Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=3\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a}\cdot\vec{c}=5$, then $|\vec{c}|$ is equal to:
Step-by-Step Solution
Key Concept: For $\vec{c}$ coplanar with $\vec{a},\vec{b}$ and perpendicular to $\vec{b}$, write $\vec{c}=\lambda(\vec{b}\times(\vec{a}\times\vec{b}))$, use BAC-CAB to expand, then apply $\vec{a}\cdot\vec{c}=5$ to find $\lambda$.
$\vec{c}=\lambda\,\vec{b}\times(\vec{a}\times\vec{b})=\lambda[(\vec{b}\cdot\vec{b})\vec{a}-(\vec{a}\cdot\vec{b})\vec{b}]$.
$\vec{b}\cdot\vec{b}=11$, $\vec{a}\cdot\vec{b}=3+2-3=2$.
$\vec{c}=\lambda(11\vec{a}-2\vec{b})=\lambda(11\hat{i}+22\hat{j}+33\hat{k}-6\hat{i}-2\hat{j}+2\hat{k})=5\lambda(\hat{i}+4\hat{j}+7\hat{k})$.
$\vec{a}\cdot\vec{c}=5\lambda(1+8+21)=5\lambda\cdot30=5 \Rightarrow \lambda=\dfrac{1}{30}$.
$\vec{c}=\dfrac{1}{6}(\hat{i}+4\hat{j}+7\hat{k})$, $|\vec{c}|=\dfrac{\sqrt{1+16+49}}{6}=\dfrac{\sqrt{66}}{6}=\sqrt{\dfrac{11}{6}}$.
Correct Answer: 1