If \vec{a}= 2\hat{i}+2\hat{j}+3\hat{k},\vec{b}= -\hat{i}+2\hat{j}+\hat{k}\, and\, \vec{c}= 3\hat{i}+\hat{j}  are such that \vec{a}+\lambda\, \vec{b} is perpendicular to \vec{c}, then find the value of \lambda \cdot

 

 

 

 
 
 
 
 

Answers (1)

\vec{a}= 2\hat{i}+2\hat{j}+3\hat{k}
\vec{b}= -\hat{i}+2\hat{j}+\hat{k}\: \: \: \left [ given \right ]
\vec{c}= 3\hat{i}+\hat{j}
a+\lambda \, b= \left ( 2\hat{i}+2\hat{j}+3\hat{k} \right )+\lambda \left ( -\hat{i}+2\hat{j}+\hat{k} \right )
              = \left ( 2-\lambda \right )\hat{i}+\left ( 2+2\lambda \right )\hat{j}+\left ( 3+\lambda \right )\hat{k}
For a+\lambda \, b  is \perp ^{r} to c
\therefore \left ( a+\lambda \, b \right )\cdot \vec{c}= 0
\left ( \left ( 2-\lambda \right )\hat{i}+\left ( 2+2\lambda \right )\hat{j}+\left ( 3+\lambda \right )\hat{k} \right )\cdot \left ( 3 \hat{i}+\hat{j} \right )= 0
\left ( 2-\lambda \right )3+\left ( 2+2\lambda \right )\cdot 1+0= 0
6-3\lambda +2+2\lambda = 0
-\lambda +8= 0
\lambda = 8

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