Let the vectors overline{a}, overline{b}, ove | vedclass

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(B) Given: $|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=4$.According to the condition,the projection of $\overline{b}$ on $\overline{a}$ is equ

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$6$

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(B) Given: $|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=4$.According to the condition,the projection of $\overline{b}$ on $\overline{a}$ is equal to the projection of $\overline{c}$ on $\overline{a}$.$\frac{\overline{b} \cdot \overline{a}}{|\overline{a}|} = \frac{\overline{c} \cdot \overline{a}}{|\overline{a}|}$$\implies \overline{b} \cdot \overline{a} = \overline{c} \cdot \overline{a}$$\implies (\overline{b} - \overline{c}) \cdot \overline{a} = 0$ ... $(i)$Also,$\overline{b}$ is perpendicular to $\overline{c}$,so $\overline{b} \cdot \overline{c} = 0$.Now,consider $|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b} - \overline{c}|^2 + 2\overline{a} \cdot (\overline{b} - \overline{c})$.From $(i)$,$\overline{a} \cdot (\overline{b} - \overline{c}) = 0$.So,$|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 - 2(\overline{b} \cdot \overline{c})$.Since $\overline{b} \cdot \overline{c} = 0$,we have:$|\overline{a} + \overline{b} - \overline{c}|^2 = (2)^2 + (4)^2 + (4)^2 - 0 = 4 + 16 + 16 = 36$.Therefore,$|\overline{a} + \overline{b} - \overline{c}| = \sqrt{36} = 6$.

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