What is the angle between #<0,9,-6 > # and #<1,3,5> #?

1 Answer
Dec 2, 2017

The angle is #=92.7^@#

Explanation:

The vectors are #vecA= < 0,9,-6># and #vecB= <1,3,5>#

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=〈0,9,-6〉.〈1,3,5〉=(0xx1)+(9xx3)+(-6xx5)=0+27-30=-3#

The modulus of #vecA#= #∥〈0,9,-6〉∥=sqrt(0+9^2+(-6)^2)=sqrt(81+36)=sqrt(117)#

The modulus of #vecB#= #∥〈1,3,5〉∥=sqrt(1^2+3^2+5^2)=sqrt(1+9+25)=sqrt35#

So,

#costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-3/(sqrt117*sqrt35)=-0.0469#

#theta=arccos(-0.0469)=92.7^@#