Let $\vec{a}=3\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times\vec{c}=2(\vec{a}\times\vec{b})+24\hat{j}-6\hat{k}$ and $(\vec{a}-\vec{b}+\hat{i})\cdot\vec{c}=-3$. Then $|\vec{c}|^2$ is equal to _____.
Step-by-Step Solution
Key Concept: Compute $\vec{a}+\vec{b}=5\hat{i}+\hat{j}+4\hat{k}$ and $2(\vec{a}\times\vec{b})$. Set up a system from the cross product equation using the determinant expansion, then use the dot product condition to solve for all components of $\vec{c}$.
$\vec{a}+\vec{b}=5\hat{i}+\hat{j}+4\hat{k}$; $2(\vec{a}\times\vec{b})=14\hat{i}-14\hat{j}-14\hat{k}$; RHS $=14\hat{i}+10\hat{j}-20\hat{k}$. Cross product system: $z-4y=14$, $4x-5z=10$, $5y-x=-20$. With $2x+3y-2z=-3$: solution $x=5,y=-3,z=2$. $|\vec{c}|^2=25+9+4=38$.
Correct Answer: 38