The scalar product of the vector with a unit vector along the sum of

The scalar product of the vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}\: \: and\: \: \vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}$ is equal to 1. Find the value of $\lambda$ and hence find the unit vector along $\vec{b}+\vec{c}.$

Answers (1)

Then , $\vec{n}= \frac{\left ( 2+\lambda \right )\hat{i}+b\hat{j}-2\hat{k}}{\sqrt{\left ( 2+\lambda \right )^{2}+36+4}}$
given, $\vec{a}\cdot \hat{n}= 1$
$\left ( \hat{i}+ \hat{j}+ \hat{k} \right )\cdot \left [ \frac{\left ( 2+\lambda \right ) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\left ( 2+\lambda \right )^{2}+40}} \right ]= 1$
$\Rightarrow \left ( 2+\lambda \right )+6\cdot 2= \sqrt{\left ( 2+\lambda \right )^{2}+90}$
$\Rightarrow \left ( 2+\lambda \right )+4= \sqrt{\left ( 2+\lambda \right )^{2}+40}$ $\left [ \because \hat{i}\cdot \hat{i}= 1,\hat{j}\cdot \hat{j}= 1,\hat{k}\cdot \hat{k}= 1 \right ]$
$\Rightarrow \lambda +6= \sqrt{\left ( 2+\lambda \right )^{2}+40}$
On squaing both sides, we get
$\left ( 6+\lambda \right )^{2}= \left ( 2+\lambda \right )^{2}+40$
$\Rightarrow 36+\lambda ^{2}+12\lambda = 4\lambda ^{2}+4\lambda +40$
$\Rightarrow 36+12\lambda - 4-4\lambda - 40= 0$
$\Rightarrow 8\lambda -8= 0$
$\Rightarrow \lambda = 1$
then, $\vec{b}+\vec{c}= \left ( 2+\lambda \right )\hat{i}+6\hat{j}+\, -2\hat{k}$
$= 3\hat{i}+6\hat{j}-2\hat{k}$
unit vector along $\left ( \vec{b}+\vec{c} \right )= \frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}$
$= \frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{49}}= \frac{3\hat{i}+6\hat{j}-2\hat{k}}{-7}$