Question

With reference to the right handed system of mutually perpendicular unit vectors $\hat{i},\hat{j}$ and $\hat{k}$. It is given that $\alpha =3\hat{i}-\hat{j},\beta =2\hat{i}+\hat{j}-3\hat{k}$. If $\beta ={\beta }_1+{\beta }_2$, where ${\beta }_1$ is parallel to $\alpha$ and ${\beta }_2$ is perpendicular to $\alpha$, then ${\beta }_1$ and ${\beta }_2$ are respectively

• A. $\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$ and $\frac{3}{2}\hat{i}-\frac{1}{2}\hat{j}$
• B. $\frac{3}{2}\hat{i}-\frac{1}{2}\hat{j}$ and $\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$
• C. $\hat{i}+\frac{1}{2}\hat{j}$ and $\hat{i}+\frac{1}{2}\hat{j}-3\hat{k}$
• D. $\hat{i}+\frac{1}{2}\hat{j}-3\hat{k}$ and $\hat{i}+\frac{1}{2}\hat{j}$

Answer 1

Answer: (B)

${\beta }_1$ is parallel to $\alpha$
$\therefore$ Let, ${\beta }_1=\lambda \alpha ,\lambda$ is a scalar, ie.
${\beta }_1=3\lambda \hat{i}-\lambda \hat{j}$
Now, ${\beta }_2=\beta -{\beta }_1=(2-3\lambda )\hat{i}+(1+\lambda )\hat{j}-3\hat{k}$
Now, since ${\beta }_2$ is to be perpendicular to $\alpha$, we should have $\alpha \cdot {\beta }_2=0$ ie $3(2-3\lambda )-(1+\lambda )=0$
or $\lambda =\frac{1}{2}$
Therefore, ${\beta }_1=\frac{3}{2}\hat{i}-\frac{1}{2}\hat{j}$
and ${\beta }_2=\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$