Question

Let the unit vectors a andb be perpendicular and the unit vector c be inclined at an angle $\theta$ to both a and b. If $\mathbf { c } = \alpha \mathbf { a } + \beta \mathbf { b } + \gamma ( \mathbf { a } \times \mathbf { b } )$, then

• A. $\alpha = \beta = \cos \theta$,$\gamma ^ { 2 } = \cos 2 \theta$
• B. $\alpha = \beta = \cos \theta , \gamma ^ { 2 } = - \cos 2 \theta$
• C. $\alpha = \cos \theta , \beta = \sin \theta , \gamma ^ { 2 } = \cos 2 \theta$
• D. None of these

Answer 1

Answer: (B)

$\alpha = \beta = \cos \theta , \gamma ^ { 2 } = - \cos 2 \theta$

Answer 2

We have, $| \mathbf { a } | = | \mathbf { b } | = 1$

$\mathbf { a } \cdot \mathbf { b } = 0$; (as

$\mathbf { c } = \alpha \mathbf { a } + \beta \mathbf { b } + \gamma ( \mathbf { a } \times \mathbf { b } )$ ......(i)

Taking dot product by $\mathbf { a }$,

̃ $| \mathbf { a } | . | \mathbf { c } | \cos \theta = \alpha .1 + 0 + 0$ ̃ 1.| $\mathbf { c } \mid \cdot \cos \theta = \alpha$

As $| \mathbf { c } | = 1$ ; $\therefore$ $\alpha = \cos \theta$

Taking dot product of (i) by $\mathbf { b }$

$\mathbf { b } \cdot \mathbf { c } = \mathbf { b } \cdot \mathbf { a } + \beta | \mathbf { b } | ^ { 2 } + \gamma [ \mathbf { b } \mathbf { a } \mathbf { a } \mathbf { b } ]$

̃

$\therefore$ $\beta = 1.1 \cdot \cos \theta = \cos \theta$

$| \mathbf { c } | ^ { 2 } = 1$ ̃ $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = 1$

̃ $\cos ^ { 2 } \theta + \cos ^ { 2 } \theta + \gamma ^ { 2 } = 1$

$\therefore$ $\gamma ^ { 2 } = 1 - 2 \cos ^ { 2 } \theta = - \cos 2 \theta$

Hence, $\alpha = \beta = \cos \theta , \gamma ^ { 2 } = - \cos 2 \theta$