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(C) Given that the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$.Since $|\overli

Vedclass · Accepted Answer

$\sqrt{14}$

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(C) Given that the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$.Since $|\overline{u}|=1$,the projection formula $\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|}$ simplifies to $\overline{v} \cdot \overline{u}$.Thus,$\overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u} \Rightarrow (\overline{v} - \overline{w}) \cdot \overline{u} = 0 \dots (i)$.Also,$\overline{v}$ and $\overline{w}$ are perpendicular,so $\overline{v} \cdot \overline{w} = 0 \dots (ii)$.We need to find $|\overline{u} - \overline{v} + \overline{w}|^2$.$|\overline{u} - \overline{v} + \overline{w}|^2 = |\overline{u}|^2 + |-\overline{v}|^2 + |\overline{w}|^2 - 2(\overline{u} \cdot \overline{v}) + 2(\overline{u} \cdot \overline{w}) - 2(\overline{v} \cdot \overline{w})$.Substituting the values $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$ and using $(i)$ and $(ii)$:$|\overline{u} - \overline{v} + \overline{w}|^2 = 1^2 + 2^2 + 3^2 - 2(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) - 2(0)$.Since $\overline{u} \cdot \overline{v} = \overline{u} \cdot \overline{w}$,the term $2(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) = 0$.Therefore,$|\overline{u} - \overline{v} + \overline{w}|^2 = 1 + 4 + 9 = 14$.Hence,$|\overline{u} - \overline{v} + \overline{w}| = \sqrt{14}$.

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