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(A) Given that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$.$\Rightarrow \frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} = \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} \Rightarrow \overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a}$.Also,$\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,so $\overrightarrow{b} \cdot \overrightarrow{c} = 0$.Let $k = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|$.Squaring both sides,we get:$k^2 = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) - 2(\overrightarrow{a} \cdot \overrightarrow{c}) - 2(\overrightarrow{b} \cdot \overrightarrow{c})$.Since $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{a} \cdot \overrightarrow{c}$ and $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,the terms cancel out:$k^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{a} \cdot \overrightarrow{c}) - 2(0) = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2$.Substituting the given magnitudes $|\overrightarrow{a}|=2, |\overrightarrow{b}|=4, |\overrightarrow{c}|=4$:$k^2 = 2^2 + 4^2 + 4^2 = 4 + 16 + 16 = 36$.Therefore,$k = \sqrt{36} = 6$.

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