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

Question

1632 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T16:19:26

#17.446^\circ#

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 · 2018-07-26T16:20:07

The angle is #=17.45^@#

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^@#