Vector Algebra
Cross Product — Finding Dot Product from Magnitude Conditions
nta_pyq_2026_jan
Grade 12
Question:
Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$, $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a}\times\vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11}$, $|\vec{c}\times\vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\dfrac{\pi}{4}$. Then $\vec{a}\cdot\vec{d}$ is equal to:
Step-by-Step Solution
Key Concept: $\vec{c}=\vec{a}\times\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$. $|\vec{c}|=3$, $|\vec{a}|=3$. From $|\vec{c}\times\vec{d}|=3$ and angle $\frac{\pi}{4}$: $|\vec{c}||\vec{d}|\sin\frac{\pi}{4}=3\Rightarrow3|\vec{d}|\cdot\frac{1}{\sqrt{2}}=3\Rightarrow|\vec{d}|=\sqrt{2}$.
$\vec{a}\cdot\vec{d}=0$.
Correct Answer: 3