Question

Two vectors$\overrightarrow{A}$and$\overrightarrow{B}$have, precisely equal magnitudes. For the magnitude of$\overrightarrow{A}+\overrightarrow{B}$to be 100times larger than the magnitude of$\overrightarrow{A}-\overrightarrow{B}$, what must be the angle between them?

Short answer

The distance between $\left|\overrightarrow{A}+\overrightarrow{B}\right|$and $\left|\overrightarrow{A}-\overrightarrow{B}\right|$ is$\theta =1.{15}^{°}$

Step-by-step solution

Definition of the magnitude of vector.

• The length of a given vector (v) is calculated using the magnitude of a vector formula, which is indicated as |v|.
• In a nutshell, this value is the distance between the vector's start and end points. We utilise the distance formula to calculate the magnitude of the vector, which we shall go over here.

Given and known values.

We know, from appendix , that the law of cosines is:

$R^2=A^2+B^2+2ABcos\theta$

Where is the magnitude of the vector $\overrightarrow{\mathbf{R}}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}},\theta$is the angle between $\overrightarrow{A}$and $\overrightarrow{B}$A and B are the magnitudes of the vectors $\overrightarrow{A}$and$\overrightarrow{B}$respectively.

We have the magnitude of $\overrightarrow{A}$and $\overrightarrow{B}$are equals, so we have the relation:

$A=B$

And we also have the magnitude of$\overrightarrow{A}+\overrightarrow{B}$istimes larger than the magnitude of$\overrightarrow{A}-\overrightarrow{B}$so we have the relation:

$$\begin{gathered} \theta ={\cos }^{-1}\frac{9999}{10001}=1.15 \\ \theta =1.15° \end{gathered}$$

To determine the angle between A→+B→ and  A→-B→

We use equation (1) for$\overrightarrow{A}+\overrightarrow{B}$, so we get:

$|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}{|}^2=A^2+B^2+2AB\cos \theta$

Using the relation $R.1$we get:

$$\begin{gathered} |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}{|}^2=2A^2+2A^2\cos \theta |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}{|}^2 \\ =2A^2(1+\cos \theta ) \end{gathered}$$

Using equation (1) for$\overrightarrow{A}-\overrightarrow{B}$, we get:

$|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}{|}^2=A^2+B^2-2AB\cos \theta$

Using the relation$R.1$, we get:

$\begin{array}{c}|\overrightarrow{\mathit{A}}-\overrightarrow{\mathit{B}}{|}^2=2A^2-2A^{2}cos\theta \\ |\overrightarrow{\mathit{A}}-\overrightarrow{\mathit{B}}{|}^2=2A^2(1-cos\theta )\end{array}$

Dividing equation (2) by (3), we get:

$\frac{|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}{|}^2}{|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}{|}^2}=(100{)}^2=\frac{1+\cos \theta }{1-\cos \theta }$

Solving for we get:

$$\begin{gathered} \theta ={\cos }^{-1}\frac{9999}{10001}=1.15 \\ \theta =1.15° \end{gathered}$$

Therefore, the distance between$\left|\overrightarrow{A}+\overrightarrow{B}\right|$and $\left|\overrightarrow{A}-\overrightarrow{B}\right|$ is $\theta ={1.15}^{°}$