Vector Algebra
Unit vector perpendicular to two given vectors with an angle constraint
nta_pyq_2025_apr
Grade 12
Question:
Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{c}=2\hat{i}+3\hat{j}-\hat{k}$, and makes an angle of $\cos^{-1}\!\left(-\dfrac{1}{3}\right)$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{a}$ makes an angle of $\dfrac{\pi}{3}$ with the vector $\hat{i}+\alpha\hat{j}+\hat{k}$, then the value of $\alpha$ is:
$\sqrt{6}$
$-\sqrt{6}$
$-\sqrt{3}$
$\sqrt{3}$
Step-by-Step Solution
Key Concept: Compute $\vec{b}\times\vec{c}$ to get the direction of $\hat{a}$, then use the given angle condition to select the correct sign, and finally apply the $\pi/3$ angle condition to find $\alpha$.
$\vec{b}\times\vec{c}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\2&3&-1\end{vmatrix}=\hat{i}(2-9)-\hat{j}(-1-6)+\hat{k}(3+4)=-7\hat{i}+7\hat{j}+7\hat{k}$.
$\hat{a}=\pm\dfrac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$.
Angle with $\hat{i}+\hat{j}+\hat{k}$: $\cos\theta=\pm\dfrac{-1+1+1}{\sqrt{3}\cdot\sqrt{3}}=\pm\dfrac{1}{3}$.
Required angle gives $\cos\theta=-\tfrac{1}{3}$, so $\hat{a}=\dfrac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}$.
$\cos\dfrac{\pi}{3}=\dfrac{1-\alpha-1}{\sqrt{3}\cdot\sqrt{\alpha^2+2}}=\dfrac{-\alpha}{\sqrt{3}\cdot\sqrt{\alpha^2+2}}=\dfrac{1}{2}$.
Since $-\alpha>0$, we need $\alpha<0$. $3(\alpha^2+2)=4\alpha^2 \Rightarrow \alpha^2=6 \Rightarrow \alpha=-\sqrt{6}$.
Correct Answer: 2