bar{a}, bar{b}, bar{c} are unit vectors. If b | vedclass

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(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,so $|\bar{a}|=1, |\bar{b}|=1, |\bar{c}|=1$. Since $\bar{a} \perp \bar{b}$,we have $\bar{a}

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$1-l^2-m^2$

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(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,so $|\bar{a}|=1, |\bar{b}|=1, |\bar{c}|=1$. Since $\bar{a} \perp \bar{b}$,we have $\bar{a} \cdot \bar{b} = 0$. Given $(\bar{a}-\bar{c}) \cdot (\bar{b}+\bar{c}) = 0$. Expanding this,we get $\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} - \bar{c} \cdot \bar{b} - |\bar{c}|^2 = 0$. Since $\bar{a} \cdot \bar{b} = 0$ and $|\bar{c}|^2 = 1$,we have $\bar{a} \cdot \bar{c} - \bar{b} \cdot \bar{c} = 1$. Given $\bar{c} = l\bar{a} + m\bar{b} + n(\bar{a} 	imes \bar{b})$. Taking dot product with $\bar{a}$: $\bar{c} \cdot \bar{a} = l(\bar{a} \cdot \bar{a}) + m(\bar{b} \cdot \bar{a}) + n(\bar{a} 	imes \bar{b}) \cdot \bar{a} = l(1) + m(0) + 0 = l$. Taking dot product with $\bar{b}$: $\bar{c} \cdot \bar{b} = l(\bar{a} \cdot \bar{b}) + m(\bar{b} \cdot \bar{b}) + n(\bar{a} 	imes \bar{b}) \cdot \bar{b} = l(0) + m(1) + 0 = m$. Substituting these into $\bar{a} \cdot \bar{c} - \bar{b} \cdot \bar{c} = 1$,we get $l - m = 1$. Also,$|\bar{c}|^2 = 1 \implies (l\bar{a} + m\bar{b} + n(\bar{a} 	imes \bar{b})) \cdot (l\bar{a} + m\bar{b} + n(\bar{a} 	imes \bar{b})) = 1$. Since $\bar{a}, \bar{b}, \bar{a} 	imes \bar{b}$ are mutually orthogonal,$|\bar{a} 	imes \bar{b}| = |\bar{a}||\bar{b}|\sin(90^{\circ}) = 1$. Thus,$l^2 + m^2 + n^2(1)^2 = 1$,which means $n^2 = 1 - l^2 - m^2$.

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

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