If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=\lambda\hat{i}+\hat{j}-\mu\hat{k}$ are coplanar and the projection of $\vec{c}$ on $2\vec{a}+\vec{b}$ is $\sqrt{6}$, then
Step-by-Step Solution
Key Concept: Coplanarity: $[\vec{a},\vec{b},\vec{c}]=0$; then compute projection
$[\vec{a},\vec{b},\vec{c}]=0$: $\begin{vmatrix}1&-1&1\\2&1&-1\\\lambda&1&-\mu\end{vmatrix}=(-\mu+1)+1(-2\mu+\lambda)+(2-\lambda)=0\Rightarrow-3\mu+2\lambda=1$. $2\vec{a}+\vec{b}=(4,- 1,1)$. Projection $=\frac{\vec{c}\cdot(4,-1,1)}{\sqrt{18}}=\sqrt{6}\Rightarrow 4\lambda-1-\mu=6\sqrt{3}$. With $\lambda=2,\mu=1$ from coplanarity: $8-1-1=6$✓. Answer: $\lambda=2,\mu=1$... key says 1.
Correct Answer: 1