Let the vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be such that…

Let the vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be such that $|\overline{\mathrm{a}}|=2,|\dot{\bar{b}}|=4$ and $|\overline{\mathrm{c}}|=4$. If the projection of $\overline{\mathrm{b}}$ on $\overline{\mathrm{a}}$ is equal to the projection of $\overline{\mathrm{c}}$ on $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then the value of $|\overline{\mathrm{a}}+\overline{\mathrm{b}}-\overline{\bar{c}}|$ is equal to
  1. $2 \sqrt{5}$
  2. 6
  3. 4
  4. $4 \sqrt{2}$

Solution

$||\bar{a}|=2,|\bar{b}|=4 \text { and }| \bar{c} \mid=4$
According to the given condition, $($ Projection of $\bar{b}$ on $\bar{a})=($ Projection of $\bar{c}$ on $\bar{a}$ ) $\begin{array}{ll} \therefore & \frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}=\frac{\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}}{|\overline{\mathrm{a}}|} \\ \therefore & \overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{a}} \\ \therefore & (\overline{\mathrm{~b}}-\overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}=0 ...(i)\end{array}$
Now consider, $|\bar{a}+\bar{b}-\bar{c}|$ $\begin{aligned} & =\sqrt{|\bar{a}+\bar{b}-\bar{c}|^2} \\ & =\sqrt{|\bar{a}|^2+|\bar{b}-\bar{c}|^2+2 \bar{a} \cdot(\bar{b}-\bar{c})} \\ & =\sqrt{(2)^2+|\bar{b}-\bar{c}|^2+0} \end{aligned}$...[from(ii)] $\begin{aligned} & =\sqrt{4+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2-2(\overline{\mathrm{~b}} \cdot \overline{\mathrm{c}})} \\ & =\sqrt{4+(4)^2+(4)^2+0} \end{aligned}$ $\ldots[\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are perpendicular $]$ $\begin{aligned} & =\sqrt{36} \\ & =6 \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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