Vectors & 3D Geometry
Cross product constraint; minimum distance
MMTS_Full_Test_22
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$ be two vectors perpendicular to each other with $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{c}\times\vec{a}=\vec{b}$. The least value of $|\vec{c}-\vec{a}|$ is
(A) 1
(B) $\dfrac{1}{2}$
(C) $\dfrac{1}{4}$
(D) $\dfrac{3}{2}$
Step-by-Step Solution
Key Concept: From $\vec{c}\times\vec{a}=\vec{b}$, write $\vec{c}$ in component form: $\vec{c}=\vec{c}_\perp+\vec{c}_\parallel$ where $\vec{c}_\parallel$ is along $\vec{a}$. Minimize $|\vec{c}-\vec{a}|$ by optimizing the perpendicular component.
$|\vec{c}-\vec{a}|_{\min}=\sqrt{\frac{9}{4}}=\frac{3}{2}$.
Correct Answer: (D) $\dfrac{3}{2}$