Vector Algebra
Triangle from Vectors — Finding $|\vec{c}|^2$
nta_pyq_2024_apr
Grade 12
Question:
Let three vectors $\vec{a}=\alpha\hat{i}+4\hat{j}+2\hat{k}$, $\vec{b}=5\hat{i}+3\hat{j}+4\hat{k}$, $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5\sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:
Step-by-Step Solution
Key Concept: $\vec{c}=\vec{a}-\vec{b}=(\alpha-5)\hat{i}+\hat{j}-2\hat{k}$. Area $=\frac{1}{2}|\vec{a}\times\vec{c}|=5\sqrt{6}\Rightarrow|\vec{a}\times\vec{c}|=10\sqrt{6}$.
$\alpha=8$. $\vec{c}=3\hat{i}+\hat{j}-2\hat{k}$. $|\vec{c}|^2=14$.
Correct Answer: 2