Question
If $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$, find a unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}$.

Answer

Given:

 $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$

Now, $\vec{\text{a}}+\vec{\text{b}}=3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}$

$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{3^2+6^2+6^2}$

$=\sqrt{9+36+36}$

$=\sqrt{81}$

$=9$

Unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}=\frac{\vec{\text{a}}+\vec{\text{b}}}{\big|\vec{\text{a}}+\vec{\text{b}}\big|}=\frac{3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}}{9}$

$=\frac{1}9\times3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)=\frac{1}3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$

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