JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Three Dimensional Geometry

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Q (0, - 1, - 3)

is the image of the point

P

in the plane

3 x - y + 4 z = 2

and

R

is the point

(3, - 1, - 2)

, then the area (in square units) of

△ PQR

is:

(A) $\frac{91}{2}$ ✓ Correct
(B) $213$
(C) $\frac{65}{2}$
(D) $\frac{91}{4}$

Answer: (A)

Solution

Let

M

be the foot of the perpendicular from

Q (0, - 1, - 3)

to the plane

3 x - y + 4 z = 2

. That distance is

M Q = \frac{∣3 ( 0 ) + 1 - 12 - 2∣}{9 + 1 + 16} = \frac{13}{26} = \frac{13}{2}

. Since

P

is the mirror image,

P Q = 2 M Q = 26

. With

R Q = (3 - 0)^{2} + 0 + (- 2 + 3)^{2} = 10

, the perpendicular from

R

to line

P Q

works out to

R M = \frac{7}{2}

, so the area of triangle

P QR

\frac{1}{2} P Q \cdot R M = \frac{1}{2} 26 \frac{7}{2} = \frac{91}{2}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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