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Question: If \(| \overrightarrow { \mathrm { A } } + \overrightarrow { \mathrm { B } } | = | \overrightarrow ...

If ∣A→+B→∣=∣A→−B→∣| \overrightarrow { \mathrm { A } } + \overrightarrow { \mathrm { B } } | = | \overrightarrow { \mathrm { A } } - \overrightarrow { \mathrm { B } } | , then the angle between will be

A

30∘30 ^ { \circ }

B

45∘45 ^ { \circ }

C

60∘60 ^ { \circ }

D

90∘90 ^ { \circ }

Answer

90∘90 ^ { \circ }

Explanation

Solution

Let θ\theta be the angle between the vectors and .

Then

∣A→−B→∣=A2+B2+2ABcos⁡θ| \overrightarrow { \mathrm { A } } - \overrightarrow { \mathrm { B } } | = \sqrt { \mathrm { A } ^ { 2 } + \mathrm { B } ^ { 2 } + 2 \mathrm { AB } \cos \theta }

∣A→+B→∣=A2+B2−2ABcos⁡θ| \overrightarrow { \mathrm { A } } + \overrightarrow { \mathrm { B } } | = \sqrt { \mathrm { A } ^ { 2 } + \mathrm { B } ^ { 2 } - 2 \mathrm { AB } \cos \theta }

According to given problem

∣A→+B→∣=∣A→−B→∣| \overrightarrow { \mathrm { A } } + \overrightarrow { \mathrm { B } } | = | \overrightarrow { \mathrm { A } } - \overrightarrow { \mathrm { B } } |

∴A2+B2+2ABcos⁡θ=A2+B2−2ABcos⁡θ\therefore \sqrt { \mathrm { A } ^ { 2 } + \mathrm { B } ^ { 2 } + 2 \mathrm { AB } \cos \theta } = \sqrt { \mathrm { A } ^ { 2 } + \mathrm { B } ^ { 2 } - 2 \mathrm { AB } \cos \theta } Squaring both sides, we get

As A

θ=π2\theta = \frac { \pi } { 2 }