Question

If \[\overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k\text{ }}\text{and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}\text{,}\]then find a unit vector parallel to the vector $\overrightarrow{a}+\overrightarrow{b}.$

Answer 1

Hint:In this question we have given two vectors and we have to find a unit vector which is parallel to sum of vectors. So, first of all we have to find the sum of the given vectors. As we know that If $\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}\text{ and }\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}.$ then sum of vectors is define as $\overrightarrow{a}+\overrightarrow{b}=({{a}_{1}}+{{b}_{1}})\widehat{i}+({{a}_{2}}+{{b}_{2}})\widehat{j}+({{a}_{3}}+{{b}_{3}})\widehat{k}$. As we have to find the unit vector parallel to a vector so we have to divide the vector by its magnitude.
The magnitude of a vector $\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}\text{ }$is define as \[\overrightarrow{\left| a \right|}=\sqrt{{{\left( {{a}_{1}} \right)}^{2}}+{{\left( {{a}_{2}} \right)}^{2}}+{{\left( {{a}_{3}} \right)}^{2}}}\]
Unit vector parallel to the vector is define as \[\widehat{n}=\dfrac{\overrightarrow{a}}{\left| a \right|}\]

Complete step by step answer:
From question we have the value of
\[\overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k\text{ }}\text{and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}\text{,}\]
As we know that if,
$\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}\text{ and }\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}.$ then sum of vectors is define as $\overrightarrow{a}+\overrightarrow{b}=({{a}_{1}}+{{b}_{1}})\widehat{i}+({{a}_{2}}+{{b}_{2}})\widehat{j}+({{a}_{3}}+{{b}_{3}})\widehat{k}$
Here we have the value of
${{a}_{1}}=4,{{a}_{2}}=-1,{{a}_{3}}=1,{{b}_{1}}=2,{{b}_{2}}=-2,{{b}_{3}}=1$
So, we can write
$\overrightarrow{a}+\overrightarrow{b}=(4+2)\widehat{i}+(-1-2)\widehat{j}+(1+1)\widehat{k}$
Hence, we can write further
$\overrightarrow{a}+\overrightarrow{b}=6\widehat{i}-3\widehat{j}+2\widehat{k}$,
Now we have to find the value of magnitude of sum of vectors, so, we can write
 $\left| \overrightarrow{a}+\overrightarrow{b} \right|=\left| 6\widehat{i}-3\widehat{j}+2\widehat{k} \right|$
As we know that
\[\overrightarrow{\left| a \right|}=\sqrt{{{\left( {{a}_{1}} \right)}^{2}}+{{\left( {{a}_{2}} \right)}^{2}}+{{\left( {{a}_{3}} \right)}^{2}}}\], so, we can write
\[\begin{align}
  & \left| \overrightarrow{a}+\overrightarrow{b} \right|=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\
 & \left| \overrightarrow{a}+\overrightarrow{b} \right|=\sqrt{36+9+4} \\
 & \left| \overrightarrow{a}+\overrightarrow{b} \right|=\sqrt{47} \\
\end{align}\]
Now we can find the value of unit vector which if parallel to vector sum of vectors that is $\overrightarrow{a}+\overrightarrow{b}$, by dividing it by magnitude of the sum of vectors that is \[\left| \overrightarrow{a}+\overrightarrow{b} \right|\], so we can write
\[\widehat{n}=\dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\]
Putting the values, we can write
\[\widehat{n}=\dfrac{6\widehat{i}-3\widehat{j}+2\widehat{k}}{\sqrt{47}}\]
We can write it as
\[\widehat{n}=\dfrac{6}{\sqrt{47}}\widehat{i}+\dfrac{-3}{\sqrt{47}}\widehat{j}+\dfrac{2}{\sqrt{47}}\widehat{k}\]

Note:
It is important to note that when we write a unit vector as $\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}\text{ }$it mean that $\left| \overrightarrow{a} \right|=1$so that we have ${{\left( {{a}_{1}} \right)}^{2}}+{{\left( {{a}_{2}} \right)}^{2}}+{{\left( {{a}_{3}} \right)}^{3}}=1$, also if we translate a vector parallel to itself its value does not change. The value of a vector only changes when either its magnitude changes or its direction or both.