Question

If $\vec{\alpha}$ and $\vec{\beta}$ be two vectors such that $|\vec{\alpha}| = 8$ , $|\vec{\beta}| = 6$ and $\vec{\alpha} \cdot \vec{\beta} = 0$, find the value of $|\vec{\alpha} \times \vec{\beta}|$

Answer 1

let alfa=x and beta=y
then let's proceed to our solution
as non of x and y is 0 and x.y=0 so angle between them should be 90°[ as cos (theta)=0=>theta=90°]
|X×Y|=|x| |y| sin(theta)
=8×6×sin (90°)=48×1=48

Answer 2

$|\vec{\alpha}| = 8$
$|\vec{\beta}| = 6$

and

$\vec{\alpha} \cdot \vec{\beta} = 0$
$\implies \cos \theta = 0$
$\implies \theta = 90 °$

so,

$| \vec{\alpha} \times \vec{\beta} | = 6 \cdot 8 \cdot \sin 90°$

or,
$| \vec{\alpha} \times \vec{\beta} | = 48$

hope you got help from my answer if yes then please mark my answer as brainliest ☺️☺️