Question

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

• A. ${\beta }_1=\frac{3}{2}i+\frac{1}{2}j$
• B. ${\beta }_1=\frac{3}{2}i-\frac{1}{2}j$
• C. ${\beta }_2=\frac{1}{2}i+\frac{3}{2}j-3k$
• D. ${\beta }_2=\frac{1}{2}i-\frac{3}{2}j-3k$

Answer 1

Answer: (B), (C)

The correct options are
B ${\beta }_1=\frac{3}{2}i-\frac{1}{2}j$
C ${\beta }_2=\frac{1}{2}i+\frac{3}{2}j-3k$

Since ${\beta }_1$ is parallel to $\alpha$,

let ${\beta }_1=\lambda \alpha$ where $\lambda$ is a scalar.

$\beta ={\beta }_1+{\beta }_2\Rightarrow {\beta }_2=\beta -{\beta }_1$ (given)

$=2i+j-3k-\lambda \alpha =2i+j-3k-\lambda (3i-j)$

$-(2-3\lambda )i+(1+\lambda )j-3k$

Since ${\beta }_2$ is perpendiculat to $\alpha$

$\therefore {\beta }_2.a=0\Rightarrow [(2-3\lambda )i+(1+\lambda )j-3k].(3i-j)=0\Rightarrow 3(2-3\lambda )-(1+\lambda )=0$

$\Rightarrow 6-9\lambda -1-\lambda =0\Rightarrow 5=10\lambda =0\Rightarrow \lambda =\frac{1}{2}$

$\therefore {\beta }_1=\lambda \alpha =\frac{1}{2}(3i-j)=\frac{3}{2}i-\frac{1}{2}j$

${\beta }_2=(2-\frac{3}{2})i+(1+\frac{1}{2})j-3k=\frac{1}{2}i+\frac{3}{2}j-3k$

$\therefore 2i+j-3k=\lambda \alpha +{\beta }_2$

Hence $\beta ={\beta }_1+{\beta }_2$