If unit vectors bar{a} and bar{b} are perpend | vedclass

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(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$. Since they are perpendicular,$\bar{a} \cdot \bar{b} = 0$.

Vedclass · Accepted Answer

$\alpha = \beta = \cos 	heta$ and $r^2 = -\cos 2	heta$

Vedclass · Answer

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$. Since they are perpendicular,$\bar{a} \cdot \bar{b} = 0$.Given $\bar{c}$ is a unit vector,$|\bar{c}| = 1$. The angle between $\bar{c}$ and $\bar{a}$ is $	heta$,so $\bar{c} \cdot \bar{a} = |\bar{c}||\bar{a}| \cos 	heta = \cos 	heta$.Substituting $\bar{c} = \alpha \bar{a} + \beta \bar{b} + r(\bar{a} 	imes \bar{b})$:$\bar{c} \cdot \bar{a} = (\alpha \bar{a} + \beta \bar{b} + r(\bar{a} 	imes \bar{b})) \cdot \bar{a} = \alpha |\bar{a}|^2 + \beta (\bar{b} \cdot \bar{a}) + r((\bar{a} 	imes \bar{b}) \cdot \bar{a}) = \alpha(1) + \beta(0) + r(0) = \alpha$.Thus,$\alpha = \cos 	heta$.Similarly,$\bar{c} \cdot \bar{b} = \beta = \cos 	heta$.Now,$\bar{c} = \cos 	heta \bar{a} + \cos 	heta \bar{b} + r(\bar{a} 	imes \bar{b})$.Since $|\bar{c}|^2 = 1$,we have:$|\cos 	heta \bar{a} + \cos 	heta \bar{b} + r(\bar{a} 	imes \bar{b})|^2 = 1$.Using the property that $\bar{a}, \bar{b}, (\bar{a} 	imes \bar{b})$ are mutually orthogonal:$\cos^2 	heta |\bar{a}|^2 + \cos^2 	heta |\bar{b}|^2 + r^2 |\bar{a} 	imes \bar{b}|^2 = 1$.Since $|\bar{a} 	imes \bar{b}| = |\bar{a}||\bar{b}| \sin 90^\circ = 1$,we get:$\cos^2 	heta + \cos^2 	heta + r^2(1) = 1$.$2 \cos^2 	heta + r^2 = 1$.$r^2 = 1 - 2 \cos^2 	heta = - (2 \cos^2 	heta - 1) = - \cos 2	heta$.

If unit vectors $\bar{a}$ and $\bar{b}$ are perpendicular to each other and a unit vector $\bar{c}$ makes an angle $\theta$ with both $\bar{a}$ and $\bar{b}$,and $\bar{c} = \alpha \bar{a} + \beta \bar{b} + r(\bar{a} \times \bar{b})$,then:

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