What is the angle between #<-3,2,6 > # and #<6,1,-5> #?

1 Answer
Truong-Son N.
May 4, 2016

I got #"2.56 rad"# or about #146.6^@#.

Let #veca = << -3,2,6 >># and #vecb = << 6,1,-5 >>#. This angle between #veca# and #vecb# is found by the following equation:

#\mathbf(veca cdot vecb = ||veca||cdot||vecb||costheta)#

where:

  • #vecacdotvecb# is the dot product between #veca# and #vecb#.
  • #||veca||# is the norm of #veca#.
  • #theta# is the angle between #veca# and #vecb#.

The dot product in #RR^n# for some vectors #veca# and #vecb# is defined as follows:

#\mathbf(vecacdotvecb)#

#= << a_1,a_2, . . . , a_n >> cdot << b_1,b_2, . . . , b_n >>#

#= \mathbf(a_1b_1 + a_2b_2 + . . . + a_nb_n)#

And the norm in #RR^n# for some vector #vecu# is defined as:

#\mathbf(||veca||)#

#= sqrt(vecacdotveca#

#= sqrt(a_1a_1 + a_2a_2 + . . . + a_n a_n)#

#= \mathbf(sqrt(a_1^2 + a_2^2 + . . . + a_n^2))#

So, we evaluate the angle as follows:

#costheta = (veca cdot vecb)/(||veca||cdot||vecb||)#

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

First, we should find the dot product between #veca# and #vecb#.

#color(green)(vecacdotvecb)#

#= << -3,2,6 >> cdot << 6,1,-5 >>#

#= -3*6 + 2*1 + 6*-5#

#= -18 + 2 - 30#

#= color(green)(-46)#

Then, we should find each norm.

#color(green)(||veca||)#

#= sqrt(veca cdot veca)#

#= sqrt(<< -3,2,6 >>cdot<< -3,2,6 >>)#

#= sqrt(-3*-3 + 2*2 + 6*6)#

#= sqrt(9 + 4 + 36)#

#= color(green)(7)#

#color(green)(||vecb||)#

#= sqrt(vecb cdot vecb)#

#= sqrt(<< 6,1,-5 >>cdot<< 6,1,-5 >>)#

#= sqrt(6*6 + 1*1 + -5*-5)#

#= sqrt(36 + 1 + 25)#

#= color(green)(sqrt62)#

Finally, let's get the angle.

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

#= arccos((-46)/(7sqrt62))#

#~~# #color(blue)("2.56 rad")# or about #color(blue)(146.6^@)#.

And you can see that from using Wolfram Alpha on this question as well.

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