If #A = <5 ,2 ,5 >#, #B = <6 ,5 ,3 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2017-07-11T05:19:19

The angle is #=92.1º#

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈5,2,5〉-〈6,5,3〉=〈-1,-3,2〉#

The angle between #vecA# and #vecC# is given by the dot product definition.

#vecA.vecC=∥vecA∥*∥vecC∥costheta#

Where #theta# is the angle between #vecA# and #vecC#

The dot product is

#vecA.vecC=〈5,2,5〉.〈-1,-3,2〉=-5-6+10=-1#

The modulus of #vecA#= #∥〈5,2,5〉∥=sqrt(25+4+25)=sqrt54#

The modulus of #vecC#= #∥(-1,-3,2〉∥=sqrt(1+9+4)=sqrt14#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-1/(sqrt54*sqrt14)=-0.036#

#theta=92.1#º

Answer 2 · 2017-07-11T05:41:01

We can start by finding #vecC#. To subtract vectors, we subtract the corresponding components. For #vecA-vecB#, we have:

#vecC = <5,2,5> - <6,5,3>#

# = <(5-6),(2-5),(5-3)>#

#=<-1,-3,2>#

The angle between two vectors can be found using this general formula:

#cos(theta)=(veca*vecb)/(|veca|*|vecb|)#

We can start by finding the dot product of A and C.

#vecA*vecC=<5,2,5> *<-1,-3,2>#

#=(5*-1)+(2*-3)+(5*2)#

#=-5-6+10#

#=-1#

Now we can find the product of the magnitude of each vector.

#|vecA|=|<5,2,5>#

#=sqrt(5^2+2^2+5^2)#

#=sqrt(54)#

#=3sqrt(6)#

#|vecB|=|<-1,-3,2>#

#=sqrt((-1)^2+(-3)^2+2^2)#

#=sqrt(14)#

So the product of the magnitudes is #3sqrt6*sqrt14=3sqrt(84)=6sqrt(21)#

We now have:

#cos(theta)=-1/(6sqrt21)=-sqrt21/126#

Solving for #theta#:

#theta=arccos(-sqrt21/126)#

#=92 ^o#