Let $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}+1}{0}$ and $\mathrm{L}_{2}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}}{0}=\frac{\mathrm{z}+4}{\alpha}, \alpha \in \mathrm{R}$, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point $\mathrm{A}(1,1,-1)$ on $\mathrm{L}_{2}$, then the value of $26 \alpha(\mathrm{~PB})^{2}$ is ______.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Three Dimensional Geometry

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)

L_{2}

, then the value of

26 α (PB)^{2}

is ______.

Answer: 216

L_{1}

has points

(1 + 3 λ, 1 - λ, - 1)

and

L_{2}

has points

(2 + 2 μ, 0, - 4 + α μ)

.

Setting these equal at the intersection point

B

1 - λ = 0 ⟹ λ = 1.

1 + 3 λ = 2 + 2 μ ⟹ 4 = 2 + 2 μ ⟹ μ = 1.

- 1 = - 4 + α μ ⟹ α = 3.

B = (4, 0, - 1)

.

Now find

P

, the foot of perpendicular from

A (1, 1, - 1)

L_{2}

(with

α = 3

, direction

(2, 0, 3)

, base point

(2, 0, - 4)

).

Vector from base point to

A

(1 - 2, 1 - 0, - 1 + 4) = (- 1, 1, 3)

.

Projection parameter:

t = \frac{( - 1 , 1 , 3 ) \cdot ( 2 , 0 , 3 )}{2 ^{2} + 0 ^{2} + 3 ^{2}} = \frac{- 2 + 0 + 9}{13} = \frac{7}{13} .

P = (2, 0, - 4) + t (2, 0, 3) = (2 + \frac{14}{13}, 0, - 4 + \frac{21}{13}) = (\frac{40}{13}, 0, - \frac{31}{13}) .

P B = B - P = (4 - \frac{40}{13}, 0, - 1 + \frac{31}{13}) = (\frac{12}{13}, 0, \frac{18}{13}) .

(P B)^{2} = (\frac{12}{13})^{2} + (\frac{18}{13})^{2} = \frac{144 + 324}{169} = \frac{468}{169} = \frac{36}{13} .

26 α (P B)^{2} = 26 \times 3 \times \frac{36}{13} = 78 \times \frac{36}{13} = 6 \times 36 = 216.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer