If the projection of bar{a} on bar{b}+bar{c} | vedclass

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(C) Let $\bar{d} = \bar{b} + \bar{c}$. The projection of $\bar{a}$ on $\bar{d}$ is given by $\frac{\bar{a} \cdot \bar{d}}{|\bar{d}|}$.The projection o

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$4 \sqrt{10}$

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(C) Let $\bar{d} = \bar{b} + \bar{c}$. The projection of $\bar{a}$ on $\bar{d}$ is given by $\frac{\bar{a} \cdot \bar{d}}{|\bar{d}|}$.The projection of $\bar{d}$ on $\bar{a}$ is given by $\frac{\bar{d} \cdot \bar{a}}{|\bar{a}|}$.According to the problem,$\frac{\bar{a} \cdot \bar{d}}{|\bar{d}|} = 2 	imes \frac{\bar{d} \cdot \bar{a}}{|\bar{a}|}$.Since $\bar{a} \cdot \bar{d} = \bar{d} \cdot \bar{a}$,we can cancel this term (assuming $\bar{a} \cdot \bar{d} 
eq 0$),leading to $\frac{1}{|\bar{d}|} = \frac{2}{|\bar{a}|}$,which implies $|\bar{a}| = 2|\bar{d}|$.Now,calculate $|\bar{d}|^2 = |\bar{b} + \bar{c}|^2 = |\bar{b}|^2 + |\bar{c}|^2 + 2|\bar{b}||\bar{c}| \cos(\frac{\pi}{4})$.$|\bar{d}|^2 = (2 \sqrt{2})^2 + 4^2 + 2(2 \sqrt{2})(4) \frac{1}{\sqrt{2}} = 8 + 16 + 16 = 40$.Thus,$|\bar{d}| = \sqrt{40} = 2 \sqrt{10}$.Finally,$|\bar{a}| = 2|\bar{d}| = 2(2 \sqrt{10}) = 4 \sqrt{10}$.

If the projection of $\bar{a}$ on $\bar{b}+\bar{c}$ is twice the projection of $\bar{b}+\bar{c}$ on $\bar{a}$,and if $|\bar{b}|=2 \sqrt{2}$,$|\bar{c}|=4$,and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$,then $|\bar{a}|=$

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