Question

If the scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda \hat{i}+2\hat{j}+3\hat{k}$ is equal to onethen thevalue of $\lambda$ is

• A. 0
• B. -1
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (D)

Let $a=\hat{i}+\hat{j}+\hat{k},b=2\hat{i}+4\hat{j}-5\hat{k}$ and $c=\lambda \hat{i}+2\hat{j}+3\hat{k}$
Now, $b+c=2\hat{i}+4\hat{j}-5\hat{k}+\lambda \hat{i}+2\hat{j}+3\hat{k}=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$
$\therefore |b+c|=\sqrt{(2+\lambda {)}^2+(6{)}^2+(-2{)}^2}$
$=\sqrt{4+{\lambda }^2+4\lambda +36+4}=\sqrt{{\lambda }^2+4\lambda +44}$
The unit vector along $(b+c)$, i.e.,
$\frac{b+c}{|b+c|}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}$
Sualar product $(\hat{i}+\hat{j}+\hat{k})$ with this unil vector is 1 .
$\therefore (\hat{i}+\hat{j}+\hat{k})\cdot \frac{b+c}{|b+c|}=1$
$\Rightarrow (\hat{i}+\hat{j}+\hat{k})\cdot \frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}=1$
$\Rightarrow \frac{1(2+\lambda )+1(6)+1(-2)}{\sqrt{{\lambda }^2+4\lambda +44}}=1$
$\Rightarrow \frac{(2+\lambda )+6+-2}{\sqrt{{\lambda }^2+4\lambda +44}}=1$
$\Rightarrow \lambda +6=\sqrt{{\lambda }^2+4\lambda +44}$
$\Rightarrow (\lambda +6{)}^2={\lambda }^2+4\lambda +44$
$\Rightarrow {\lambda }^2+12\lambda +36={\lambda }^2+4\lambda +44$
$\Rightarrow 8\lambda =8$
$\Rightarrow \lambda =1$
Hence, the value of $\lambda$ is 1 .