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