# if the

Question:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k}$, find a unit vector parallel to the vector $2 \vec{a}-\vec{b}+3 \vec{c}$.

Solution:

We have,

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k}$

$2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k})$

$=2 \hat{i}+2 \hat{j}+2 \hat{k}-2 \hat{i}+\hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k}$

$=3 \hat{i}-3 \hat{j}+2 \hat{k}$

$|2 \vec{a}-\vec{b}+3 \vec{c}|=\sqrt{3^{2}+(-3)^{2}+2^{2}}=\sqrt{9+9+4}=\sqrt{22}$

Hence, the unit vector along $2 \vec{a}-\vec{b}+3 \vec{c}$ is

$\frac{2 \vec{a}-\vec{b}+3 \vec{c}}{|2 \vec{a}-\vec{b}+3 \vec{c}|}=\frac{3 \hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{22}}=\frac{3}{\sqrt{22}} \hat{i}-\frac{3}{\sqrt{22}} \hat{j}+\frac{2}{\sqrt{22}} \hat{k}$