question_answer

Find \[\lambda ,\] when projection of \[\vec{a}=\lambda \hat{i}+\hat{j}+4\hat{k}\] on \[\vec{b}=2\hat{i}+6\hat{j}+3\hat{k}\] is 4 units.

Answer:

Given, \[\vec{a}=\lambda \hat{i}+\hat{j}+4\hat{k},\] \[\overrightarrow{b}=2\hat{i}+6\hat{j}+3\hat{k}\] and projection of \[\vec{a}\]on \[\overrightarrow{b}\]= 4 units \[\Rightarrow \]\[\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=4\]      [\[\because \]projection of \[\vec{a}\] on \[\overrightarrow{b}\]\[=\frac{\overrightarrow{a}\cdot \vec{b}}{|\overrightarrow{b}|}\] ] \[\Rightarrow \] \[\frac{(\lambda \,\hat{i}+\hat{j}+4\hat{k})\cdot (2\,\hat{i}+6\hat{j}+3\hat{k})}{\sqrt{{{(2)}^{2}}+{{(6)}^{2}}+{{(3)}^{2}}}}=4\] \[\Rightarrow \]   \[\frac{2\lambda +6+12}{\sqrt{49}}=4\] \[\Rightarrow \]   \[2\lambda +18=28\] \[\Rightarrow \]   \[\lambda =5\]