Let $\overline{\mathrm{A}}, \overline{\mathrm{B}}, \overline{\mathrm{C}}$ be vectors of lengths 3 units, 4…

Let $\overline{\mathrm{A}}, \overline{\mathrm{B}}, \overline{\mathrm{C}}$ be vectors of lengths 3 units, 4 units, 5 units respectively. let $\bar{A}$ be perpendicular to $\overline{\mathrm{B}}+\overline{\mathrm{C}}, \overline{\mathrm{B}}$ be perpendicular to $\overline{\mathrm{C}}+\overline{\mathrm{A}}$ and $\overline{\mathrm{C}}$ be perpendicular to $\overline{\mathrm{A}}+\overline{\mathrm{B}}$, then the length of vector $\overline{\mathrm{A}}+\overline{\mathrm{B}}+\overline{\mathrm{C}}$ is
  1. $2 \sqrt{5}$
  2. $\sqrt{30}$
  3. $\sqrt{45}$
  4. $5 \sqrt{2}$

Solution

$\begin{aligned} & |\overline{\mathrm{a}}|=3,|\overline{\mathrm{~b}}|=4 \text { and }|\overrightarrow{\mathrm{c}}|=5 \\ & \overline{\mathrm{a}} \cdot(\overline{\mathrm{~b}}+\overline{\mathrm{c}})=0, \overline{\mathrm{~b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})=0 \text { and } \overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0 \\ & \Rightarrow 2(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \end{aligned}$
Now, $\begin{aligned} & |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=9+16+25+0 \quad \ldots[\text { From }(\mathrm{i}) \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=50 \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=5 \sqrt{2} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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