Question

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

Short answer

The angle between two vector is$2{\tan }^{-1}\left(\frac{1}{n}\right)$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The twovectors$\overrightarrow{A}$and$\overrightarrow{B}$have precisely equal magnitudes.

The magnitude of $\overrightarrow{A}+\overrightarrow{B}$to be larger than the magnitude of $\overrightarrow{A}-\overrightarrow{B}$by the factor n.

Diagram according to the question

Two vectors with the same magnitude of A are $\overrightarrow{A}$and$\overrightarrow{B}$ . R is the magnitude of the two vectors $\overrightarrow{A}$and $\overrightarrow{B}$added together. The flowing graphic depicts a vector diagram for the addition of two vectors.

Here, is the angle$\theta$ made by vector $\overrightarrow{B}$.

Application of law of cosines in A→+B→

After the application of law of cosines to find the magnitudes of $\overrightarrow{A}+\overrightarrow{B}$.

$\begin{array}{rcl}R& =& |\overrightarrow{A}+\overrightarrow{B}|\\ & =& \sqrt{A^2+B^2+2AB\cos \theta }\\ & =& \sqrt{A^2+A^2+2A\left(A\right)\cos \theta }(As,A=B)\\ & =& \sqrt{2A^2+2A^2\cos \theta }\\ R& =& A\sqrt{2(1+\cos \theta )}\end{array}$

$\begin{array}{rcl}R& =& A\sqrt{2\left(2{\cos }^2\frac{\theta }{2}\right)}\left(As,1+\cos \theta =2{\cos }^2\frac{\theta }{2}\right)\\ & =& 2A\cos \frac{\theta }{2}\end{array}$

Magnitude and angle of vector

Here D is the magnitude of difference between $\overrightarrow{A}$and $\overrightarrow{B}$.is the angle made by$\overrightarrow{B}$ vector.

Application of law of cosines in A→-B→

After the application of law of cosines to find the mignitudes of $\overrightarrow{A}-\overrightarrow{B}$.

$\begin{array}{rcl}D& =& |\overrightarrow{A}-\overrightarrow{B}|\\ & =& \sqrt{A^2+B^2-2AB\cos \theta }\\ & =& \sqrt{A^2+A^2-2A\left(A\right)\cos \theta }(As,A=B)\\ & =& \sqrt{2A^2-2A^2\cos \theta }\\ & =& A\sqrt{2(1-\cos \theta )}\\ & =& A\sqrt{2\left(2{\sin }^2\frac{\theta }{2}\right)}\left(As,1-\cos \theta =2{\sin }^2\frac{\theta }{2}\right)\\ & =& 2A\sin \frac{\theta }{2}\end{array}$

Ratio between A→-B→ and A→+B→

After the substitution of values, The ratio of magnitude of $\overrightarrow{A}+\overrightarrow{B}$and $\overrightarrow{A}-\overrightarrow{B}$will be got:

$\frac{R}{D}=n$

Substitute all the value in the above equation.

$\frac{2A\cos \frac{\theta }{2}}{2A\sin \frac{\theta }{2}}=n$

$\begin{array}{rcl}\tan \frac{\theta }{2}& =& \frac{1}{n}\\ \frac{\theta }{2}& =& {\tan }^{-1}\left(\frac{1}{n}\right)\\ \theta & =& 2{\tan }^{-1}\left(\frac{1}{n}\right)\end{array}$

The angle between two vector is$2{\tan }^{-1}\left(\frac{1}{n}\right)$