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Question: If the given planes \(ax + by + cz + d = 0\) and \(a'x + b'y + c'z + d' = 0\) be mutually perpendic...

If the given planes ax+by+cz+d=0ax + by + cz + d = 0 and

ax+by+cz+d=0a'x + b'y + c'z + d' = 0 be mutually perpendicular, then

A

aa=bb=cc\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}

B

aa+bb+cc=0\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} = 0

C

aa+bb+cc+dd=0aa' + bb' + cc' + dd' = 0

D

aa+bb+cc=0aa' + bb' + cc' = 0

Answer

aa+bb+cc=0aa' + bb' + cc' = 0

Explanation

Solution

It is a fundamental concept.