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 $|\bar{a}|=2 ;|\vec{b}|=4$ and $|\bar{c}|=4$. If the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\bar{c}$ on $\bar{a}$ and $\bar{b}$ is perpendicular to $\bar{c}$, then the value of $|\bar{a}+\bar{b}-\bar{c}|$ is
  1. $2 \sqrt{5}$
  2. 6
  3. 4
  4. $4 \sqrt{2}$

Solution

$|\overrightarrow{\mathrm{a}}|=2,|\overline{\mathrm{~b}}|=4,|\overrightarrow{\mathrm{c}}|=4$
According to the given condition, Projection of $\overline{\mathrm{b}}$ on $\overline{\mathrm{a}}=$ projection of $\overline{\mathrm{c}}$ on $\overline{\mathrm{a}}$ $\begin{aligned} & \Rightarrow \frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}}{|\overline{\mathrm{a}}|}=\frac{\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}}{|\overline{\mathrm{a}}|} \\ & \Rightarrow \overline{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{a}} \\ & \Rightarrow(\overline{\mathrm{~b}}-\overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}=0...(i) \end{aligned}$
Now, $|\bar{a}+\bar{b}-\bar{c}|=\sqrt{|\bar{a}+\bar{b}-\bar{c}|^2}$ $=\sqrt{|\bar{a}|^2+|\bar{b}-\bar{c}|^2+2 \bar{a} \cdot(\bar{b}-\bar{c})}$ $=\sqrt{(2)^2+|\bar{b}-\bar{c}|^2+2(0)}$ ...[From (i)] $=\sqrt{4+|\vec{b}-\bar{c}|^2}$ $=\sqrt{4+|\bar{b}|^2+|\bar{c}|^2-2(\bar{b} \cdot \bar{c})}$ $\begin{aligned} & =\sqrt{4+4^2+4^2-2(0)} \\ & =\sqrt{36}=6 \end{aligned}$ ...[ $\bar{b}$ is perpendicular to $\bar{c}]$

Asked in: MHT CET 2024 (16 May Shift 1)

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