Q82 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics

If $\bar{x}=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \bar{y}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ and $\overline{\mathrm{z}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ where $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar vectors, then value of $\bar{x} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})+\bar{y} \cdot(\overline{\mathrm{~b}}+\overline{\mathrm{c}})+\overline{\mathrm{z}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})$ is

  • Correct answer
    A
    3
  • B
    1
  • C
    -1
  • D
    0

Solution

$\begin{aligned} & \bar{x} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})+\bar{y} \cdot(\overline{\mathrm{~b}}+\overline{\mathrm{c}})+\overline{\mathrm{z}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}}) \\ = & (\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot \bar{x}+(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot \bar{y}+(\overline{\mathrm{c}}+\overline{\mathrm{a}}) \cdot \overline{\mathrm{z}} \\ & =\frac{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}}]+0+[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}}]+0}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]} \\ = & 3\end{aligned}$

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

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