Let the unit vectors →a and →b be perpendicular to each other and the unit vector →c be inclined at an angle θ to both →a and →b. If →c=x→a+y→b+z(→a×→b), then
A
x=cosθ,y=sinθ,z=cos2θ
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B
x=sinθ,y=cosθ,z=−cos2θ
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C
x=y=cosθ,z2=cos2θ
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D
x=y=cosθ,z2=−cos2θ
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Solution
The correct option is Dx=y=cosθ,z2=−cos2θ We have c=ax+by+z(a×b)
⇒c.a=x
and c.b=y⇒x=y=cosθ
Now, c.c=|c|2
⇒(ax+by+z(a×b)).(ax+by+z(a×b))=|c|2
⇒2x2+z2{|a|2|b|2−(a.b)2}=1
⇒2x2+z2{1−0}=1 since a and b are perpendicular to each other.