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  • 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)

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}

Posted by

Ravindra Pindel

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