Question

Answer the question in the following attachment​

Answer 1

Given :

|a + b| = |a - b|

To find :

The angle between a and b

Solution :

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,

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

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

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

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

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

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

$\sf cos \theta = 0$

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

$\sf \theta = 90 \degree$

Thus, the angle between a and b is 90°.

Answer 2

Hola ⚘⚘

Given:

• |A + B| = |A| - |B|

Tofind:

• angle between A and b

Calculation:

>> √a²+b²+2abcosθ = √a²+b²-2abcosθ

>> a²+b²+2abcosθ = a²+b²-2abcosθ

>> 2abcosθ + 2abcosθ = 0

>> 4ab cosθ = 0

>> cosθ = 0 = cos 90°

>> θ = 90°

The angle between a and b is 90°