Question

if |a+b| = |a-b| prove that the angle between a and b is 90°​

Answer 1

Explanation:

|A+B|=|A-B|

Squaring both sides,

|A+B|^2=|A−B|^2

Since A.A=|A|^2

|A+B|^2=|A−B|^2

(A+B).(A+B)=(A−B).(A−B)

(A.A)+(A.B)+(B.A)+(B.B)=(A.A)−(A.B)−(B.A)+(B.B) Using distributive property)

|A|^2+2A.B+|B|^2=|A|^2−2A.B+|B|^2

4(A.B)=0

A.B=0

|A||B|cos(theta)=0

Since A and B are non-zero vectors, A and B must be perpendicular as cos (90°) = 0.

$hope \: it \: helps \boxed:$

Answer 2

Given :

|a + b| = |a - b|

To prove :

The angle between a and b is 90°

Proof :

we know that,

If two vectors of magnitudes are acting at an angle, then the magnitude of their resultant is given by parallelogram law that is,

$\sf |a + b| = \sqrt{ {a}^{2} + {b}^{2} + 2abcos \theta }$

similarly,

$\sf |a - b| = \sqrt{ {a}^{2} + {b}^{2} - 2abcos \theta }$

According to Question,

$\sf |a + b| = |a - b|$

$\sf \sqrt{ {a}^{2} + {b}^{2} + 2abcos \theta } = \sqrt{ {a}^{2} + {b}^{2} - 2abcos \theta }$

$\sf {a}^{2} + {b}^{2} + 2abcos \theta = {a}^{2} + {b}^{2} - 2abcos \theta$

$\sf \not{a}^{2} + \not {b}^{2} + 2abcos \theta - \not{a}^{2} - \not{b}^{2} + 2abcos \theta = 0$

$\sf 2ab \: cos \theta + 2ab \: cos \theta = 0$

$\sf 4ab \: cos \theta = 0$

$\sf cos \theta = 0$

$\sf cos \theta = cos \: 90 \degree$

$\sf \theta = 90 \degree$

Hence proved !