Question

Let $\vec{\alpha}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is

• A. 9
• B. 11
• C. 7
• D. 6

Answer 1

Answer: (C)

Let $\vec{\beta}_1=\lambda \vec{\alpha}$
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$=(\hat{i}+2 \hat{ j }-4 \hat{ k })-\lambda(4 \hat{ i }+3 \hat{ j }+5 \hat{ k })$
$=(1-4 \lambda) \hat{ i }+(2-3 \lambda) \hat{ j }-(5 \lambda+4) \hat{ k }$
$ \vec{\beta}_2 \cdot \vec{\alpha}=0$
$\Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 $
$ \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0$
$ \Rightarrow 50 \lambda=-10$
$\Rightarrow \lambda=\frac{-1}{5}$
$ \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{ i }+\left(2+\frac{3}{5}\right) \hat{ j }-(-1+4) \hat{ k } $
$ \vec{\beta}_2=\frac{9}{5} \hat{ i }+\frac{13}{5} \hat{ j }-3 \hat{ k }$
$ 5 \vec{\beta}_2=9 \hat{ i }+13 \hat{ j }-15 \hat{ k }$
$5 \vec{\beta}_2 \cdot(\hat{ i }+\hat{ j }+\hat{ k })=9+13-15=7$