Vector Algebra
Coplanar Vector Perpendicular to Given Vector
Grade 12
Question:
<p>Let \(\vec{a}=\hat{i}+\hat{j}+2\hat{k}\), \(\vec{b}=2\hat{i}-\hat{j}+\hat{k}\),
\(\vec{c}=3\hat{i}-\hat{k}\). A vector \(\vec{v}\) is coplanar with \(\vec{a}\)
and \(\vec{b}\), perpendicular to \(\vec{c}\), and satisfies
\(\vec{v}\cdot(\hat{i}+2\hat{j}+\hat{k})=8\). Find \(|\vec{v}|^2\).</p>
Step-by-Step Solution
Key Concept: Write v = sa + tb (coplanar condition). Apply v \perp c to eliminate one parameter, then use the dot-product condition to find the other.
$\vec{v}=s\vec{a}+t\vec{b}=(s+2t)\hat{i}+(s-t)\hat{j}+(2s+t)\hat{k}$.
$\vec{v}\perp\vec{c}=(3,0,-1)$: $3(s+2t)-(2s+t)=0\Rightarrow s+5t=0\Rightarrow s=-5t$.
$\vec{v}=(-3t,-6t,-9t)=-3t(1,2,3)$.
$\vec{v}\cdot(1,2,1)=-3t(1+4+3)=-24t=8\Rightarrow t=-\tfrac{1}{3}$.
$\vec{v}=(1,2,3)\Rightarrow|\vec{v}|^2=1+4+9=\boxed{14}$.
Correct Answer: C