Question

If ${\bar{A}}=3\hat{i}+2\hat{j}$ and ${\bar{B}}=2\hat{i}+3\hat{j}-\hat{k},$ then find a unit vector along $\left( {\bar{A}}-{\bar{B}} \right).$

Answer 1

Hint: A unit vector is a vector of length one. We will find the magnitude of the given vector by taking the square root of the sum of the squares of the components of the vector. Then we will divide the vector with the magnitude to find the unit vector along the vector.

Complete step by step answer:
We are asked to find the unit vector along the vector ${\bar{A}}-{\bar{B}}\text{.}$
Let us consider the given vectors, vector $\text{A}$ and vector $\text{B}\text{.}$
So, ${\bar{A}}=3\hat{i}+2\hat{j}$ and ${\bar{B}}=2\hat{i}+3\hat{j}-\hat{k}.$
Since ${\bar{A}}$ and ${\bar{B}}$ are vectors, their difference ${\bar{A}}-{\bar{B}}$ is also a vector.
To find this vector, we need to subtract ${\bar{B}}$ from ${\bar{A}}$ by the component wise subtraction.
So, we will get ${\bar{A}}-{\bar{B}}=3\hat{i}+2\hat{j}-\left( 2\hat{i}+3\hat{j}-\hat{k} \right)=3\hat{i}+2\hat{j}-2\hat{i}-3\hat{j}+\hat{k}=\hat{i}-\hat{j}+\hat{k}.$
Now, since we got the difference vector by vector subtraction, we need to find the magnitude of the obtained vector so as to divide the vector by the magnitude to get the unit vector along this vector.
To find the magnitude of the vector, we need to find the square root of the sum of the squares of the three components.
So, the magnitude of the difference vector is given by $\left| {\bar{A}}-{\bar{B}} \right|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}=\sqrt{3}.$
Now, we divide the vector ${\bar{A}}-{\bar{B}}$ by $\sqrt{3}$ to get the unit vector along ${\bar{A}}-{\bar{B}}.$
Now, we will get $\dfrac{1}{\sqrt{3}}\left( {\bar{A}}-{\bar{B}} \right)=\dfrac{1}{\sqrt{3}}\left( i-j+k \right).$
Therefore, the unit vector along the vector $\left( {\bar{A}}-{\bar{B}} \right)=\dfrac{1}{\sqrt{3}}\left( i-j+k \right)=\dfrac{1}{\sqrt{3}}i-\dfrac{1}{\sqrt{3}}j+\dfrac{1}{\sqrt{3}}k.$

Hence the unit vector \[\hat{n}=\dfrac{{\bar{A}}-{\bar{B}}}{\left| {\bar{A}}-{\bar{B}} \right|}=\dfrac{1}{\sqrt{3}}i-\dfrac{1}{\sqrt{3}}j+\dfrac{1}{\sqrt{3}}k.\]

Note: Generally, the unit vector along a vector ${\bar{P}}$ is a vector that represents only the direction of the vector ${\bar{P}}$ and has a magnitude $1.$ It is obtained as $\hat{n}=\dfrac{{{\bar{P}}}}{\left| {{\bar{P}}} \right|}.$ In the above problem, the magnitude e of the unit vector $\left| {\hat{n}} \right|=\sqrt{{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \dfrac{-1}{\sqrt{3}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}=\sqrt{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=\sqrt{\dfrac{3}{3}}=\sqrt{1}=1.$