Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
is $30^{\circ} .$ Then $\vec{\alpha}$ is
Solution:
2774 Upvotes
Verified Answer
The correct answer is:
$\pm 2(\vec{\beta} \times \vec{\gamma})$
$\vec{\alpha}=\lambda(\vec{\beta} \times \vec{\gamma})=\lambda\left(|\vec{\beta}||\vec{\gamma}|\left|\sin 30^{\circ}\right|\right.$
$\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$
$\begin{array}{lc}\Rightarrow & 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2} \\ \Rightarrow & |\lambda|=2 \\ \rightarrow & \lambda-\pm 2\end{array}$
$\therefore \quad \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$
$\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$
$\begin{array}{lc}\Rightarrow & 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2} \\ \Rightarrow & |\lambda|=2 \\ \rightarrow & \lambda-\pm 2\end{array}$
$\therefore \quad \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.