If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that…

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})+\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})+\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0$ and $|\overline{\mathrm{a}}|=1$, $|\overline{\mathrm{b}}|=8$ and $|\overline{\mathrm{c}}|=4$, then $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|$ has the value
  1. $81$
  2. $9$
  3. $5$
  4. $4$

Solution

$\begin{aligned} & |\bar{a}|=1,|\bar{b}|=8,|\bar{c}|=4, \text { and } \\ & \bar{a} \cdot(\bar{b}+\bar{c})+\bar{b} \cdot(\bar{c}+\bar{a})+\bar{c} \cdot(\bar{a}+\bar{b})=0 \\ & \Rightarrow 2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=0 \end{aligned}$ Now, $\begin{aligned} & |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1+64+16+0 \quad \ldots[\text { From (i) }] \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=81 \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=9 \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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