Question

find the angle between two vectors a= 2i-4j+6k and b=3i+j+2k

Answer 1

Hi ..dear...
here is your solution.......the angle between vector is 60°

Answer 2

Answer : The angle between a and b is 60.

Explanation:

We have two vectors

$\vec{a}=2i-4j+6k$

$\vec{b}=3i+j+2k$

We know that

$\vec{a}.\vec{b}=|a||b|\ cos\theta$

$|a|=\sqrt{(2)^2+(-4)^2+(6)^2}=\sqrt{56}$

$|b|=\sqrt{(3)^2+(1)^2+(2)^2}=\sqrt{14}$

$\cos\theta=\dfrac{(2i-4j+6k)(3i+j+2k)}{\sqrt{56}\sqrt{14}}$

$cos\theta=\dfrac{1}{2}$

$\theta=60^0$

Hence, this is the required solution.