Question

The equation of the plane passing through the point (1, 2, –3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is ______.

Options

• 3x – 10y – 2z + 11 = 0

• 6x – 5y – 2z – 2 = 0

• 11x + y + 17z + 38 = 0

• 6x – 5y + 2z + 10 = 0

Answer 1

The equation of the plane passing through the point (1, 2, –3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is 11x + y + 17z + 38 = 0.

Explanation:

Given: A plane is passing through point (1, 2, –3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7.

Plane P₁: 3x + y – 2z = 5

Plane P₂: 2x – 5y – z = 7.

Normal to P₁ has direction ratios (3, 1, –2)

Direction ratios of normal to P₂ are (2, –5, –1).

Let `veca = 3hati + hatj - 2hatk`

and `vecb = 2hati - 5hatj - hatk`

The direction ratios of required plane's normal are parallel to `veca xxvecb` i.e.,

`veca xx vecb = |(hati, hatj, hatk),(3, 1, -2),(2, -5, -1)|`

= `hati(-1 - 10)-hatj(-3 + 4) + hatk(-15 - 2)`

= `-11hati - hatj - 17hatk`

So, Direction ratios of normal to required plane are (–11, –1, –17)

Equation of plane passing through (x₁, y₁, z₁) and direction ratios of normal are (a, b, c) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

So, required plane is –11(x – 1) – 1(y – 2) – 17(z + 3) = 0

⇒ –11x + 11 – y + 2 – 17z – 51 = 0

⇒ 11x + y + 17z + 38 = 0