Question

The magnitude of scalar and vector products of two vectors are 48root3 and 144 respectively.What is angle b/w the two vectors??? Give detailed ans plzzz.

Answer 1

$\boxed {<strong>Vectors }$

Final Answer : 60°

Steps:

1) We have,

Scalar Product = $48\sqrt{3}$

Vector Product : $144$

It is pretty Obvious to say, that

$\tan(\theta) = \frac{|Vector\:Product |}{|Scalar\:Product| }</strong></p><p></p><p><strong>$

where $\theta$ is angle betweem two vectors.

2) Using above,

$\tan(\theta) = \frac{144}{ 48 \sqrt{3}}</strong></p><p></p><p><strong>\\ </strong></p><p></p><p><strong>=>\tan( \theta ) = \sqrt{3} \\ </strong></p><p></p><p><strong>=> \theta = 60 \degree$

Hence, angle between two vectors is 60° .

Answer 2

HELLO DEAR,

GIVEN THAT:-

scalar product = 48√3 = A.B.cosØ

Vector product = 144 = A.BsinØ

and hope we know $\bold{\large{tan\theta = \frac{VECTOR\;\; PRODUCT}{SCALAR\;\; PRODUCT}}}$

So, $\bold{\large{tan\theta = \frac{144}{48\sqrt{3}}}}$

$\bold{\large{tan\theta = \sqrt{3}}}$

Hence, $\bold{\boxed{\theta = 60\degree}}$

I HOPE ITS HELP YOU DEAR,
THANKS