If #veca = <8 ,1 ,-5 >#, #vecb = <6 ,-2 ,-8 ># and #vecc=veca-vecb#, what is the angle between #veca# and #vecc#?

1 Answer
Truong-Son N.
May 4, 2016

I get about #84.84^@# or #"1.48 rad"#.

The angle between #veca# and #vecc# is ultimately determined from the following relationship:

#\mathbf(vecacdotvecc = || veca ||cdot|| vecc || costheta)#

Therefore, we will need to find #vecc# by subtracting #veca# and #vecb#.

#color(green)(veca - vecb)#

#= << 8,1,-5 >> - << 6,-2,-8 >>#

#= << 8-6, 1-(-2), -5-(-8) >>#

#= color(green)(<< 2,3,3 >> = vecc)#

Next, the angle between #veca# and #vecc# is gotten like so:

#costheta = (vecacdotvecc)/(|| veca ||cdot|| vecc ||)#

#color(green)(theta = arccos((vecacdotvecc)/(|| veca ||cdot|| vecc ||)))#

Afterwards, we do not yet know #veca cdot vecc#, so we'll need to evaluate the dot product between #veca# and #vecc#, which is just a sum of product terms consisting of corresponding vector coordinates.

#color(green)(veca cdot vecc)#

#= << 8,1,-5 >> cdot << 2,3,3 >>#

#= 8*2 + 1*3 + -5*3#

#= 16 + 3 - 15 = color(green)(4)#

And lastly, we'll need to know the product of the norms of #veca# and #vecc#. The norm of some arbitrary vector #vecv# is

#\mathbf(|| vecv || = sqrt(vecvcdotvecv)),#

so we'll need to dot #veca# and #vecc# with themselves and take the square root to get #|| veca ||# and #|| vecc ||#, respectively.

#color(green)(|| veca ||) = sqrt(veca cdot veca)#

#= sqrt(<< 8,1,-5 >>cdot<< 8,1,-5 >>)#

#= sqrt(8*8 + 1*1 + (-5)*(-5))#

#= sqrt(64 + 1 + 25)#

#= color(green)(sqrt(90))#

And now for #vecc#:

#color(green)(|| vecc ||) = sqrt(vecc cdot vecc)#

#= sqrt(<< 2,3,3 >>cdot<< 2,3,3 >>)#

#= sqrt(2*2 + 3*3 + 3*3)#

#= sqrt(4+9+9)#

#= color(green)(sqrt(22))#

Finally, we can combine all this to get the angle between #veca# and #vecc# as:

#color(blue)(theta) = arccos((vecacdotvecc)/(|| veca ||cdot|| vecc ||))#

#= arccos(4/(sqrt(90)*sqrt(22)))#

#= arccos(4/(3sqrt(10)*sqrt(22)))#

#= arccos(4/(3sqrt(220)))#

#= arccos(cancel(4)^(2)/(3*cancel(2)sqrt(55)))#

#= arccos(2/(3*sqrt(55)))#

#~~# #color(blue)("1.48 rad")# or about #color(blue)(84.84^@)#.

Indeed, that should be the answer, as we see the result here to be:

#arccos(2/(3sqrt55))#.

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