Question

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 one. Find the value of $\lambda$.

Answer 1

Solution :-

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=2\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{c}=\lambda \hat{i}+2\hat{j}+3\hat{k}$

$(\overrightarrow{b}+\overrightarrow{c})=(2+\lambda )\hat{i}+(4+2)\hat{j}+(-5+3)\hat{k}$

$=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$

Let $\overrightarrow{r}$ be unit vector along $(\overrightarrow{b}+\overrightarrow{c})$

$\hat{r}=\frac{1}{|\overrightarrow{b}+\overrightarrow{c}|}\times (\overrightarrow{b}\times \overrightarrow{c})$

$\hat{r}=\frac{1}{\sqrt{(}2+\lambda {)}^2+36+4}\times \left[\right(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}]$

$\hat{r}=\frac{1}{{\lambda }^2+4\lambda +44}\times \left[\right(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}]$

Now $\overrightarrow{a}.\overrightarrow{r}=1$ (given)

$(\hat{i}+\hat{y}+\hat{k}).(\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}\times (2+\lambda )\hat{i}+6\hat{j}-2\hat{k}))=1$

$(\hat{i}+\hat{j}+\hat{k})\left(\right(2+\lambda )\hat{i}+6\hat{y}-2\hat{k})=\sqrt{{\lambda }^2+4\lambda +44}$

$(\lambda +2)+1.6+1.(-2)=\sqrt{{\lambda }^2+4\lambda +44}$

$\lambda +2+6-2=\sqrt{{\lambda }^2+4\lambda +44}$

$(\lambda +6{)}^2={\lambda }^2+4\lambda +44$

${\lambda }^2+12\lambda +36={\lambda }^2+4\lambda +44$

$8\lambda =8$

$\lambda =1$