Question

What is the angle between `<2,9,-2>` and `<0,3?0 >`?

Answer 1

`17.446^\circ`

Explanation:

The angle `\theta` between the vectors `<2, 9, -2>` & `<0, 3, 0>` is given as

`\cos\theta=\frac{(2i+9j-2k)\cdot (0i+3j+0k)}{\sqrt{2^2+9^2+(-2)^2}\sqrt{0^2+3^2+0^2}}`

`\cos\theta=\frac{0+27+0}{\sqrt{89}\cdot 3}`

`\cos\theta=\frac{9}{\sqrt{89}}`

`\theta=\cos^{-1}(9/\sqrt89)`

`\theta=17.446^\circ`

Answer 2

The angle is `=17.45^@`

Explanation:

The angle between `vecA` and `vecB` is given by the dot product definition.

`vecA.vecB=∥vecA∥*∥vecB∥costheta`

Where `theta` is the angle between `vecA` and `vecB`

The dot product is

`vecA.vecB=〈2,9,-2〉.〈0,3,0〉=0+27-0=27`

The modulus of `vecA`= `∥〈2,9,-2〉∥=sqrt(4+81+4)=sqrt89`

The modulus of `vecB`= `∥〈0,3,0〉∥=sqrt(9)=3`

So,

`costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=27/(sqrt89*3)=0.954`

`theta=arccos(0.954)=17.45^@`