Question

if |A+B| = |A-B| then angle between these vectors is?

Answer 1

Answer:

|A+B| = |A-B|

Square both sides:

|A+B|^2 = |A-B|^2

The magnitude of a vector V is the square root of the dot product with itself, i.e.

|V| = sqrt(V*V), so:

(A+B)*(A+B) = (A-B)*(A-B)

A*A + 2A*B + B*B = A*A - 2A*B + B*B

or 2A*B = -2A*B

Thus, A*B = 0, making the angle between them 90 degrees.

Answer 2

Answer :

Given :

$\dag\sf\:|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$

To Find :

Angle between both vectors.

Solution :

★ As per parallelogram law of vector addition, magnitude of resultant vector R of two vectors A and B inclined at an angle θ is given by

$\sf:\implies\:|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+B^2+2ABcos\theta}$

★ Formula of vector subtraction :

$\sf:\implies\:|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{A^2+B^2-2ABcos\theta}$

ATQ,

$:\implies\sf\:|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$

$\sf:\implies\:\sqrt{A^2+B^2+2ABcos\theta}=\sqrt{A^2+B^2-2ABcos\theta}$

$\sf:\implies\:A^2+B^2+2ABcos\theta=A^2+B^2-2ABcos\theta$

$:\implies\sf\:4ABcos\theta=0$

$:\implies\sf\:cos\theta=0$

$:\implies\:\underline{\boxed{\bf{\purple{\theta=90^{\circ}}}}}$