JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Three Dimensional Geometry

Let

P_{1}

be the plane

3 x - y - 7 z = 11

and

P_{2}

be the plane passing through the points

(2, - 1, 0)

(2, 0, - 1)

, and

(5, 1, 1)

. If the foot of the perpendicular drawn from the point

(7, 4, - 1)

on the line of intersection of the planes

P_{1}

and

P_{2}

(α, β, γ)

, then

α + β + γ

is equal to ______.

Answer: 11

Solution

First, find the equation of plane

P_{2}

using the determinant form with the three given points. This gives

P_{2} : x - y - z = 3

.

The normal vectors are

n_{1} = (3, - 1, - 7)

and

n_{2} = (1, - 1, - 1)

.

The direction vector of the line of intersection is:

n = n_{1} \times n_{2} = (6, 4, 2)

To find a point on the line, set

z = 0

in both plane equations and solve:

x = 4

y = 1

.

The line of intersection is:

\frac{x - 4}{6} = \frac{y - 1}{4} = \frac{z - 0}{2} = λ

Any point on the line is

F = (6 λ + 4, 4 λ + 1, 2 λ)

.

For F to be the foot of perpendicular from

A = (7, 4, - 1)

A F \cdot n = 0

(6 λ - 3, 4 λ - 3, 2 λ + 1) \cdot (6, 4, 2) = 0

36 λ - 18 + 16 λ - 12 + 4 λ + 2 = 0

56 λ = 28 \Rightarrow λ = \frac{1}{2}

Thus

F = (7, 3, 1)

, and

α + β + γ = 11

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer