JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let the point $A$ be the foot of perpendicular drawn from the point $P (a, b, 0)$ on the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - α}{3}$ . If the midpoint of the line segment $P A$ is $(0, \frac{3}{4}, \frac{- 1}{4})$ , then the value of $a^{2} + b^{2} + α^{2}$ is equal to :

A
$1$
B
$2$
C
$6$
D
$9$

Solution

Parametrise $A$ on the line as $A = (2 r + 1, r + 2, 3 r + α)$ for some $r$ .

Midpoint of $P A$ with $P = (a, b, 0)$ gives the three equations: $\frac{2 r + 1 + a}{2} = 0$ , $\frac{r + 2 + b}{2} = \frac{3}{4}$ , and $\frac{3 r + α}{2} = - \frac{1}{4}$ .

These simplify to $2 r + a = - 1$ , $r + b = - \frac{1}{2}$ , and $3 r + α = - \frac{1}{2}$ .

Perpendicularity of $P A$ with the line's direction $(2, 1, 3)$ provides the fourth condition. Combining with the midpoint relations yields $a + b = 0$ , and then $r = - \frac{1}{2}$ , giving $a = 0$ , $b = 0$ , $α = 1$ .

$∴ a^{2} + b^{2} + α^{2} = 1$ .

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