Mathematics - Vectors Question with Solution | TestHub

Let $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{k}$.

The given equation is $\vec{a}\times \vec{b}=\vec{c}$.

We observe that $\hat{j}\times \vec{b}=\hat{j}\times (\hat{i}+2\hat{j}+\hat{k})=-\hat{k}+\hat{i}=\vec{c}$.

Thus, $\vec{a}\times \vec{b}=\hat{j}\times \vec{b}$, which implies $(\vec{a}-\hat{j})\times \vec{b}=\vec{0}$.

This means $\vec{a}-\hat{j}$ is parallel to $\vec{b}$.

So, $\vec{a}-\hat{j}=\lambda \vec{b}$ for some scalar $\lambda \in R$.

$\vec{a}=\hat{j}+\lambda (\hat{i}+2\hat{j}+\hat{k})=\lambda \hat{i}+(1+2\lambda )\hat{j}+\lambda \hat{k}$.

The final answer is $\boxed{\lambda \hat{i}+(2\lambda +1)\hat{j}+\lambda \hat{k},\lambda \in R}$.