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

A.
<3, 2, 1>
B.
<2, 3, 6>
C.
<6, 3, 2>
D.
<1, 2, 3>

Solution:

Concept:The equation to the plane passing through P(x1, y1, z1) and having direction ratios (a, b, c) for its normal isa(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 = 2The 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 = 6The direction ratios of normal to the plane are <3, 2, 1>? The direction ratios of normal to the plane are <3, 2, 1>

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