Q71 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}, \vec{c}$ and angle between $\vec{b}, \vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$

  • A
    6
  • B
    14
  • Correct answer
    C
    38
  • D
    26

Solution

$\vec{a} \perp \vec{b}, \vec{a} \perp \vec{c}$ Angle $\mathrm{b} / \mathrm{w} \vec{b} \& \vec{c}=\frac{2 \pi}{3}$ $\begin{aligned} & |\vec{a}+3 \vec{b}-4 \vec{c}|^2=(\vec{a}+3 \vec{b}-4 \vec{c}) \cdot(\vec{a}+3 \vec{b}-4 \vec{c}) \\ = & |\vec{a}|^2+3 \vec{a} \cdot \vec{b}-4 \vec{a} \cdot \vec{c}+3 \vec{b} \cdot \vec{a}+9|\vec{b}|^2 \\ & -12 \vec{b} \cdot \vec{c}-4 \vec{c} \cdot \vec{a}-12 \vec{b} \cdot \vec{c}+16|\vec{c}|^2 \\ = & 1+9+16-24(1)(1) \cos \frac{2 \pi}{3}=38\end{aligned}$

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

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