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 Let a¯,b¯ and c¯ be the unit vectors such that a¯ and b¯ are mutually perpendicular and c¯ is  equally inclined to a¯ and b¯ at angle θ . If c¯=xa¯+yb¯+z(a¯×b¯), then 

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a
z2=1−2x2
b
z2=1−x2+y2
c
z2=1+2y2
d
z2=1+2x2

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detailed solution

Correct option is A

Given that   a¯=b¯=c¯=1 and,  a¯.b¯=0⇒a¯,b¯=90∘As   c¯=xa¯+y b¯+z a¯×b¯ ⇒a¯c¯cosθ=x+y0+z0       by  taking   dot product with a   ⇒cosθ=x  Similarly,  cosθ=yc¯=xa¯+y b¯+z a¯×b¯  Squaring on both sidesc¯2=x2a−2+y2b−2+z2a¯×b¯2+2xya¯.b¯+2yz b¯.a¯×b¯+2xz a¯.a¯×b¯1=x2+y2+z2+0+0+0 ⇒1=cos2θ+cos2θ+z2 ⇒1−2cos2θ=z2 ⇒1−2x2=z2∴z2=1−2x2


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