Question

Prove the |vector A×vector B |whole square + |A•B|whole square =(AB) whole square

Answer 1

Refer the attachment!

Answer 2

$|A\times B|^2+|A{\cdot} B|^2=(AB)^2$ is proved.

Explanation:

In this case we need to proof the following expression :

$|A\times B|^2+|A{\cdot} B|^2=(AB)^2$

We know that,

$A\times B=|A||B|\ sin\theta$

$(A\times B)^2=A^2B^2\ sin^2\theta$...........(1)

We know that,

$|A{\cdot} B|=|A||B|\ cos\theta$

$|A{\cdot} B|=A^2B^2\ cos^2\theta$.........(2)

Adding equation (1) and (2) as :

= $A^2B^2\ sin^2\theta+A^2B^2\ cos^2\theta$  

= $A^2B^2(sin^2\theta+cos^2\theta)$

Since,

$(sin^2\theta+cos^2\theta)=1$

So,

$|A\times B|^2+|A{\cdot} B|^2=(AB)^2$

Hence, proved

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