What are the direction ratios of the normal to the plane ?

Concept:

The equation to the plane passing through P(x₁, y₁, z₁) and having direction ratios (a, b, c) for its normal is

a(x - x₁) + b(y - y₁) + c (z - z₁) = 0.

Calculation:

Since the plane \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\) pass through the point (2, 3, -6);

it satisfies the given equation.

⇒ \(\displaystyle \frac{2\times 2}{k} + \frac{2\times3}{3} + \frac{(-6)}{3} = 2\)

⇒ k = 2

The given equation of the plane is \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\)

Putting the value of k = 2, we get,

\(\displaystyle \frac{2x}{2} + \frac{2y}{3} + \frac{z}{3} = 2\)

⇒ 3x + 2y + z = 6

The direction ratios of normal to the plane are <3, 2, 1>

∴ The direction ratios of normal to the plane are <3, 2, 1>