Question

If vectors P=3i-aj+2k and Q=2i+j-2k are mutually perpendicular, then the value of 'a' is?

Answer 1

howdy dude ,

your answer is ------

Given, P = 3i-aj+2k and Q = 2i+j-2k
now
P.Q = PQ cos theta ...(1)

now P.Q = 3×2+(-a)×1+2×-2
= 6-a-4 = 2-a
=> P.Q= 2-a

put this value in equation (1), we get
2-a = PQcos theta

since , it is given that P and Q are mutually perpendicular to each other

so, 2-a = PQ cos(90°)

=> 2-a = 0. [ since cos(90°) = 0]
=> a = 2


hence , answer is a = 2


hope it help you