Let the image of the point $P(1,2,3)$ in the plane $2x-y+z=9$ be $Q$. If the coordinates of the point $R$ are $(6,10,7)$, then the square of the area of the triangle PQR is

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Three Dimensional Geometry

Let the image of the point

P (1, 2, 3)

in the plane

2 x - y + z = 9

Q

. If the coordinates of the point

R

are

(6, 10, 7)

, then the square of the area of the triangle PQR is

Answer: 594

To find the image Q of point P(1,2,3) in the plane

2 x - y + z = 9

:

The normal vector to the plane is

n = (2, - 1, 1)

.

The line through P perpendicular to the plane is:

\frac{x - 1}{2} = \frac{y - 2}{- 1} = \frac{z - 3}{1} = λ

The foot of the perpendicular lies on the plane. Substituting into the plane equation and solving gives

λ = 1

.

The image Q is at

λ = 2

Q (5, 0, 5)

.

Now,

P Q = (4, - 2, 2)

and

P R = (5, 8, 4)

.

The area of triangle PQR is:

Area = \frac{1}{2} ∣ P Q \times P R ∣

P Q \times P R = \hat{i} 45 \hat{j} - 2 8 \hat{k} 24 = - 24 \hat{i} - 6 \hat{j} + 42 \hat{k}

∣ P Q \times P R ∣ = 576 + 36 + 1764 = 2376 = 2594

Area

= \frac{1}{2} \cdot 2594 = 594

Square of area

= 594

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer