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 $2\vec{a}-\vec{b}+3\vec{c}.$

Answer 1

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{3i^-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}.