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Question

Let a=3^i+2^j+2^kb=^i+2^j2^k Then a unit vector perpendicular to both ab and a+b is

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Solution

a=3^i+2^j+2^k
b=^i+2^j2^k
veca+b=4^i+4^j
vecab=2^i+4^k
unit vector perpendicular to both ab and a+b
c=(ab)×(a+b)
c=∣ ∣ ∣^i^j^k204440∣ ∣ ∣=^i(16)^j(16)+^k(8)
=16^i+16^j+8^k
Now
unit vector indirection of c is
c|c|
|c|=162+162+82=24
Desired unit vector = 124(16^i+16^j+8^k)
=13(2^i+2^j+^k)

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