Q. 4 C5.0( 1 Vote )

Determine the value of λ for which the following planes are perpendicular to each other.

3x – 6y – 2z = 7 and 2x + y – λz = 5

Answer :

Given planes, 3x – 6y – 2z = 7 and 2x + y – λz = 5


We know that planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are at right angles if,


a1a2 + b1b2 + c1c2 = 0 ……(a)


We have, a1 = 3, b1 = – 6, c1 = – 2 and a2 = 2, b2 = 1, c2 = – λ


Using (a) we have,


a1a2 + b1b2 + c1c2 = (3)(2) + (– 6)(1) + (– 2)(– λ) = 0


6 – 6 + 2λ = 0


0 = – 2λ


0 = λ


For λ = 0 the given planes are perpendicular to each other.


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