If $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2 \hat{k}, \bar{c}=x \hat{i}+(x-2)…
If $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2 \hat{k}, \bar{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$ and $\bar{c}$ is linear combination of $\bar{a}$ and $\bar{b}$, then $x$ has the value
1
-2
0
-4
Solution
$\begin{aligned} & \vec{c}=\vec{b}+\lambda \vec{a} \\ & \Rightarrow x \hat{i}+(x-2) \hat{j}-\widehat{k}=(\hat{i}-\hat{j}+2 \widehat{k})+\lambda(\hat{i}+\hat{j}+\widehat{k}) \\ & \Rightarrow x=1+\lambda, x-2=-1+\lambda \text { and }-1=2+\lambda \\ & \Rightarrow \lambda=-3 \text { and } x=-2\end{aligned}$