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}$