Question

Angle between Vector A( 2i + 3j + k) and Vector B (-3i + 6k)

Answer 1

Hey There!!

We are given two vectors:

$\vec{a}=2\hat{\imath}+3\hat{\jmath}+\hat{k} \\ \\ \vec{b}=-3\hat{\imath}+6\hat{k}$


Let the angle between them be $\theta$

To find the angle, we use the dot product.


$\vec{a} . \vec{b} = \mid \vec{a} \mid \, \mid \vec{b} \mid \, \cos \theta \\ \\ \\ \implies (2\hat{\imath}+3\hat{\jmath}+\hat{k}).(-3\hat{\imath}+6\hat{k}) =\mid \vec{a} \mid \, \mid \vec{b} \mid \, \cos \theta \\ \\ \\ \implies -6+6 = \mid \vec{a} \mid \, \mid \vec{b} \mid \, \cos \theta \\ \\ \\ \implies 0 = \mid \vec{a} \mid \, \mid \vec{b} \mid \, \cos \theta \\ \\ \\ \implies \cos \theta = 0 \\ \\ \\ \implies \boxed{\theta=90^{\circ}}$


Thus, both vectors are perpendicular.



Hope it helps
Purva
Brainly Community

Answer 2

Answer:

it is 90 degree fgfdhjfdsfkkfxch