Q85 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

$\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}$ are three vectors such that $|\overline{\mathrm{u}}|=1$, $|\vec{v}|=2,|\bar{w}|=3$. If the projection of $\bar{v}$ along $\bar{u}$ is equal to projection of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other, then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=$

  • A
    4
  • B
    $\sqrt{7}$
  • Correct answer
    C
    $\sqrt{14}$
  • D
    2

Solution

$$ |\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3 $$ According to the given condition, (Projection of $\bar{v}$ along $\bar{u}$ ) $$ =(\text { Projection of } \bar{w} \text { along } \bar{u}) $$ $$ \begin{array}{ll} \therefore & \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \\ \therefore & \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}} \\ \therefore & (\overline{\mathrm{w}}-\overline{\mathrm{v}}) \cdot \overline{\mathrm{u}}=0 \end{array} $$ Now consider, $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=\sqrt{|\overline{\mathrm{u}}+\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2}$ $$ \begin{aligned} & =\sqrt{|\overline{\mathrm{u}}|^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+2 \overline{\mathrm{u}} \cdot(\overline{\mathrm{w}}-\overline{\mathrm{v}})} \\ & =\sqrt{(1)^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+0} \\ & =\sqrt{1+|\overline{\mathrm{w}}|^2+|\overline{\mathrm{v}}|^2-2(\overline{\mathrm{w}} \cdot \overline{\mathrm{v}})} \\ & =\sqrt{1+9+4+0} \end{aligned} $$ $\ldots[\because \overline{\mathrm{w}}$ and $\overline{\mathrm{v}}$ are perpendicular $]$ $$ =\sqrt{14} $$

Hence, the correct answer is option: 'C'.

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