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(B) Given that $a$ and $b$ are unit vectors and are perpendicular,so $|a| = 1, |b| = 1$ and $a \cdot b = 0$.Since $c$ is a unit vector,$|c| = 1$.The v

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$\alpha = \beta = \cos 	heta, \gamma^2 = -\cos 2	heta$

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(B) Given that $a$ and $b$ are unit vectors and are perpendicular,so $|a| = 1, |b| = 1$ and $a \cdot b = 0$.Since $c$ is a unit vector,$|c| = 1$.The vector $c$ is inclined at an angle $	heta$ to both $a$ and $b$,so $c \cdot a = |c||a| \cos 	heta = \cos 	heta$ and $c \cdot b = |c||b| \cos 	heta = \cos 	heta$.Given $c = \alpha a + \beta b + \gamma (a 	imes b)$.Taking the dot product with $a$: $c \cdot a = \alpha (a \cdot a) + \beta (b \cdot a) + \gamma ((a 	imes b) \cdot a) = \alpha (1) + \beta (0) + 0 = \alpha$. Thus,$\alpha = \cos 	heta$.Taking the dot product with $b$: $c \cdot b = \alpha (a \cdot b) + \beta (b \cdot b) + \gamma ((a 	imes b) \cdot b) = \alpha (0) + \beta (1) + 0 = \beta$. Thus,$\beta = \cos 	heta$.Since $c$ is a unit vector,$c \cdot c = 1$.$c \cdot c = (\alpha a + \beta b + \gamma (a 	imes b)) \cdot (\alpha a + \beta b + \gamma (a 	imes b)) = \alpha^2 |a|^2 + \beta^2 |b|^2 + \gamma^2 |a 	imes b|^2 = 1$.Since $|a 	imes b| = |a||b| \sin 90^\circ = 1$,we have $\alpha^2 + \beta^2 + \gamma^2 = 1$.Substituting $\alpha = \beta = \cos 	heta$,we get $2 \cos^2 	heta + \gamma^2 = 1$.Therefore,$\gamma^2 = 1 - 2 \cos^2 	heta = - (2 \cos^2 	heta - 1) = - \cos 2	heta$.

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

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