Question

Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z = 0.

Answer 1

The equation of the plane passing through the point (-1,3,2) is:
a(x+1) + b(y-3) + c(z-2) = 0     ...(1)

Where a, b, c are the direction ratios of normal to the plane.
It is known that two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0, are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0

Plane (1) is perpendicular to plane, x + 2y + 3z = 5
∴a.1 + b.2 + c.3 = 0 ⇒ a + 2b + 3c = 0         ...(2)

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0
∴a.3 + b.3 + c.3 = 0
⇒ 3a + 3b + c = 0         ...(3)

From equation (2) and (3), we obtain
$\frac {a}{2 × 1-3 × 3}$= $\frac {b}{3 × 3-1 × 1}$ = $\frac {c}{1 × 3 -2 × 3}$

⇒ $\frac {a}{-7} =\frac { b}{8} = \frac {c}{-3} =  k \ (say)$

⇒ $a = -7k, b = 8k, c = -3k$

Substituting the values of a, b, and c in equation (1), we obtain

-7k(x+1) + 8k(y-3) - 3k(z-2) = 0

⇒ (-7x-7) + (8y-24) - 3z + 6 = 0

⇒ -7x + 8y- 3z - 25 = 0

⇒ 7x - 8y + 3z + 25 = 0

This is the required equation of the plane.