If the mirror image of the point ( $1,3,5$ ) with respect to the plane $4 x-5 y+2 z=8$ is ( $\alpha, \beta, \gamma$ ) then $5(\alpha+\beta+\gamma)$ equals :

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Three Dimensional Geometry

If the mirror image of the point (

1, 3, 5

) with respect to the plane

4 x - 5 y + 2 z = 8

is (

α, β, γ

) then

5 (α + β + γ)

equals :

(A)41
(B)39
(C)47✓ Correct
(D)43

Answer: (C)

Solution

The plane

4 x - 5 y + 2 z = 8

has normal

(4, - 5, 2)

, and at the point

(1, 3, 5)

4 (1) - 5 (3) + 2 (5) - 8 = - 9, 4^{2} + (- 5)^{2} + 2^{2} = 45

The image is reached by stepping twice the signed perpendicular distance along the normal:

(α, β, γ) = (1, 3, 5) - \frac{2 ( - 9 )}{45} (4, - 5, 2) = (1, 3, 5) + \frac{2}{5} (4, - 5, 2) = (\frac{13}{5}, 1, \frac{29}{5})

Then

α + β + γ = \frac{13}{5} + 1 + \frac{29}{5} = \frac{47}{5}

, so

5 (α + β + γ) = 47

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option

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