The distance of the point $Q(0,2,-2)$ from the line passing through the point $P(5,-4,3)$ and perpendicular to the lines $\vec{r}=(-3\hat{i}+2\hat{k})+\lambda(2\hat{i}+3\hat{j}+5\hat{k}), \lambda \in \mathbb{R}$ and $\vec{r}=(\hat{i}-2\hat{j}+\hat{k})+\mu(-\hat{i}+3\hat{j}+2\hat{k}), \mu \in \mathbb{R}$ is

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Three Dimensional Geometry

The distance of the point

Q (0, 2, - 2)

from the line passing through the point

P (5, - 4, 3)

and perpendicular to the lines

r = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k}), λ \in R

and

r = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k}), μ \in R

Answer: (D)

A direction vector for the required line is perpendicular to both given lines' directions,

(2, 3, 5)

and

(- 1, 3, 2)

, so it is along their cross product:

(2, 3, 5) \times (- 1, 3, 2) = (3 (2) - 5 (3), 5 (- 1) - 2 (2), 2 (3) - 3 (- 1)) = (- 9, - 9, 9)

which is along

(1, 1, - 1)

. The required line through

P (5, - 4, 3)

r = (5, - 4, 3) + t (1, 1, - 1)

A general point on it is

(5 + t, - 4 + t, 3 - t)

. The vector from

Q (0, 2, - 2)

to this point is

(5 + t, t - 6, 5 - t)

. For the foot of the perpendicular, this vector must be perpendicular to the line's direction

(1, 1, - 1)

(5 + t) (1) + (t - 6) (1) + (5 - t) (- 1) = 0 ⟹ 3 t - 6 = 0 ⟹ t = 2

Foot of perpendicular:

(7, - 2, 1)

. Distance from

Q (0, 2, - 2)

QF = (7 - 0)^{2} + (- 2 - 2)^{2} + (1 - (- 2))^{2} = 49 + 16 + 9 = 74

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option