Question

Vectors $\overrightarrow{\mathbf{A}}$and $\overrightarrow{\mathbf{B}}$have equal magnitudes of 5.00 The sum of $\overrightarrow{\mathbf{A}}$and$\overrightarrow{\mathbf{B}}$ is the vector role="math" localid="1663656292910" $\mathbf{6}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}$. Determine the angle between $\overrightarrow{\mathbf{A}}$and$\overrightarrow{\mathbf{B}}$

Short answer

The angle between the vectors$\overrightarrow{A}\text{and}\overrightarrow{B}$ is$\varphi =2\theta =106°$ .

Step-by-step solution

Define the component of vector

In a two-dimensional coordinate system, the projection of vector along the x-axis and y-axis are commonly considered as the components of a vector. The magnitude of the components can be given as-

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\mathrm{Vcos}\theta \\ {\mathrm{v}}_{\mathrm{y}}=\mathrm{Vsin}\theta \end{gathered}$$

with V denoting the vector.

Given Data

The magnitude of$\overrightarrow{A}\text{and}\overrightarrow{B}$ is-

$\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|=5.00$

Draw a diagram according to question

Using law of vector addition,

Let$\stackrel{\rightharpoonup }{\mathrm{A}}=\left({\mathrm{A}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{A}}_{\mathrm{y}}\hat{\mathrm{j}}\right)$

$\overrightarrow{B}=\left(B_x\hat{i}+B_y\hat{j}\right)$

$\stackrel{\rightharpoonup }{\mathrm{A}}+\stackrel{\rightharpoonup }{\mathrm{B}}=6.00\hat{\mathrm{j}}$

$\left({\mathrm{A}}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{x}}\right)\hat{\mathrm{i}}+\left({\mathrm{A}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{y}}\right)\hat{\mathrm{j}}=0\hat{\mathrm{i}}+6.00\hat{\mathrm{j}}$ …………. (1)

Let$\theta =$ the angle between $\overrightarrow{A}$or$\overrightarrow{B}$ and the y-axis.

Now simplify the above vectors to get the angle.

$\begin{array}{rcl}{A}_x+B_x& =& 0\\ A_x& =& -B_x\end{array}$ ……… (2)

$A_y+B_y=\text{6.00}$ ……… (3)

We have given that

$\begin{array}{rcl}\left|\overrightarrow{A}\right|& =& \left|\overrightarrow{B}\right|\\ \left|\overrightarrow{A}\right|& =& 5.00\\ A_x^2+A_y^2& =& B_x^2+B_y^2\\ A_x^2+A_y^2& =& 5.{00}^2\end{array}$

From equation (2) and (1), we have

$A_x^2=B_x^2$

$A_y^2=B_y^2$

and$A_y=B_y$

From equation (3), we have-

$\begin{array}{rcl}2B_y& =& 6.00\\ B_y& =& 3.00\end{array}$

So, $A_y=3.00$

The angel made by the vector with its component is given as-

$\begin{array}{rcl}\cos \theta & =& \frac{A_y}{A}\\ & =& \frac{B_y}{B}\\ & =& \frac{3.00}{5.00}\\ & =& 0.600\end{array}$

Solving further,

$\begin{array}{rcl}\theta & =& {\cos }^{-1}0.6\\ & =& 53.1°\end{array}$

Therefore, the angle between the vectors is

$\begin{array}{rcl}\varphi & =& 2\theta \\ & =& 2\times 53.1°\\ & =& 106.2°\end{array}$

The angle between the vectors is $106.2°$.