |$

$= \frac{1}{2}$ $| - 10 \hat{i} - 14 \hat{j} - 28 \hat{k} |$

$= \frac{1}{2} \sqrt{100 + 196 + 784} = 3 \sqrt{30} sq . units$ .

Q 40 :

If the square of the shortest distance between the lines $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3}$ and $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}$ is $\frac{m}{n}$ , where m, n are coprime numbers, then m + n is equal to : [2025]

6
14
21
9

1

2

3

4

(4)

We have, ${\vec{a}}_{1} = 2 \hat{i} + \hat{j} - 3 \hat{k}, {\vec{b}}_{1} = \hat{i} + 2 \hat{j} - 3 \hat{k}$

${\vec{a}}_{2} = - \hat{i} - 3 \hat{j} - 5 \hat{k}, {\vec{b}}_{2} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$

Now, ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} |$ $= 2 \hat{i} - \hat{j}$

and ${\vec{a}}_{2} - {\vec{a}}_{1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k}$

$∴$ The shortest distance (d) between given lines

$= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| ({\vec{b}}_{1} \times {\vec{b}}_{2}) |} \Rightarrow d = \frac{2}{\sqrt{5}} \Rightarrow d^{2} = \frac{4}{5}$

$∴ m = 4, n = 5 \Rightarrow m + n = 9$

Topic Question Set

Q 11 :

Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]

Q 12 :

The square of the distance of the image of the point (6, 1, 5) in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is _________. [2024]

Q 14 :

The lines $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16}$ and $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1}$ intersect at the point P. If the distance of P from the line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is $l$ , then $14 l^{2}$ is equal to _________. [2024]

Q 15 :

A line with direction ratios 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively at the points P and Q. If the length of the perpendicular from the point (1, 2, 12) to the line PQ is $l$ , then $l^{2}$ is __________. [2024]

Q 19 :

Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If $∠ Q P R$ is a right angle, then $12 a^{2}$ is equal to __________. [2024]

Q 20 :

A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __________. [2024]

Q 21 :

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is $\frac{m}{n}$ then: [2025]

Q 23 :

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is $Q (α, β, γ)$ , then $α + β + γ$ is equal to : [2025]

Q 26 :

The distance of the point (7, 10, 11) from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is [2025]

Q 27 :

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $3 \sqrt{30}$ . Then the positive value of $5 α + β$ is [2025]

Q 31 :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

Q 32 :

Let the line L pass through (1, 1, 1) and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then which of the following points lies on the line L? [2025]

Q 34 :

Consider the lines $L_{1}$ : x – 1 = y – 2 = z and $L_{2}$ : x – 2 = y = z – 1. Let the feet of the perpendiculars from the point P(5, 1, –3) on the lines $L_{1}$ and $L_{2}$ be Q and R respectively. If the area of the triangle PQR is A, then $4 A^{2}$ is equal to : [2025]

Q 36 :

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_{1}$ and $L_{2}$ ? [2025]

Q 37 :

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point P(2, –10, 1) is: [2025]

Q 38 :

Let a line pass through two distinct pointsP(–2, –1, 3) and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of $△$ PQR is equal to : [2025]

Q 39 :

Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ . Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

Q 40 :

If the square of the shortest distance between the lines $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3}$ and $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}$ is $\frac{m}{n}$ , where m, n are coprime numbers, then m + n is equal to : [2025]

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Topic Question Set

Q 11 : Let P(α,β,γ) be the image of the point Q(1, 6, 4) in the line x1=y–12=z–23. Then 2α+β+γ is equal to __________. [2024]

Q 12 : The square of the distance of the image of the point (6, 1, 5) in the line x–13=y2=z–24, from the origin is _________. [2024]

Q 13 : Let the line of the shortest distance between the lines L1:r→=(i^+2j^+3k^)+λ(i^–j^+k^) and L2:r→=(4i^+5j^+6k^)+μ(i^+j^–k^) intersect L1 and L2 at P and Q respectively. If (α,β,γ) is the mid point of the line segment PQ, then 2(α+β+γ) is equal to __________. [2024]

Q 14 : The lines x–22=y–2=z–716 and x+34=y+23=z+21 intersect at the point P. If the distance of P from the line x+12=y–13=z–11 is l, then 14l2 is equal to _________. [2024]

Q 15 : A line with direction ratios 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively at the points P and Q. If the length of the perpendicular from the point (1, 2, 12) to the line PQ is l, then l2 is __________. [2024]

Q 16 : Let O be the orgin, and M and N be the points on the lines x–54=y–41=z–53 and x+812=y+25=z+119 respectively such that MN is the shortest distance between the given lines. Then OM→·ON→ is equal to __________. [2024]

Q 17 : If d1 is the shortest distance between the lines x + 1 = 2y = –12z, x = y + 2 = 6z – 6 and d2 is the shortest distance between the lines x–12=y+8–7=z–45,x–12=y–21=z–6–3, then the value of 323d1d2 is ___________. [2024]

Q 18 : Let a line passing through the point (–1, 2, 3) intersect the lines L1:x–13=y–22=z+1–2 at M(α,β,γ) and L2:x+2–3=y–2–2=z–14 at N(a, b, c). Then the value of (α+β+γ)2(a+b+c)2 equals __________. [2024]

Q 19 : Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If ∠QPR is a right angle, then 12a2 is equal to __________. [2024]

Q 20 : A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __________. [2024]

Q 21 : Let the vertices Q and R of the triangle PQR lie on the line x+35=y–12=z+43, QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is mn then: [2025]

Q 22 : The line L1 is parallel to the vector a→=–3i^+2j^+4k^ and passes through the point (7, 6, 2) and the line L2 is parallel to the vector b→=2i^+j^+3k^ and passes through the point (5,3, 4). The shortest distance between the lines L1 and L2 is : [2025]

Q 23 : If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q(α,β,γ), then α+β+γ is equal to : [2025]

Q 25 : Let a line passing through the point (4, 1, 0) intersect the line L1:x–12=y–23=z–34 at the point A(α,β,γ) and the line L2:x–6=y=–z+4 at the point B(a, b, c). Then |101αβγabc| is equal to [2025]

Q 26 : The distance of the point (7, 10, 11) from the line x–41=y–40=z–23 along the line x–92=y–133=z–176 is [2025]

Q 27 : Let the shortest distance between the lines x–33=y–α–1=z–31 and x+3–3=y+72=z–β4 be 330. Then the positive value of 5α+β is [2025]

Q 28 : Let A and B be two distinct points on the line L:x–63=y–72=z–7–2. Both A and B are at a distance 217 from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then OA→·OB→ is equal to [2025]

Q 29 : Let A be the point of intersection of the lines L1:x–71=y–50=z–3–1 and L2:x–13=y+34=z+75. Let B and C be the points on the lines L1 and L2 respectively such that AB=AC=15. Then the square of the area of the triangle ABC is [2025]

Q 30 : Let the values of p, for which the shortest distance between the lines x+13=y4=z5 and r→=(pi^+2j^+k^)+λ(2i^+3j^+4k^) is 16, be a, b, (a < b). Then the length of the latus rectum of the ellipse x2a2+y2b2=1 is : [2025]

Q 31 : If the shortest distance between the lines x–12=y–23=z–34 and x1=yα=z–51 is 56, then the sum of all possible values of α is [2025]

Q 32 : Let the line L pass through (1, 1, 1) and intersect the lines x–12=y+13=z–14 and x–31=y–42=z1. Then which of the following points lies on the line L? [2025]

Q 33 : If the equation of the line passing through the point (0,–12,0) and perpendicular to the lines r→=λ(i^+aj^+bk^) and r→=(i^–j^–6k^)+μ(–bi^+aj^+5k^) is x–1–2=y+4d=z–c–4, then a + b + c + d is equal to : [2025]

Q 34 : Consider the lines L1 : x – 1 = y – 2 = z and L2 : x – 2 = y = z – 1. Let the feet of the perpendiculars from the point P(5, 1, –3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to : [2025]

Q 35 : Let the values of λ for which the shortest distance between the lines x–12=y–23=z–34 and x–λ3=y–44=z–55 is 16 be λ1 and λ2. Then the radius of the circle passing through the points (0, 0), (λ1,λ2) and (λ2,λ1) is [2025]

Q 36 : Let L1:x–12=y–23=z–34 and L2:x–23=y–44=z–55 be two lines. Then which of the following points lies on the line of the shortest distance between L1 and L2? [2025]

Q 37 : The perpendicular distance, of the line x–12=y+2–1=z+32 from the point P(2, –10, 1) is: [2025]

Q 38 : Let a line pass through two distinct pointsP(–2, –1, 3) and Q, and be parallel to the vector 3i^+2j^+2k^. If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of △PQR is equal to : [2025]

Q 39 : Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line x–37=y–2–1=z+1–2. Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

Q 40 : If the square of the shortest distance between the lines x–21=y–12=z+3–3 and x+12=y+34=z+5–5 is mn, where m, n are coprime numbers, then m + n is equal to : [2025]

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Q 11 :

Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]

Q 12 :

The square of the distance of the image of the point (6, 1, 5) in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is _________. [2024]

Q 14 :

The lines $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16}$ and $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1}$ intersect at the point P. If the distance of P from the line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is $l$ , then $14 l^{2}$ is equal to _________. [2024]

Q 15 :

A line with direction ratios 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively at the points P and Q. If the length of the perpendicular from the point (1, 2, 12) to the line PQ is $l$ , then $l^{2}$ is __________. [2024]

Q 19 :

Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If $∠ Q P R$ is a right angle, then $12 a^{2}$ is equal to __________. [2024]

Q 20 :

A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __________. [2024]

Q 21 :

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is $\frac{m}{n}$ then: [2025]

Q 23 :

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is $Q (α, β, γ)$ , then $α + β + γ$ is equal to : [2025]

Q 26 :

The distance of the point (7, 10, 11) from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is [2025]

Q 27 :

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $3 \sqrt{30}$ . Then the positive value of $5 α + β$ is [2025]

Q 31 :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{\sqrt{6}}$ , then the sum of all possible values of $α$ is [2025]

Q 32 :

Let the line L pass through (1, 1, 1) and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then which of the following points lies on the line L? [2025]

Q 34 :

Consider the lines $L_{1}$ : x – 1 = y – 2 = z and $L_{2}$ : x – 2 = y = z – 1. Let the feet of the perpendiculars from the point P(5, 1, –3) on the lines $L_{1}$ and $L_{2}$ be Q and R respectively. If the area of the triangle PQR is A, then $4 A^{2}$ is equal to : [2025]

Q 36 :

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_{1}$ and $L_{2}$ ? [2025]

Q 37 :

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point P(2, –10, 1) is: [2025]

Q 38 :

Let a line pass through two distinct pointsP(–2, –1, 3) and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of $△$ PQR is equal to : [2025]

Q 39 :

Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ . Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

Q 40 :

If the square of the shortest distance between the lines $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3}$ and $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}$ is $\frac{m}{n}$ , where m, n are coprime numbers, then m + n is equal to : [2025]