Vector Algebra
Minimum of $27|\vec{c}-\vec{a}|^2$ with Cross Product Constraint
nta_pyq_2024_apr
Grade 12
Question:
Consider three vectors $\vec{a},\vec{b},\vec{c}$. Let $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{a}=\vec{b}\times\vec{c}$. If $\alpha\in\left[0,\dfrac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:
Step-by-Step Solution
Key Concept: $|\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2\vec{a}\cdot\vec{c}=|\vec{c}|^2+4-0$ (since $\vec{a}=\vec{b}\times\vec{c}\Rightarrow\vec{a}\perp\vec{c}$). $|\vec{a}|=|\vec{b}||\vec{c}|\sin\alpha=2\Rightarrow|\vec{c}|=\frac{2}{3\sin\alpha}=\frac{2}{3}\csc\alpha$.
$|\vec{c}|_{\min}=4/(3\sqrt{3})$ at $\alpha=\pi/3$. $27|\vec{c}-\vec{a}|^2_{\min}=124$.
Correct Answer: 2