Question

Let the unit vectors $A$ and $B$ be perpendicular and the unit vector $C$ be inclined at $\theta$ to both $A$ and $B$. If $C=\alpha A+\beta B+\gamma (A\times B)$ then

• A. $\alpha =\beta$
• B. ${\gamma }^2=1-2{\alpha }^2$
• C. ${\gamma }^2=-\cos 2\theta$
• D. ${\beta }^2=\frac{1+\cos 2\theta }{2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A ${\gamma }^2=1-2{\alpha }^2$
B $\alpha =\beta$
C ${\beta }^2=\frac{1+\cos 2\theta }{2}$
D ${\gamma }^2=-\cos 2\theta$

Since they are unit vectors, we have $\left|A\right|=\left|B\right|=1\left|C\right|$.
It is also given that the angle $\theta$ between $a$ and $c$ equals that between $b$ and $c$
i.e $\frac{A.B}{\left|A\right|\left|C\right|}=A.C=\cos \theta =\frac{B.C}{\left|B\right|\left|C\right|}=B.C$
Since $A.B=0$, we get from the given value of $C$
$A.C=\alpha A.A+\beta A.B+\gamma A.(A\times B)=\alpha$
i.e $\alpha =\cos \theta$ and similarly $B.C=\cos \theta =\beta$
Thus $\alpha =\beta =\cos \theta$ so option (A) is correct
Next we have
$1=C.C={2\alpha }^2+{\gamma }^2{|A\times B|}^2={2\alpha }^2+[{\left|A\right|}^2{\left|B\right|}^2-{(A\times B)}^2]={2\alpha }^2+{\gamma }^2\Rightarrow {\gamma }^2=1-{2\alpha }^2=1-2{\cos }^2\theta =-\cos \theta$
$\Rightarrow {\alpha }^2={\beta }^2=\frac{1-{\gamma }^2}{2}=\frac{1+\cos 2\theta }{2}$
Which proves (B),(C) and (D)