Question

The d.r.s of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle `pi/4` with plane x + y = 3, are ______.

Options

• `1, sqrt(2), 1`

• `1, 1, sqrt(2)`

• 1, 1, 2

• `sqrt(2), 1, 1`

Answer 1

The d.r.s of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle `pi/4` with plane x + y = 3, are `1, 1, sqrt(2)`.

Explanation:

Let the d.r.s of the normal to the plane be proportional to a, b, c.

It passes through (1, 0, 0)

∴ The equation of the plane is

a(x – 1)+ b(y – 0) + c(z – 0) = 0   ......(i)

Also, the plane passes through (0, 1, 0).

∴ a(–1) + b(1) + c(0) = 0

⇒ a = b   ......(ii)

Now, the angle between the required plane and the plane x + y = 3 is `pi/4`.

∴ `cos  pi/4 = ("a"(1) + "b"(1) + "c"(0))/(sqrt("a"^2 + "b"^2 + "c"^2) sqrt(1 + 1))`

⇒ `1/sqrt(2) = ("a" + "b")/(sqrt("a"^2 + "b"^2 + "c"^2) sqrt(2))`

Squaring both sides, we get

⇒ a² + b² + c² = a² + b² + 2ab

⇒ c² = 2ab   ......(iii)

From (ii) and (iii), we get

a : b : c = `"a" : "a" : sqrt(2)"a" = 1 : 1 : sqrt(2)`