Home > Trigonometry > Vectors

If #A = <5 ,2 ,8 >#, #B = <2 ,5 ,6 ># and #C=A-B#, what is the angle between A and C?

Answer 1

Daniel Acosta

The angle is #=56.4#º

et's start by calculating

#vecC=vecA-vecB#

#vecC=〈5,2,8〉-〈2,5,6〉=〈3,-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=〈5,2,8〉.〈3,-3,2〉=15-6+16=25#

The modulus of #vecA#= #∥〈5,2,8〉∥=sqrt(25+4+64)=sqrt93#

The modulus of #vecC#= #∥〈3,-3,2〉∥=sqrt(9+9+4)=sqrt22#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=25/(sqrt93*sqrt22)=0.55#

#theta=56.4#º

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

If #A = &lt;5 ,2 ,8 &gt;#, #B = &lt;2 ,5 ,6 &gt;# and #C=A-B#, what is the angle between A and C?

If #A = <5 ,2 ,8 >#, #B = <2 ,5 ,6 ># and #C=A-B#, what is the angle between A and C?