Three Dimensional Geometry jee mains questions with solutions

Q 21 :

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1, 4, 0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is: [2025]

$\sqrt{15}$
$\sqrt{17}$
$\sqrt{13}$
$\sqrt{14}$

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(4)

Let the parallel line is $\frac{x - 1}{1} = \frac{y - 4}{2} = \frac{z - 0}{3} = λ$

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

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

$\Rightarrow x = 2 t + 2, y = 3 t + 6, z = 4 t + 3$

So, their point of intersection is

$(λ + 1, 2 λ + 4, 3 λ) = (2 t + 2, 3 t + 6, 4 t + 3)$

$\Rightarrow λ = 2 t + 1$

and $2 λ + 4 = 3 t + 6$ [ $∵ λ = 2 t + 1$ ]

$\Rightarrow t = 0, λ = 1$

So, point of intersection is (2, 6, 3).

So, distance $= \sqrt{{(2 - 1)}^{2} + {(6 - 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{14}$ .

Q 22 :

Let the line passing through the points (–1, 2, 1) and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

$5 \sqrt{5}$
$5 \sqrt{6}$
5
10

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2

3

4

(1)

Equation of line passing through (–1, 2, 1) and parallel to $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ is $\frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4}$ .

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

Coordinates of P are $(2 λ - 1, 3 λ + 2, 4 λ + 1)$ .

As point P also lies on

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

$∴$ Coordinates of point P are $(3 μ - 2, 2 μ + 3, μ + 4)$

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

On solving, we get $λ = μ = 1$

$∴$ The point P is (1, 5, 5)

$∴$ Required distance $Q P = \sqrt{{(4 - 1)}^{2} + {(- 5 - 5)}^{2} + {(1 - 5)}^{2}}$

$= \sqrt{9 + 100 + 16} = 5 \sqrt{5}$

Q 23 :

Let in a $△$ ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△$ ABC is : [2025]

56
17
42
21

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(4)

Let BM be the height of the triangle ABC.

Direction ratios of AC = 3, 2, –2

Coordinates of $M = (3 λ + 6, 2 λ + 7, - 2 λ + 7)$

Direction ratios of BM $= (3 λ + 6 - 1, 2 λ + 7 - 2, - 2 λ + 7 - 3)$

$= (3 λ + 5, 2 λ + 5, - 2 λ + 4)$

$∵ B M ⊥ A C$

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

$\Rightarrow 9 λ + 15 + 4 λ + 10 + 4 λ - 8 = 0$

$\Rightarrow 17 λ + 17 = 0 \Rightarrow λ = - 1$

$∴$ Coordinates of M = (3, 5, 9)

$∴ B M = \sqrt{{(3 - 1)}^{2} + {(5 - 2)}^{2} + {(9 - 3)}^{2}} = 7$

Area of $△ A B C = = \frac{1}{2} \times 7 \times 6 = 21 sq . units$ .

Q 24 :

If the image of the point (4, 4, 3) in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(α, β, γ)$ , then $α + β + γ$ is equal to [2025]

9
7
12
8

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4

(1)

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

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

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

Since, $\vec{P M}$ is perpendicular to the given line

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

$\Rightarrow λ = 1$

$∴$ The coordinates of point M is (3, 3, 4).

Let Q be the image of the point P. Then, M be the mid-point of PQ.

$∴ (\frac{4 + α}{2}, \frac{4 + β}{2}, \frac{3 + γ}{2}) \equiv (3, 3, 4)$

$\Rightarrow$ $α = 2, β = 2, γ = 5$ $∴$ $α + β + γ = 9$

Q 25 :

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is : [2025]

66
41
44
54

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(1)

Equation of line passing through the point $(\frac{15}{7}, \frac{32}{7}, 7)$ having direction ratios 1, 4 and 7 is given by

$\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = λ (say)$

$∴ x = λ + \frac{15}{7}, y = 4 λ + \frac{32}{7} and z = 7 λ + 7$

It lies on the given line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$

i.e., $\frac{λ + \frac{15}{7} + 1}{3} = \frac{7 λ + 7 + 5}{7}$

$\Rightarrow 7 λ + 22 = 21 λ + 36$

$\Rightarrow λ = - 1$

$∴ (x, y, z) \equiv (\frac{8}{7}, \frac{4}{7}, 0)$

Square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ and $(\frac{8}{7}, \frac{4}{7}, 0)$

$= {(\frac{15}{7} - \frac{8}{7})}^{2} + {(\frac{32}{7} - \frac{4}{7})}^{2} + {(7 - 0)}^{2}$

= 1 + 16 + 49 = 66.

Q 26 :

Let A, B, C be three points in xy-plane, whose position vector are given by $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $a \hat{i} + (1 - a) \hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors $\vec{O A}$ and $\vec{O B}$ is $\frac{9}{\sqrt{2}}$ , then the sum of all the possible values of a is : [2025]

1
0
9/2
2

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4

(1)

Equation of line OA is $x - \sqrt{3} y = 0$

Equation of line OB is $\sqrt{3} x - y = 0$

Equation of angle bisector is $\frac{x - \sqrt{3} y}{\sqrt{1 + 3}} + \frac{\sqrt{3} x - y}{\sqrt{3 + 1}} = 0$

$\Rightarrow x - y = 0$

$| \frac{a - (1 - a)}{\sqrt{2}} |$ $= \frac{9}{\sqrt{2}} \Rightarrow a = 5 or - 4$

Required sum = 5 + (–4) = 1.

Q 27 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k} and \vec{b} = 2 \hat{i} + 7 \hat{j} + 3 \hat{k}$ .

Let $L_{1} : \vec{r} = (- \hat{i} + 2 \hat{j} + \hat{k}) + λ \vec{a}, λ \in R$ and $L_{2} : \vec{r} = (\hat{j} + \hat{k}) + μ \vec{b}, μ \in R$ be two lines. if the line $L_{3}$ passes through the point of intersection of $L_{1}$ and $L_{2}$ , and is parallel to $\vec{a} + \vec{b}$ , then $L_{3}$ passes through the point: [2025]

(2, 8, 5)
(–1, –1, 1)
(5, 17, 4)
(8, 26, 12)

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(4)

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

Point of intersection of $L_{1}$ & $L_{2}$ is given by

$(- 1 + λ) \hat{i} + (2 + 2 λ) \hat{j} + (1 + λ) \hat{k} = 2 μ \hat{i} + (1 + 7 μ) \hat{j} + (1 + 3 μ) \hat{k}$

$\Rightarrow - 1 + λ = 2 μ, 2 + 2 λ = 1 + 7 μ, 1 + λ = 1 + 3 μ$

$\Rightarrow μ = 1 and λ = 3$

$∴$ Position vector of point of intersection of $L_{1}$ and $L_{2}$ is $2 \hat{i} + 8 \hat{j} + 4 \hat{k}$ .

Hence, $L_{3}$ is given by

$2 \hat{i} + 8 \hat{j} + 4 \hat{k} + γ (\vec{a} + \vec{b}) = 2 \hat{i} + 8 \hat{j} + 4 \hat{k} + γ (3 \hat{i} + 9 \hat{j} + 4 \hat{k})$

For $γ = 2,$ line $L_{3}$ passes through point (8, 26, 12).

Q 28 :

Let $L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}$ and $L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}$ be two lines.

Let $L_{3}$ be a line passing through the point $(α, β, γ)$ and be perpendicular to both $L_{1}$ and $L_{2}$ . If $L_{3}$ intersects $L_{1}$ , then $| 5 α - 11 β - 8 γ |$ equals : [2025]

20
25
16
18

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(2)

Let D.r.s. of $L_{3}$ be a, b, c

Now, $L_{3}$ is $⊥^{r}$ to $L_{1}$ and $L_{2}$ = a – b + 2c = 0 ... (i)

and –a + 2b + c = 0 ... (ii)

Solving (i) and (ii), we get $\frac{a}{- 5} = \frac{b}{- 3} = \frac{c}{1}$

Equation of line $L_{3}$ is $\frac{x - α}{- 5} = \frac{y - β}{- 3} = \frac{z - γ}{1} = k$

Any point on $L_{3}$ is $(- 5 k + α, - 3 k + β, k + γ)$

Now, $L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2} = λ$

Any point on $L_{1}$ is $(λ + 1, - λ + 2, 2 λ + 1)$

Now $L_{1}$ and $L_{3}$ intersects.

$∴ - 5 k + α = λ + 1, - 3 k + β = - λ + 2, k + γ = 2 λ + 1$

$\Rightarrow α = 5 k + λ + 1, β = 3 k - λ + 2, γ = - k + 2 λ + 1$

$∴ | 5 α - 11 β - 8 γ |$

$= | 25 k + 5 λ + 5 - 33 k + 11 λ - 22 + 8 k - 16 λ - 8 | = | - 25 | = 25$ .

Q 29 :

Let ABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of $△$ ABC in the line 3x + 6y – 53 = 0. Then $h^{2} + k^{2} + h k$ is equal to : [2025]

40
36
47
37

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(4)

Centroid of $△ A B C = (\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}) = (\frac{17}{3}, \frac{28}{3})$

Let image of centroid with respect to line mirror is (h, k).

$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = \frac{- 2 (3 \times \frac{17}{3} + 6 \times \frac{28}{3} - 53)}{3^{2} + 6^{2}}$

Solving, we get h = 3 and k = 4

$∴ h^{2} + k^{2} + h k = 37$ .

Q 30 :

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ . Let the line $\vec{r} = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}), λ \in R$ , intersect the line L at Q. Then $2 {(P Q)}^{2}$ is equal to : [2025]

25
19
27
29

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(3)

We have, $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2} = μ$

$\Rightarrow Point is P (μ + 1, - μ - 1, 2 μ + 2)$

$∴ D . r' s of A P = (μ, - μ - 3, 2 μ)$

Now, $μ (1) + (- μ - 3) (- 1) + 2 (2 μ) = 0$

$\Rightarrow μ + μ + 3 + 4 μ = 0$

$\Rightarrow 6 μ + 3 = 0 \Rightarrow μ = \frac{- 1}{2}$

$∴ P \equiv (\frac{1}{2}, \frac{- 1}{2}, 1)$

Now, line L intersect the other line at Q, i.e.,

$(- 1 + λ, 1 - λ, - 2 + λ)$

$∴ μ + 1 = - 1 + λ, - μ - 1 = 1 - λ and 2 μ + 2 = - 2 + λ$

$\Rightarrow μ = λ - 2, μ = λ - 2 and 2 μ + 2 = - 2 + λ$

$\Rightarrow 2 (λ - 2) + 2 = - 2 + λ$

$\Rightarrow 2 λ - 4 + 2 = - 2 + λ \Rightarrow λ = 0 and μ = - 2$

$∴ Q \equiv (- 1, 1, - 2)$

So, ${(P Q)}^{2} = {(\frac{- 3}{2})}^{2} + {(\frac{3}{2})}^{2} + {(- 3)}^{2} = \frac{9}{4} + \frac{9}{4} + 9 = 9 + \frac{9}{2} = \frac{27}{2}$

Hence, $2 P Q^{2} = 27$ .

Q 31 :

Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$ and $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}$ . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

$\sqrt{10}$
2
3
$2 \sqrt{3}$

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4

(3)

Vector parallel to line $L =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 3 & 4 \end{matrix} |$

$= 10 \hat{i} - 10 \hat{j} + 5 \hat{k} = 5 (2 \hat{i} - 2 \hat{j} + \hat{k})$

Equation of line L passing through point P(2, –1, 3) and parallel to vector $(2 \hat{i} - 2 \hat{j} + \hat{k})$ , is

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

$∴ x = 2 λ + 2, y = - 2 λ - 1, z = λ + 3$

$∵$ Line L intersects the yz-plane

$∴ 2 λ + 2 = 0 \Rightarrow λ = - 1$

Hence, point Q is (0, 1, 2)

Distance between point P(2, –1, 3) and Q(0, 1, 2)

$= \sqrt{{(0 - 2)}^{2} + {(1 + 1)}^{2} + {(2 - 3)}^{2}} = \sqrt{9} = 3$ .

Q 32 :

Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

(56)

$L_{1} : x + 2 = y - 1 = z = λ (say)$

$L_{2} : \frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1} = μ (say)$

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

Any point of line $L_{1}, L_{2} and L_{3}$ is given by $(λ - 2, λ + 1, λ)$ , $(5 μ + 3, - μ, μ + 1)$ and $(- 3 δ, 3 δ + 3, δ + 2)$ respectively.

Point of intersection of $L_{1}$ and $L_{2}$ is given by

$\begin{matrix} λ - 2 = 5 μ + 3 \\ λ + 1 = - μ \\ λ = μ + 1 \end{matrix}} \Rightarrow λ = 0, μ = - 1$

$∴$ Point of intersection is A(–2, 1, 0).

Point of intersection of $L_{2}$ and $L_{3}$ is given by

$\begin{matrix} 5 μ + 3 = - 3 δ \\ - μ = 3 δ + 3 \\ μ + 1 = δ + 2 \end{matrix}} \Rightarrow μ = 0, δ = - 1$

$∴$ Point of intersection is B(3, 0 , 1).

Point of intersection of $L_{1}$ and $L_{3}$ is given by

$\begin{matrix} λ - 2 = - 3 δ \\ λ + 1 = 3 δ + 3 \\ λ = δ + 2 \end{matrix}} \Rightarrow λ = 2, δ = 0$

$∴$ Point of intersection is C(0, 3, 2).

Now, Area of $△ A B C = \frac{1}{2}$ $| \vec{A B} \times \vec{B C} |$

$∴ A = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 1 & - 1 \\ - 3 & 3 & 1 \end{matrix} | |$

$= \frac{1}{2} | 4 \hat{i} + 8 \hat{j} - 12 \hat{k |}$

$= \frac{1}{2} \sqrt{16 + 64 + 144} = \sqrt{56}$

$∴ A^{2} = 56$ .

Q 33 :

Let $L_{1} : \frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0}$ and $L_{2} : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{α}, α \in R$ be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and $L_{2}$ , then the value of $26 α {(P B)}^{2}$ is __________. [2025]

(216)

Point $B \equiv (3 μ + 1, - μ + 1, - 1) \equiv (2 λ + 2, 0, α λ - 4)$

$\Rightarrow 3 μ + 1 = 2 λ + 2, - μ + λ = 0, λ α - 4 = - 1$

$∴ μ = 1 \Rightarrow λ = 1 and α = 3$

So, point B(4, 0, –1).

Let point P is (2k + 2, 0, 3k – 4).

So, Dr's of AP is < 2k + 1, –1, 3k – 3 >

Since, $L_{2} ⊥ A P$

$\Rightarrow 2 (2 k + 1) + 0 (- 1) + 3 (3 k - 3) = 0$

$\Rightarrow 4 k + 2 + 9 k - 9 = 0 \Rightarrow k = \frac{7}{13}$

$∴ Point P \equiv (\frac{40}{13}, 0, \frac{- 31}{13})$

$∴ {(P B)}^{2} = {(\frac{40}{13} - 4)}^{2} + {(0 - 0)}^{2} + {(\frac{- 31}{13} + 1)}^{2} = \frac{468}{169}$

$∴ 26 α {(P B)}^{2} = 26 \times 3 \times \frac{468}{169} = 216$ .

Q 34 :

Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

(957)

Given, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be on the line.

Here P be the image of point.

Since, R is on the line L, then

$R \equiv (2 λ + 1, 3 λ - 1, 4 λ) = (5, p, q) [Given]$

$\Rightarrow 2 λ + 1 = 5$

$\Rightarrow 2 λ = 4 \Rightarrow λ = 2$

$∴ R \equiv (5, 5, 8)$

Since, T is also on the line L, then

$T \equiv (2 μ + 1, 3 μ - 1, 4 μ)$

Now, $\vec{Q T} = (2 μ - 6) \hat{i} + (3 μ + 1) \hat{j} + (4 μ - 5) \hat{k}$

and $\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ (Normal)

Taking $\vec{Q T} \cdot \vec{b} = 0$

$\Rightarrow 4 μ - 12 + 9 μ + 3 + 16 μ - 20 = 0 \Rightarrow 29 μ - 29 = 0$

$\Rightarrow μ = 1$

$∴ T \equiv (3, 2, 4)$

$\Rightarrow Q T = \sqrt{{(7 - 3)}^{2} + {(- 2 - 2)}^{2} + {(5 - 4)}^{2}}$

$= \sqrt{16 + 16 + 1} = \sqrt{33} \Rightarrow P Q = 2 \sqrt{33}$

Similarly, $R T = \sqrt{29}$

$\Rightarrow Area of △ P Q R = \frac{1}{2} \times \sqrt{29} \times 2 \sqrt{33} = \sqrt{29} \times \sqrt{33}$

$\Rightarrow {(Area of △ P Q R)}^{2} = {(\sqrt{29} \times \sqrt{33})}^{2} = 29 \times 33 = 957$ .

Q 35 :

The shortest distance between the lines $\frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3}$ and $\frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$ is [2023]

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

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4

(1)

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

$Here, x_{1} = 4, y_{1} = - 2, z_{1} = - 3, a_{1} = 4, b_{1} = 5, c_{1} = 3$

$x_{2} = 1, y_{2} = 3, z_{2} = 4, a_{2} = 3, b_{2} = 4, c_{2} = 2$

$Shortest distance,$ $d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} |$

$= \frac{| \begin{matrix} 1 - 4 & 3 + 2 & 4 + 3 \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{matrix} |}{\sqrt{{(4 \times 4 - 3 \times 5)}^{2} + {(5 \times 2 - 4 \times 3)}^{2} + {(3 \times 3 - 4 \times 2)}^{2}}}$

$= \frac{18}{\sqrt{6}} = 3 \sqrt{6}$

Q 36 :

The shortest distance between the lines $\frac{x + 2}{1} = \frac{y}{- 2} = \frac{z - 5}{2}$ and $\frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0}$ is [2023]

9
6
8
7

1

2

3

4

(1)

Q 37 :

Let S be the set of all values of $λ$ , for which the shortest distance between the lines $\frac{x - λ}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}$ and $\frac{x + λ}{3} = \frac{y}{- 4} = \frac{z - 6}{0}$ is 13. Then $8$ $| \sum_{λ \in S} λ |$ is equal to [2023]

302
306
308
304

1

2

3

4

(2)

$Shortest distance between lines$

$\frac{x - λ}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}$ and $\frac{x + λ}{3} = \frac{y}{- 4} = \frac{z - 6}{0}$ is 13

So, $| \frac{| \begin{matrix} λ + λ & 3 - 0 & - 6 - 6 \\ 0 & 4 & 1 \\ 3 & - 4 & 0 \end{matrix} |}{\sqrt{{(0 + 4)}^{2} + {(3 - 0)}^{2} + {(0 - 12)}^{2}}} |$ $= 13$

$\Rightarrow$ $| \frac{| \begin{matrix} 2 λ & 3 & - 12 \\ 0 & 4 & 1 \\ 3 & - 4 & 0 \end{matrix} |}{13} |$ $= 13$

$\Rightarrow$ $| 2 λ (0 + 4) - 3 (- 3) - 12 (0 - 12) |$ $= 13 \times 13$

$\Rightarrow$ $| 8 λ + 9 + 144 |$ $= 169$ $\Rightarrow$ $| 8 λ + 153 |$ $= 169$

$\Rightarrow 8 λ + 153 = 169$ and $8 λ + 153 = - 169 \Rightarrow λ = 2, \frac{- 161}{4}$

$∴$ $8$ $| \sum_{λ \in S} λ |$ $= 8$ $| \frac{- 161}{4} + 2 |$ $= 8$ $| \frac{- 153}{4} |$ $= 153 \times 2 = 306$

Q 38 :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is [2023]

$7 \sqrt{3}$
$6 \sqrt{3}$
$4 \sqrt{3}$
$5 \sqrt{3}$

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4

(2)

Q 39 :

Consider the lines $L_{1}$ and $L_{2}$ given by $L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2}, L_{2} : \frac{x - 2}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} .$ A line $L_{3}$ having direction ratios $1, - 1, - 2$ intersects $L_{1}$ and $L_{2}$ at the points P and Q respectively. Then the length of line segment PQ is [2023]

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

1

2

3

4

(3)

$P lies on \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} = λ$

$P \equiv (2 λ + 1, λ + 3, 2 λ + 2)$

$Q lies on \frac{x - 2}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} = μ$

$Q \equiv (μ + 2, 2 μ + 2, 3 μ + 3)$

$D.R.'s of P Q = (2 λ - μ - 1, λ - 2 μ + 1, 2 λ - 3 μ - 1)$

$which are proportional to 1, - 1, - 2$

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

$- 2 λ + μ + 1 = λ - 2 μ + 1 \Rightarrow λ = μ$

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

$P = (7, 6, 8), Q = (5, 8, 12)$

$P Q^{2} = 2^{2} + 2^{2} + 4^{2} = 24 \Rightarrow P Q = 2 \sqrt{6}$

Q 40 :

The shortest distance between the lines $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is [2023]

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

1

2

3

4

(3)

Q 41 :

The foot of perpendicular of the point (2, 0, 5) on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then, which of the following is not correct? [2023]

$\frac{γ}{α} = \frac{5}{8}$
$\frac{β}{γ} = - 5$
$\frac{α β}{γ} = \frac{4}{15}$
$\frac{α}{β} = - 8$

1

2

3

4

(2)

$D.r's of line A B < α - 2, β, γ - 5 >$

Since $A B$ is perpendicular to

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

$∴$ $2 (α - 2) + 5 β - 1 (γ - 5) = 0$

$\Rightarrow 2 α + 5 β - γ + 1 = 0$ ...(i)

Also, $α, β, γ$ lies on the given line.

$∴$ $\frac{α + 1}{2} = \frac{β - 1}{5} = \frac{γ + 1}{- 1}$ ...(ii)

From (ii), (taking the first two and last two terms)

$α = \frac{2 β - 7}{5}, γ = \frac{- β - 4}{5}$

Putting these in (i), we get $4 β - 14 + 25 β + β + 4 + 5 = 0$

$\Rightarrow β = \frac{1}{6}$

which gives $α = \frac{- 4}{3}$ and $γ = \frac{- 5}{6}$ $\Rightarrow \frac{γ}{α} = \frac{\frac{- 5}{6}}{\frac{- 4}{3}} = \frac{5}{8}$ ,

$\frac{β}{γ} = \frac{1}{6} \times \frac{- 6}{5} = \frac{- 1}{5} \Rightarrow \frac{α β}{γ} = \frac{4}{15}, \frac{α}{β} = - 8, \frac{β}{γ} \neq - 5$

Q 42 :

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is [2023]

10
28
22
16

1

2

3

4

(2)

$Let two lines L_{1} and L_{2} intersect at point P .$ Then we have the points on line $L_{1}$ are as $= (λ + 1, 2 λ + 2, λ - 3)$ and we have the points on line $L_{2}$ are as $= (2 μ + a, 3 μ - 2, μ + 3)$

${∵ L_{1} = \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1} = λ (say) and L_{2} = \frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1} = μ (say)}$

$Now if L_{1} and L_{2} intersect, then we must have$

$λ - 3 = μ + 3 \Rightarrow λ = μ + 6 ...(i)$

and $2 λ + 2 = 3 μ - 2 \Rightarrow 2 λ = 3 μ - 4 ...(ii)$

On solving (i) and (ii), we get $μ = 16$ and $λ = 22$

So, the coordinates of $P$ must be $(23, 46, 19)$ $\Rightarrow a = - 9$

Hence, the distance of the point $P$ from $z = - 9$ is $28 .$

Q 43 :

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

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

1

2

3

4

(4)

$Since two lines L_{1} and L_{2} are given as$

$L_{1} \equiv \frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}, L_{2} \equiv \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$

$For line L_{1}, we have \vec{a} = \hat{i} - 8 \hat{j} + 4 \hat{k}$

$For line L_{2}, we have \vec{b} = \hat{i} + 2 \hat{j} + 6 \hat{k}$

$\vec{p} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}$ and $\vec{q} = 2 \hat{i} + \hat{j} - 3 \hat{k}$

$So, we have$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} |$

$= \hat{i} (21 - 5) - \hat{j} (- 6 - 10) + \hat{k} (2 + 14)$

$= 16 \hat{i} + 16 \hat{j} + 16 \hat{k} = 16 (\hat{i} + \hat{j} + \hat{k})$

$The shortest distance, d$

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

$=$ $| \frac{[\hat{i} (1 - 1) + \hat{j} (- 8 - 2) + \hat{k} (4 - 6)] \cdot 16 (\hat{i} + \hat{j} + \hat{k})}{\sqrt{16^{2} + 16^{2} + 16^{2}}} |$

$=$ $| \frac{(- 10 \hat{j} - 2 \hat{k}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k})}{16 \sqrt{3}} |$

$=$ $| \frac{(- 10) (16) + (- 2) (16)}{16 \sqrt{3}} |$ $=$ $| \frac{192}{16 \sqrt{3}} |$ $=$ $| \frac{- 12}{\sqrt{3}} |$ $= 4 \sqrt{3}$

Q 44 :

Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z$ be $2 \sqrt{6}$ . If $(α, β, γ)$ lies on L, then which of the following is NOT possible? [2023]

$α + 2 γ = 24$
$2 α + γ = 7$
$α - 2 γ = 19$
$2 α - γ = 9$

1

2

3

4

(1)

$Since shortest distance between lines L and L_{1} is 2 \sqrt{6} .$

$Here, L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$

$and L_{1} : \frac{x + 1}{1} = \frac{y - 1}{1} = \frac{z - 4}{- 1}$

$Shortest distance between L and L_{1} is given by$

$d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2} + {(a_{1} b_{2} - a_{2} b_{1})}^{2}}} |$

$∴$ $2 \sqrt{6} =$ $| \frac{| \begin{matrix} - 1 - 5 & 1 - λ & 4 + λ \\ - 2 & 0 & 1 \\ 1 & 1 & - 1 \end{matrix} |}{\sqrt{{(0 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 0)}^{2}}} |$

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

$=$ $| \frac{6 + λ - 1 - 2 λ - 8}{\sqrt{6}} |$ $=$ $| \frac{- λ - 3}{\sqrt{6}} |$

$\Rightarrow \frac{- 3 - λ}{\sqrt{6}} = \pm 2 \sqrt{6} \Rightarrow \frac{- 3 - λ}{\sqrt{6}} = 2 \sqrt{6} and \frac{- 3 - λ}{\sqrt{6}} = - 2 \sqrt{6}$

$- 3 - λ = 12 \Rightarrow λ = - 15$ and $- 3 - λ = - 12 \Rightarrow λ = 9$

$Since λ \geq 0, therefore λ = 9$

$Now, from line L, \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1} = r (say)$

$\Rightarrow x = - 2 r + 5, y = 0 + λ, z = r - λ$

$\Rightarrow x = - 2 r + 5, y = 9, z = r - 9$

$Since (α, β, γ) lies on L,$ then $α = - 2 r + 5, β = 9, γ = r - 9$

$Now, α + 2 γ = - 2 r + 5 + 2 r - 18 = - 13$

$2 α + γ = - 4 r + 10 + r - 9 = - 3 r + 1, r \in R$

$α - 2 γ = - 2 r + 5 - 2 r + 18 = - 4 r + 23, r \in R$

$2 α - γ = - 4 r + 10 - r + 9 = - 5 r + 19, r \in R$

$Hence, α + 2 γ = 24 is not possible.$

Q 45 :

If the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of $8 α β$ is _______ . [2023]

(18)

$Let \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α} = λ (say)$ ...(i)

$and \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β} = μ (say)$ ...(ii)

$∴$ $Any point on line (i) and (ii) are of the forms (2 λ + 1, 3 λ + 2, α λ + 3)$ $and (5 μ + 4, 2 μ + 1, β μ) respectively .$

$The lines are intersecting.$

$∴$ $2 λ + 1 = 5 μ + 4, 3 λ + 2 = 2 μ + 1, α λ + 3 = β μ$

$Solving first two equations, we get λ = - 1 and μ = - 1$

$From third equation, we have$

$- α + 3 = - β \Rightarrow α - 3 = β$ ...(iii)

$Let y = 8 α β = 8 α (α - 3)$

$y' = 8 α + 8 (α - 3) = 16 α - 24$

$For maxima/minima, 16 α - 24 = 0 \Rightarrow α = \frac{3}{2}$

$Now, y'' = 16 > 0$

$∴$ $8 α β is minimum at α = \frac{3}{2}$

$So, minimum value of$ $| 8 α β |$ $=$ $| 8 \times \frac{3}{2} \times \frac{- 3}{2} |$ $= 18$

Q 46 :

If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2 A + y \sin 2 B + z \sin 2 C - 9,$ where A, B, C are the angles of a triangle ABC, then $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$ is equal to _________ . [2023]

(5)

$Any point on the given line \frac{x}{1} = \frac{y}{1} = \frac{z}{1} = λ is (λ, λ, λ)$

$If it intersects the given lines then it must satisfy them.$

$\Rightarrow λ (\sin A + \sin B + \sin C) = 2 \times 3^{2}$ ...(i)

$and λ (\sin 2 A + \sin 2 B + \sin 2 C) = 3^{2}$ ...(ii)

$On dividing equation (ii) by (i), we get$

$\frac{\sin 2 A + \sin 2 B + \sin 2 C}{\sin A + \sin B + \sin C} = \frac{1}{2}$ $\Rightarrow \frac{4 \sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$

$\Rightarrow 8 [\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}] = \frac{1}{2} \Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$

$∴$ $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 80 \times \frac{1}{16} = 5$

Q 47 :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______ . [2023]

(14)

$L_{1} : \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$

$L_{2} : \frac{x - 6}{3} = \frac{y - 1}{- 2} = \frac{z + 8}{0}$

$Shortest distance =$ $\frac{| | \begin{matrix} 6 - 2 & 1 - (- 1) & - 8 - 6 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}$

$=$ $\frac{| | \begin{matrix} 4 & 2 & - 14 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| 4 \hat{i} + 6 \hat{j} - 12 \hat{k} |}$ $= \frac{4 (0 + 4) - 2 (0 - 6) - 14 (- 6 - 6)}{\sqrt{4^{2} + 6^{2} + 12^{2}}}$

$= \frac{16 + 12 + 14 \times 12}{14} = 14 units.$

Q 48 :

If the shortest distance between the lines $\frac{x + \sqrt{6}}{2} = \frac{y - \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$ and $\frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5}$ is $6$ , then the square of the sum of all possible values of $λ$ is ________ . [2023]

(384)

$Given, shortest distance between the lines$

$\frac{x + \sqrt{6}}{2} = \frac{y - 2 \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$

$and \frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5} is 6 .$

$Vector along line of shortest distance$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$ $= \hat{i} (15 - 16) - \hat{j} (10 - 12) + \hat{k} (8 - 9)$

$= - \hat{i} + 2 \hat{j} - \hat{k} (its magnitude is \sqrt{6})$

$Now, \frac{1}{\sqrt{6}}$ $| \begin{matrix} λ + \sqrt{6} & \sqrt{6} & - 3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$ $= \pm 6$ $\Rightarrow λ = - 2 \sqrt{6}, 10 \sqrt{6}$

$So, square of sum of these values: {(10 \sqrt{6} - 2 \sqrt{6})}^{2} = {(8 \sqrt{6})}^{2} = 384$

Q 49 :

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______ . [2023]

(18)

$The equation of the line passing through (1, 2, 3) and (2, 3, 4) is$

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

and vector form of equation of given second line is

$\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} - \hat{j}) (∵ \vec{r} = \vec{b} + μ \vec{q})$

$Now,$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{matrix} |$ $= \hat{i} + 2 \hat{j} - 3 \hat{k}$

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

$=$ $| \frac{(- 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 3 \hat{k})}{\sqrt{14}} |$ $=$ $| \frac{- 6 + 3}{\sqrt{14}} |$ $= \frac{3}{\sqrt{14}} = α = \frac{3}{\sqrt{14}}$

$Now, 28 α^{2} = 28 \times \frac{9}{14} = 18$

Q 50 :

Let the co-ordinates of one vertex of $Δ A B C$ be $A (0, 2, α)$ and the other two vertices lie on the line $\frac{x + α}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} .$ For $α \in Z$ , if the area of $Δ A B C$ is $21$ sq. units and the line segment $B C$ has length $2 \sqrt{21}$ units, then $α^{2}$ is equal to _________ . [2023]

Topic Question Set

Q 21 :

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1, 4, 0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is: [2025]

Q 22 :

Let the line passing through the points (–1, 2, 1) and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

Q 23 :

Let in a $△$ ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△$ ABC is : [2025]

Q 24 :

If the image of the point (4, 4, 3) in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(α, β, γ)$ , then $α + β + γ$ is equal to [2025]

Q 25 :

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is : [2025]

Q 29 :

Let ABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of $△$ ABC in the line 3x + 6y – 53 = 0. Then $h^{2} + k^{2} + h k$ is equal to : [2025]

Q 32 :

Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

Q 34 :

Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

Q 35 :

The shortest distance between the lines $\frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3}$ and $\frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$ is [2023]

Q 36 :

The shortest distance between the lines $\frac{x + 2}{1} = \frac{y}{- 2} = \frac{z - 5}{2}$ and $\frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0}$ is [2023]

(1)

Q 37 :

Let S be the set of all values of $λ$ , for which the shortest distance between the lines $\frac{x - λ}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}$ and $\frac{x + λ}{3} = \frac{y}{- 4} = \frac{z - 6}{0}$ is 13. Then $8$ $| \sum_{λ \in S} λ |$ is equal to [2023]

Q 38 :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is [2023]

(2)

Q 40 :

The shortest distance between the lines $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is [2023]

(3)

Q 41 :

The foot of perpendicular of the point (2, 0, 5) on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then, which of the following is not correct? [2023]

Q 42 :

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is [2023]

Q 43 :

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

Q 44 :

Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z$ be $2 \sqrt{6}$ . If $(α, β, γ)$ lies on L, then which of the following is NOT possible? [2023]

Q 45 :

If the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of $8 α β$ is _______ . [2023]

Q 46 :

If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2 A + y \sin 2 B + z \sin 2 C - 9,$ where A, B, C are the angles of a triangle ABC, then $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$ is equal to _________ . [2023]

Q 47 :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______ . [2023]

Q 48 :

If the shortest distance between the lines $\frac{x + \sqrt{6}}{2} = \frac{y - \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$ and $\frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5}$ is $6$ , then the square of the sum of all possible values of $λ$ is ________ . [2023]

Q 49 :

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______ . [2023]

(9)

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Topic Question Set

Q 21 : The distance of the line x–22=y–63=z–34 from the point (1, 4, 0) along the line x1=y–22=z+33 is: [2025]

Q 22 : Let the line passing through the points (–1, 2, 1) and parallel to the line x–12=y+13=z4 intersect the line x+23=y–32=z–41 at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

Q 23 : Let in a △ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line x–63=y–72=z–7–2. Then the area (in sq. units) of△ ABC is : [2025]

Q 24 : If the image of the point (4, 4, 3) in the line x–12=y–21=z–13 is (α,β,γ), then α+β+γ is equal to [2025]

Q 25 : The square of the distance of the point (157,327,7) from the line x+13=y+35=z+57 in the direction of the vector i^+4j^+7k^ is : [2025]

Q 27 : Let a→=i^+2j^+k^ and b→=2i^+7j^+3k^. Let L1:r→=(–i^+2j^+k^)+λa→,λ∈R and L2:r→=(j^+k^)+μb→,μ∈R be two lines. if the line L3 passes through the point of intersection of L1 and L2, and is parallel to a→+b→, then L3 passes through the point: [2025]

Q 28 : Let L1:x–11=y–2–1=z–12 and L2:x+1–1=y–22=z1 be two lines. Let L3 be a line passing through the point (α,β,γ) and be perpendicular to both L1 and L2. If L3 intersects L1, then |5α–11β–8γ| equals : [2025]

Q 29 : Let ABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of △ABC in the line 3x + 6y – 53 = 0. Then h2+k2+hk is equal to : [2025]

Q 30 : Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L:x–11=y+1–1=z–22. Let the line r→=(–i^+j^–2k^)+λ(i^–j^+k^), λ∈R, intersect the line L at Q. Then 2(PQ)2 is equal to : [2025]

Q 31 : Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines x–12=y+11=z–3–2 and x–31=y–23=z+24. If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

Q 32 : Let the area of the triangle formed by the lines x + 2 = y – 1 = z, x–35=y–1=z–11 and x–3=y–33=z–21 be A. Then A2 is equal to __________. [2025]

Q 33 : Let L1:x–13=y–1–1=z+10 and L2:x–22=y0=z+4α, α∈R be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and L2, then the value of 26α(PB)2 is __________. [2025]

Q 34 : Let P be the image of the point Q(7, –2, 5) in the line L:x–12=y+13=z4 and R(5, p, q) be a point on L. Then the square of the area of ∆PQR is __________. [2025]

Q 35 : The shortest distance between the lines x-44=y+25=z+33 and x-13=y-34=z-42 is [2023]

Q 36 : The shortest distance between the lines x+21=y-2=z-52 and x-41=y-12=z+30 is [2023]

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Q 37 : Let S be the set of all values of λ, for which the shortest distance between the lines x-λ0=y-34=z+61 and x+λ3=y-4=z-60 is 13. Then 8|∑λ∈Sλ| is equal to [2023]

Q 38 : The shortest distance between the lines x-51=y-22=z-4-3 and x+31=y+54=z-1-5 is [2023]

(2)

Q 39 : Consider the lines L1 and L2 given by L1:x-12=y-31=z-22,L2:x-21=y-22=z-33. A line L3 having direction ratios 1,-1,-2 intersects L1 and L2 at the points P and Q respectively. Then the length of line segment PQ is [2023]

Q 40 : The shortest distance between the lines x+1=2y=-12z and x=y+2=6z-6 is [2023]

(3)

Q 41 : The foot of perpendicular of the point (2, 0, 5) on the line x+12=y-15=z+1-1 is (α,β,γ). Then, which of the following is not correct? [2023]

Q 42 : If the lines x-11=y-22=z+31 and x-a2=y+23=z-31 intersect at the point P, then the distance of the point P from the plane z=a is [2023]

Q 43 : The shortest distance between the lines x-12=y+8-7=z-45 and x-12=y-21=z-6-3 is [2023]

Q 44 : Let the shortest distance between the lines L:x-5-2=y-λ0=z+λ1,λ≥0 and L1:x+1=y-1=4-z be 26. If (α,β,γ) lies on L, then which of the following is NOT possible? [2023]

Q 45 : If the lines x-12=2-y-3=z-3α and x-45=y-12=zβ intersect, then the magnitude of the minimum value of 8αβ is _______ . [2023]

Q 46 : If the line x=y=z intersects the line xsinA+ysinB+zsinC-18=0=xsin2A+ysin2B+zsin2C-9, where A, B, C are the angles of a triangle ABC, then 80(sinA2sinB2sinC2) is equal to _________ . [2023]

Q 47 : The shortest distance between the lines x-23=y+12=z-62 and x-63=1-y2=z+80 is equal to ______ . [2023]

Q 48 : If the shortest distance between the lines x+62=y-63=z-64 and x-λ3=y-264=z+265 is 6, then the square of the sum of all possible values of λ is ________ . [2023]

Q 49 : If the shortest distance between the line joining the points (1,2,3) and (2,3,4), and the line x-12=y+1-1=z-20 is α, then 28α2 is equal to ______ . [2023]

Q 50 : Let the co-ordinates of one vertex of ΔABC be A(0,2,α) and the other two vertices lie on the line x+α5=y-12=z+43. For α∈Z, if the area of ΔABC is 21 sq. units and the line segment BC has length 221 units, then α2 is equal to _________ . [2023]

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Q 21 :

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1, 4, 0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is: [2025]

Q 22 :

Let the line passing through the points (–1, 2, 1) and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

Q 23 :

Let in a $△$ ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△$ ABC is : [2025]

Q 24 :

If the image of the point (4, 4, 3) in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(α, β, γ)$ , then $α + β + γ$ is equal to [2025]

Q 25 :

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is : [2025]

Q 29 :

Let ABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of $△$ ABC in the line 3x + 6y – 53 = 0. Then $h^{2} + k^{2} + h k$ is equal to : [2025]

Q 32 :

Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

Q 34 :

Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

Q 35 :

The shortest distance between the lines $\frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3}$ and $\frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$ is [2023]

Q 36 :

The shortest distance between the lines $\frac{x + 2}{1} = \frac{y}{- 2} = \frac{z - 5}{2}$ and $\frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0}$ is [2023]

Q 37 :

Let S be the set of all values of $λ$ , for which the shortest distance between the lines $\frac{x - λ}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}$ and $\frac{x + λ}{3} = \frac{y}{- 4} = \frac{z - 6}{0}$ is 13. Then $8$ $| \sum_{λ \in S} λ |$ is equal to [2023]

Q 38 :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is [2023]

Q 40 :

The shortest distance between the lines $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is [2023]

Q 41 :

The foot of perpendicular of the point (2, 0, 5) on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then, which of the following is not correct? [2023]

Q 42 :

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is [2023]

Q 43 :

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

Q 44 :

Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z$ be $2 \sqrt{6}$ . If $(α, β, γ)$ lies on L, then which of the following is NOT possible? [2023]

Q 45 :

If the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of $8 α β$ is _______ . [2023]

Q 46 :

If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2 A + y \sin 2 B + z \sin 2 C - 9,$ where A, B, C are the angles of a triangle ABC, then $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$ is equal to _________ . [2023]

Q 47 :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______ . [2023]

Q 48 :

If the shortest distance between the lines $\frac{x + \sqrt{6}}{2} = \frac{y - \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$ and $\frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5}$ is $6$ , then the square of the sum of all possible values of $λ$ is ________ . [2023]

Q 49 :

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______ . [2023]