Question
Find $\lambda$ when the projection of $\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ on $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$ is 4 units.
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Answer
We know
$\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is $\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Given that
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=4$
$\Rightarrow\frac{\big(\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$\Rightarrow\frac{2\lambda+6+12}{\sqrt{4+36+9}}=4$
$\Rightarrow\frac{2\lambda+18}{7}=4$
$\Rightarrow2\lambda+18=28$
$\Rightarrow2\lambda=10$
$\therefore\lambda=5$
$\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is $\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Given that
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=4$
$\Rightarrow\frac{\big(\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$\Rightarrow\frac{2\lambda+6+12}{\sqrt{4+36+9}}=4$
$\Rightarrow\frac{2\lambda+18}{7}=4$
$\Rightarrow2\lambda+18=28$
$\Rightarrow2\lambda=10$
$\therefore\lambda=5$
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