If $\mathbf{a}=4 \hat{i}+6 \hat{j}, \mathbf{b}=3 \hat{j}+4 \hat{k}$ and $\mathbf{c}$. is the projection…
If $\mathbf{a}=4 \hat{i}+6 \hat{j}, \mathbf{b}=3 \hat{j}+4 \hat{k}$ and $\mathbf{c}$. is the projection vector of $\mathbf{a}$ on $\mathbf{b}$, then $\mathbf{c}$ and $|\mathbf{c}|$ respectively are
$\frac{18}{25}$ b, $\frac{18}{5}$
$\frac{18}{5} b, 18$
$\frac{25}{18} b, \frac{18}{5}$
$\frac{5}{18} b, \frac{5}{18}$
Solution
$\mathbf{a}=4 \hat{i}+6 \hat{j}, \mathbf{b}=3 \hat{j}+4 \hat{k}$
$\because \mathbf{c}$ is the projection vector of $\mathbf{a}$ on $\mathbf{b}$
$\begin{aligned} & \Rightarrow \mathbf{c}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}=\frac{(4 \hat{i}+6 \hat{j}) \cdot(3 \hat{j}+4 \hat{k})}{\sqrt{3^2+4^2}} \\ & \Rightarrow \mathbf{c}=\frac{18}{5} \hat{j} \Rightarrow|\mathbf{c}|=\frac{18}{5}\end{aligned}$
Now, $|\mathbf{b}|=5$
Then, $\frac{|c|}{|b|}=\frac{18}{25}$
$\Rightarrow \quad|\mathbf{c}|=\frac{18}{25}|\mathbf{b}|=0 \Rightarrow|\mathbf{c}|=\frac{18}{25} \times 5=\frac{18}{5}$
$\because \quad \mathbf{c}$ and $\mathbf{b}$ are collinear vectors.
$\Rightarrow \quad c=\frac{18}{25} b$