Consider the three points: A=(9, 8), B=(7, 10), C=(10, 6). What is the angle between AB and AC?

I thought you had to use the distance formula and dot product, but nothing seems to be working and I'm just generally confused on how to go about this. Thanks in advance!

2 Answers
Nov 12, 2017

theta ~~ 2.819842" radians" or " 161.565^@θ2.819842 radiansor161.565

Explanation:

The vector vec(AB)AB is:

vec(AB) = (B_x-A_x)hati + (B_y-A_y)hatjAB=(BxAx)ˆi+(ByAy)ˆj

vec(AB) = (7-9)hati + (10-8)hatjAB=(79)ˆi+(108)ˆj

vec(AB) = -2hati + 2hatjAB=2ˆi+2ˆj

The vector vec(AC)AC is:

vec(AC) = (C_x-A_x)hati + (C_y-A_y)hatjAC=(CxAx)ˆi+(CyAy)ˆj

vec(AC) = (10-9)hati + (6-8)hatjAC=(109)ˆi+(68)ˆj

vec(AC) = hati + -2hatjAC=ˆi+2ˆj

To compute the dot product, we multipy the hatiˆi coefficients and add to it the product of the hatjˆj coefficients:

vec(AB)*vec(AC) = (AB_(hati))(AC_(hati))+(AB_(hatj))(AC_(hatj))ABAC=(ABˆi)(ACˆi)+(ABˆj)(ACˆj)

vec(AB)*vec(AC) = (-2)(1)+(2)(-2)ABAC=(2)(1)+(2)(2)

vec(AB)*vec(AC) = -6ABAC=6

Another definition for the dot product is:

vec(AB)*vec(AC) = |vec(AB)||vec(AC)|cos(theta)" [1]"ABAC=ABACcos(θ) [1]

where || indicates the magnitude of the vector and thetaθ is the smallest angle between the two vectors.

Compute |vec(AB)|AB and |vec(AC)|AC:

|vec(AB)| = sqrt(AB_(hati)^2+AB_(hatj)^2)AB=AB2ˆi+AB2ˆj

|vec(AB)| = sqrt((-2)^2+2^2)AB=(2)2+22

|vec(AB)| = sqrt8AB=8

|vec(AB)| = 2sqrt2AB=22

|vec(AC)| = sqrt(AC_(hati)^2+AC_(hatj)^2)AC=AC2ˆi+AC2ˆj

|vec(AC)| = sqrt(1^2+(-2)^2)AC=12+(2)2

|vec(AC)| = sqrt5AC=5

Substituting the known information into equation [1]:

-6 = 2sqrt2sqrt5cos(theta)6=225cos(θ)

Solve for thetaθ:

cos(theta) = -3/sqrt10cos(θ)=310

theta = cos^-1(-3/sqrt10)θ=cos1(310)

theta ~~ 2.819842" radians" or " 161.565^@θ2.819842 radiansor161.565

Nov 12, 2017

arc cos(-3/sqrt10)=pi-arc cos(3/sqrt10).arccos(310)=πarccos(310).

Explanation:

Let, thetaθ be the angle between AB and AC.ABandAC.

Given that, A=A(9,8), B=B(7,10) and C=C(10,6).A=A(9,8),B=B(7,10)andC=C(10,6).

Then, by the Distance Formula,

AB^2=(9-7)^2+(8-10)^2=4+4=8.AB2=(97)2+(810)2=4+4=8.

BC^2=(7-10)^2+(10-6)^2=9+16=25.BC2=(710)2+(106)2=9+16=25.

AC^2=(9-10)^2+(8-6)^2=1+4=5.AC2=(910)2+(86)2=1+4=5.

By the Cosine Formula, then, we have,

costheta=(AB^2+AC^2-BC^2)/(2*AB*AC),cosθ=AB2+AC2BC22ABAC,

=(8+5-25)/(2*sqrt8*sqrt5),=8+525285,

=-12/(4sqrt10).=12410.

rArr theta=arc cos(-3/sqrt10)=pi-arc cos(3/sqrt10).θ=arccos(310)=πarccos(310).