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 .