Question

If |a+b| = |a - b| prove that the angle between a and b is 90°. where a and b are vectors​

Answer 1

Answer:

Hlo Mate...

Hope this is helpful..

Answer 2

Given:

• | a + b | = | a - b | , where a and b are vectors.

To prove:

Angle between a and b is 90°.

Proof:

Let the angle between $\vector{a}$ & $\vector{b}$ be $\theta$.

Now, we know that (as per question):

$| \vec{a} + \vec{b}| = | \vec{a} - \vec{b}|$

$\implies \: \sqrt{ {a}^{2} + {b}^{2} + 2.a.b. \cos( \theta) } = \sqrt{ {a}^{2} + {b}^{2} + 2.a.b . \cos( {180}^{ \circ} - \theta ) }$

• $\cos(180^{\circ}-\theta)=-\cos(\theta)$

$\implies \: \sqrt{ {a}^{2} + {b}^{2} + 2.a.b. \cos( \theta) } = \sqrt{ {a}^{2} + {b}^{2} - 2.a.b . \cos( \theta) }$

• Squaring both sides:

$\implies \: {a}^{2} + {b}^{2} + 2.a.b. \cos( \theta) = {a}^{2} + {b}^{2} - 2.a.b . \cos( \theta)$

$\implies \: 2.a.b. \cos( \theta) = - 2.a.b . \cos( \theta)$

$\implies \: \cos( \theta) = - \cos( \theta)$

$\implies \: 2\cos( \theta) = 0$

$\implies \: \cos( \theta) = 0$

$\implies \: \theta = {90}^{ \circ}$

[Hence proved]