Question

Vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are shown. What is the magnitude of a vector $\overrightarrow{C}$ if $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$?

$\text{A}\text{. }40\sqrt{3}$
$\text{B}\text{. }40\sqrt{2}$
$\text{C}\text{. }40$
$\text{D}\text{. 2}0$

Answer 1

Draw the vector $\left( -\overrightarrow{B} \right)$ and find the angle between the vectors $\overrightarrow{A}$ and $\left( -\overrightarrow{B} \right)$. The magnitude of vector $\left( -\overrightarrow{B} \right)$ is equal to the magnitude of vector $\overrightarrow{B}$. However, both the vectors are opposite directions. Then use the formula for the magnitude of the resultant of two vectors.

Formula used: $R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$

Complete step by step answer:
The vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are given. We are asked to find the magnitude of the vector $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$. This vector addition can also be written as $\overrightarrow{C}=\overrightarrow{A}+\left( -\overrightarrow{B} \right)$.
Let us first draw the vector $\left( -\overrightarrow{B} \right)$. Vector $\left( -\overrightarrow{B} \right)$ is a vector whose magnitude is equal to the magnitude of vector $\overrightarrow{B}$ but its direction is opposite to the direction of $\overrightarrow{B}$ (as shown).

From the figure we get to know that the angle between the vectors $\overrightarrow{A}$ and $\left( -\overrightarrow{B} \right)$ is ${{120}^{\circ }}$.
Let us know use the formula for the magnitude of the resultant of the two vectors, i.e. $R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$ ….. (i), where are R is the magnitude of the resultant, A and B are the magnitudes of the two vectors and $\theta$ is the angle between the two vectors.
In this case, R = C, A = 40.
The magnitude of $\left( -\overrightarrow{B} \right)$ is equal to the magnitude of $\overrightarrow{B}$. Therefore, B = 40.
And $\theta ={{120}^{\circ }}$.
Substitute the values in (i).
$\Rightarrow C=\sqrt{{{40}^{2}}+{{40}^{2}}+2(40)(40)\cos {{120}^{\circ }}}$.
The value of $\cos {{120}^{\circ }}=-\dfrac{1}{2}$.
$\Rightarrow C=\sqrt{{{40}^{2}}+{{40}^{2}}+2(40)(40)\left( -\dfrac{1}{2} \right)}=\sqrt{{{40}^{2}}+{{40}^{2}}-{{40}^{2}}}=\sqrt{{{40}^{2}}}=40$.
This means that the magnitude of $\overrightarrow{C}$ is 40.

So, the correct answer is “Option C”.

Note: Alternative solution:
The magnitude of resultant of $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$, is given as $C=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }$ ….. (1), where $\theta$ is the angle between the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$.
Therefore, A = B = 40 and $\theta ={{60}^{\circ }}$.
Substitute the values in (1).
$\Rightarrow C=\sqrt{{{40}^{2}}+{{40}^{2}}-2(40)(40)\cos {{60}^{\circ }}}=\sqrt{{{40}^{2}}+{{40}^{2}}-2(40)(40)\left( \dfrac{1}{2} \right)}=\sqrt{{{40}^{2}}+{{40}^{2}}-{{40}^{2}}}=\sqrt{{{40}^{2}}}=40$.