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Question

Two vectors a and b are perpendicular to each other. A vector c is inclined at an angle θ to both a and b. If |a|=2,|b|=3, |c|=2 and c=pa+qb+r(a×b), then

A
p2=cos2θ
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B
r2=cos2θ9
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C
r2=cos2θ9
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D
q2=4cos2θ9
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Solution

The correct option is D q2=4cos2θ9
Here |a|=2,|b|=3 and |c|=2
cosθ=a.c4=b.c6
c=pa+qb+r(a×b) (1)

Taking dot product with a on both sides in equation (1)
a.c=p(a.a)+q(a.b)+ra.(a×b)
4cosθ=4p cosθ=p

Similarly taking dot product with b on both sides in equation (1)
b.c=p(a.b)+q(b.b)+rb.(a×b)
6cosθ=9q
q=23p

Squaring eqn (1), we get
|c|2=p2|a|2+q2|b|2+r2|(a×b)|2 +2pq(a.b)+2pr(a).(a×b)+2qr(b).(a×b)

4=4p2+9q2+r2|(a×b)|2
4=8p2+r2|a|2b|2×1
4=8p2+36r2




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