Question

Given 2 vectors $\overrightarrow{A}=4.00\hat{i}+3.00\hat{j}$ and $\overrightarrow{B}=5.00\hat{i}-2.00\hat{j}$how do you find the magnitude & direction of the vector difference$\overrightarrow{A}-\overrightarrow{B}$? \[\]

Answer 1

Hint: We recall the component wise representation of a vector with unit orthogonal vectors$\hat{i},\hat{j}$. We use the fact that if $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}$ is component wise representation then its magnitude is given by $\left| \overrightarrow{a} \right|=a=\sqrt{a_{1}^{2}+a_{2}^{2}}$ and direction is given by the ray joining from origin to $\left( {{a}_{1}},{{a}_{2}} \right)$. We find $\overrightarrow{A}-\overrightarrow{B}$ by subtracting component wise and then find the magnitude.

Complete step by step answer:
We know that $\hat{i}$,$\hat{j}$ are unit vectors(vectors with magnitude 1) along $x,y$ axes in plane respectively. So the magnitude of these vectors $\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=1$. The vectors just like their axes are perpendicular to each other which means angle between $\hat{i}$,$\hat{j}$ is ${{90}^{\circ }}.$ We can represent any vector $\overrightarrow{a}$with component in $x-$axis as ${{a}_{1}}$ and component in $y-$axis as ${{a}_{2}}$as
\[\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}\]
We know that magnitude of above vector is given by given by $\left| \overrightarrow{a} \right|=a=\sqrt{a_{1}^{2}+a_{2}^{2}}$ and direction is given by the ray joining from origin to $\left( {{a}_{1}},{{a}_{2}} \right)$. We are given two vector with component wise representation $\overrightarrow{A}=4.00\hat{i}+3.00\hat{j}$ and $\overrightarrow{B}=5.00\hat{i}-2.00\hat{j}$ . Let us subtract the respective components of $\overrightarrow{B}$ from $\overrightarrow{A}$ to get $\overrightarrow{A}-\overrightarrow{B}$ as
\[\begin{align}
  & \overrightarrow{A}-\overrightarrow{B}=4.00\hat{i}+3.00\hat{j}-\left( 5.00\hat{i}-2.00\hat{j} \right) \\
 & \Rightarrow \overrightarrow{A}-\overrightarrow{B}=\left( 4.00-5.00 \right)\hat{i}+\left( 3.00+2.00 \right)\hat{j} \\
 & \Rightarrow \overrightarrow{A}-\overrightarrow{B}=-\hat{i}+5.00\hat{j} \\
\end{align}\]
So let us find the magnitude
\[\left| \overrightarrow{A}-\overrightarrow{B} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( 5 \right)}^{2}}}=\sqrt{1+25}=\sqrt{26}\]
The direction of $\overrightarrow{A}-\overrightarrow{B}$ will be ray joining the origin to the point $\left( -1,5 \right)$. We show it in the following figure. \[\]


Note:
We can also find the angle the vector forms with positive $x-$axis as ${{90}^{\circ }}+\theta $ where $\theta ={{\tan }^{-1}}\dfrac{1}{5}$.We knows that the vector formed by two vectors is called resultant vector. There are two laws to add vectors excluding the component wise addition: triangle law and parallelogram law. We have showed the parallelogram law of addition of vectors too as shown in the above figure as $\overrightarrow{A}-\overrightarrow{B}=\overrightarrow{A}+\left( -\overrightarrow{B} \right)$.