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=p→a+q→b+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 Dq2=4cos2θ9 Here |→a|=2,|→b|=3 and |→c|=2 ⇒cosθ=→a.→c4=→b.→c6 →c=p→a+q→b+r(→a×→b)⋯(1)
Taking dot product with →a on both sides in equation (1) ⇒→a.→c=p(→a.→a)+q(→a.→b)+r→a.(→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)+r→b.(→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)