Physics, asked by Hemanth230304, 11 months ago

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

Answers

Answered by NehaAlfassa
16

Answer:

Hlo Mate...

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Answered by nirman95
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]

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