Vector Algebra
Cross Product Magnitude — Parameter Finding
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}=\hat{i}+\alpha\hat{j}+\beta\hat{k}$, $\alpha,\beta\in\mathbb{R}$. Let $\vec{b}$ be a vector such that the angle between $\vec{a}$ and $\vec{b}$ is $\dfrac{\pi}{4}$ and $|\vec{b}|^2=6$. If $\vec{a}\cdot\vec{b}=3\sqrt{2}$, then the value of $(\alpha^2+\beta^2)|\vec{a}\times\vec{b}|^2$ is equal to:
Step-by-Step Solution
Key Concept: From $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos(\pi/4)$: $3\sqrt{2}=|\vec{a}|\sqrt{6}\cdot\frac{1}{\sqrt{2}}\Rightarrow|\vec{a}|^2=6\Rightarrow1+\alpha^2+\beta^2=6\Rightarrow\alpha^2+\beta^2=5$. Then $|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2\sin^2\theta=6\cdot6\cdot\frac{1}{2}=18$.
$|\vec{a}|^2\cdot|\vec{b}|^2\cos^2\theta=(3\sqrt{2})^2=18\Rightarrow|\vec{a}|^2=18/(6\cdot\frac{1}{2})=6$. $\alpha^2+\beta^2=5$. $|\vec{a}\times\vec{b}|^2=6\cdot6\cdot\frac{1}{2}=18$. Answer $=5\times18=90$.
Correct Answer: 1