# 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 $\vec{b}+\vec{c}.$

Here, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ $\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}\:, \vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}$

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

Then, $\hat{n}= \frac{(2+\lambda)\hat{i}+ 6 \hat{j} - 2\hat{k}}{\sqrt { (2+\lambda)^2+36+4}}$

Given $\vec{a}\cdot \hat{n}=1$

$\left ( \hat { i}+\hat { j}+\hat{k} \right )\cdot \left [ \frac{(2+\lambda )\hat{i} + 6 \hat{j} - 2 \hat{k}} {\sqrt{ (2+\lambda )^2+40}} \right ]=1$

$\Rightarrow (2+\lambda)+6-2=\sqrt{(2+\lambda )^2+40}$

$\Rightarrow (2+\lambda)+4=\sqrt{(2+\lambda )^2+40}\: \: \: \: \left [ \because \hat\hat {i}\hat{i}=1, \hat{j} \cdot \hat{j}=1,\hat{k} \cdot \hat{k}=1 \right ]$

$\Rightarrow (\lambda + 6)=\sqrt{(2+\lambda )^2+40}$

On squaring both sides, we get :

$\Rightarrow (6+\lambda)^2=(2+\lambda )^2+40$

$\Rightarrow 36+\lambda^2+12\lambda=4+\lambda^2+4\lambda+40$

$\Rightarrow 36+12\lambda-4-4\lambda-40=0$

$\Rightarrow 8\lambda-8=0$

$\Rightarrow \lambda=1$

Then, $\vec{b}+\vec{c}=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}$

$=3\hat{i}+6\hat{j}-2\hat{k}$

unit vector along $\left ( \vec{b}+\vec{c} \right )=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{49}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$

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