What is the angle between #<0,-7,6> # and #<-1,9,-3>#?

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Answer 1 · 2016-10-19T23:26:56

#theta=arccos(-81/sqrt(7735))~~2.7414#

We'll use the fact that the dot product #vec(u)*vec(v)# can be calculated in two ways:

#vec(u)*vec(v) = u_1v_1+u_2v_2+u_3v_3 = ||vecu||*||vecv||cos(theta)#

Dividing by #||vecu||*||vecv||#, we get

#cos(theta) = (u_1v_1+u_2v_2+u_3v_3)/(||vecu||*||vecv||)#

or

#theta = arccos((u_1v_1+u_2v_2+u_3v_3)/(||vecu||*||vecv||))#

In our case, we have #vecu = < 0, -7, 6 ># and #vecv = < -1, 9, -3 >#.

If we calculate their magnitudes, we get

#||vecu|| = sqrt(0^2+(-7)^2+6^2) = sqrt(85)#
and
#||vecv|| = sqrt((-1)^2+9^2+(-3)^2) = sqrt(91)#

So, the formula we derived gives us

#theta = arccos((0(-1)+(-7)(9)+6(-3))/(sqrt(85)*sqrt(91)))#

#=arccos(-81/sqrt(7735))#

#~~2.7414#