Q94 / 155 · WBJEE 2018 · Mathematics

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be the three unit vectors such that $\vec{\alpha} \cdot \vec{\beta}=\vec{\alpha} \cdot \vec{\gamma}=0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$ is $30^{\circ} .$ Then $\vec{\alpha}$ is

  • A
    $2(\vec{\beta} \times \vec{\gamma})$
  • B
    $-2(\vec{\beta} \times \vec{y})$
  • Correct answer
    C
    $\pm 2(\vec{\beta} \times \vec{\gamma})$
  • D
    $(\vec{\beta} \times \vec{y})$

Solution

$\vec{\alpha}=\lambda(\vec{\beta} \times \vec{\gamma})=\lambda\left(|\vec{\beta}||\vec{\gamma}|\left|\sin 30^{\circ}\right|\right.$ $\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$ $\begin{array}{lc}\Rightarrow & 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2} \\ \Rightarrow & |\lambda|=2 \\ \rightarrow & \lambda-\pm 2\end{array}$ $\therefore \quad \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$

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

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