First slide

Planes in 3D

Question

The image of the point ( -1,3,4) in the plane x-2y -0 is

Moderate

A

$(- \frac{17}{3}, - \frac{19}{3}, 4)$
B

$(15, 11, 4)$
C

$(- \frac{17}{3}, - \frac{19}{3}, 1)$
D

$(\frac{9}{5}, - \frac{13}{5}, 4)$

Solution

$Let P (α, β, γ) be the image of the point Q (- 1, 3, 4) .$
$Midpoint of P Q lies on x - 2 y = 0 . Then,$
$\begin{array}{l} \frac{α - 1}{2} - 2 (\frac{β + 3}{2}) = 0 \\ o r α - 1 - 2 β - 6 = 0 or α - 2 β = 7 - - - - (i) \end{array}$
Also PQ rs perpendicular to the plane. Then,
$\frac{α + 1}{1} = \frac{β - 3}{- 2} = \frac{γ - 4}{0} - - - - - (i i)$
Solving (i) and (ii), we get
$α = \frac{9}{5}, β = - \frac{13}{5}, γ = 4$
Therefore, image is
$(\frac{9}{5}, - \frac{13}{5}, 4)$
Alternate method:
For image,
$\frac{α - (- 1)}{1} = \frac{β - 3}{- 2} = \frac{γ - 4}{0} = \frac{- 2 (- 1 - 2 (3))}{(1)^{2} + (- 2)^{2}}$
$or α = \frac{9}{5}, β = - \frac{13}{5}, γ = 4$

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