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 eq

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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 $\overline{a}$. $\frac{\overline{b} \cdot \overline{a}}{|\overline{a}|} = \frac{\overline{c} \cdot \overline{a}}{|\overline{a}|} \Rightarrow \overline{b} \cdot \overline{a} = \overline{c} \cdot \overline{a} \Rightarrow (\overline{b} - \overline{c}) \cdot \overline{a} = 0 \dots (i)$. Also,$\overline{b}$ is perpendicular to $\overline{c}$,so $\overline{b} \cdot \overline{c} = 0$. Now,$|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b} - \overline{c}|^2 + 2 \overline{a} \cdot (\overline{b} - \overline{c})$. Using $(i)$,$2 \overline{a} \cdot (\overline{b} - \overline{c}) = 0$. So,$|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 - 2(\overline{b} \cdot \overline{c})$. Substituting the values: $|\overline{a} + \overline{b} - \overline{c}|^2 = (2)^2 + (4)^2 + (4)^2 - 2(0) = 4 + 16 + 16 = 36$. Therefore,$|\overline{a} + \overline{b} - \overline{c}| = \sqrt{36} = 6$.

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