Q62 / 150 · MHT CET 2021 (20 Sep Shift 2) · Mathematics

$\bar{a}, \bar{b}, \bar{c}$ are vectors such that $|\bar{a}|=5,|\bar{b}|=4,|\bar{c}|=3$ and each is perpendicular to the sum of the other two, then $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=$

  • A
    60
  • B
    12
  • C
    47
  • Correct answer
    D
    50

Solution

We have $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})=0 \cdot \overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})=0$ and $\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0$ $$ \begin{aligned} \therefore \quad & \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0 \\ & \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0 \\ & \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0 \end{aligned} $$ From (1), (2) and (3), we get $$ \begin{aligned} & 2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \\ & \text { Now }|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & \therefore|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=(5)^2+(4)^2+(3)^2+2(0)=50 \end{aligned} $$

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

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