Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(1,6,4)$ in the line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}$. Then $2 \alpha+\beta+\gamma$ is equal to

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Three Dimensional Geometry

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

Answer: 11

Solution

Line:

(t, 1 + 2 t, 2 + 3 t)

, direction

b = (1, 2, 3)

. Foot of perpendicular

A

from

Q (1, 6, 4)

Q A = (t - 1, 2 t - 5, 3 t - 2)

Q A \cdot b = 0

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

A = (\frac{17}{14}, \frac{48}{14}, \frac{79}{14})

Image

P = 2 A - Q = (\frac{10}{7}, \frac{6}{7}, \frac{51}{7})

2 α + β + γ = \frac{20 + 6 + 51}{7} = \frac{77}{7} = 11

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer

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