Let vec{a} and vec{b} be two unit vectors suc | vedclass

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(C) Given $|\vec{a}| = |\vec{b}| = 1$ and $\vec{a} \cdot \vec{b} = 0$.Since $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} 	imes \vec{b})$,we have $\vec{c

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$3$

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(C) Given $|\vec{a}| = |\vec{b}| = 1$ and $\vec{a} \cdot \vec{b} = 0$.Since $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} 	imes \vec{b})$,we have $\vec{c} \cdot \vec{a} = x$ and $\vec{c} \cdot \vec{b} = y$.Given that $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$,we have $\vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos \alpha = 2(1) \cos \alpha = 2 \cos \alpha$ and $\vec{c} \cdot \vec{b} = 2 \cos \alpha$.Thus,$x = y = 2 \cos \alpha$.Now,$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (x\vec{a} + y\vec{b} + (\vec{a} 	imes \vec{b})) \cdot (x\vec{a} + y\vec{b} + (\vec{a} 	imes \vec{b}))$.Since $\vec{a}, \vec{b}, \vec{a} 	imes \vec{b}$ are mutually orthogonal,$|\vec{c}|^2 = x^2 + y^2 + |\vec{a} 	imes \vec{b}|^2$.Since $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 90^{\circ} = 1$,we have $4 = x^2 + y^2 + 1$.Substituting $x = y = 2 \cos \alpha$,we get $4 = (2 \cos \alpha)^2 + (2 \cos \alpha)^2 + 1$.$4 = 4 \cos^2 \alpha + 4 \cos^2 \alpha + 1$.$3 = 8 \cos^2 \alpha$.Thus,the value of $8 \cos^2 \alpha$ is $3$.

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