Question

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!

Answer 1

`theta ~~ 2.819842" radians" or " 161.565^@`

Explanation:

The vector `vec(AB)` is:

`vec(AB) = (B_x-A_x)hati + (B_y-A_y)hatj`

`vec(AB) = (7-9)hati + (10-8)hatj`

`vec(AB) = -2hati + 2hatj`

The vector `vec(AC)` is:

`vec(AC) = (C_x-A_x)hati + (C_y-A_y)hatj`

`vec(AC) = (10-9)hati + (6-8)hatj`

`vec(AC) = hati + -2hatj`

To compute the dot product, we multipy the `hati` coefficients and add to it the product of the `hatj` coefficients:

`vec(AB)*vec(AC) = (AB_(hati))(AC_(hati))+(AB_(hatj))(AC_(hatj))`

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

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

Another definition for the dot product is:

`vec(AB)*vec(AC) = |vec(AB)||vec(AC)|cos(theta)" [1]"`

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

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

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

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

`|vec(AB)| = sqrt8`

`|vec(AB)| = 2sqrt2`

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

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

`|vec(AC)| = sqrt5`

Substituting the known information into equation [1]:

`-6 = 2sqrt2sqrt5cos(theta)`

Solve for `theta`:

`cos(theta) = -3/sqrt10`

`theta = cos^-1(-3/sqrt10)`

`theta ~~ 2.819842" radians" or " 161.565^@`

Answer 2

`arc cos(-3/sqrt10)=pi-arc cos(3/sqrt10).`

Explanation:

Let, `theta` be the angle between `AB and AC.`

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

Then, by the Distance Formula,

`AB^2=(9-7)^2+(8-10)^2=4+4=8.`

`BC^2=(7-10)^2+(10-6)^2=9+16=25.`

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

By the Cosine Formula, then, we have,

`costheta=(AB^2+AC^2-BC^2)/(2*AB*AC),`

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

`=-12/(4sqrt10).`

`rArr theta=arc cos(-3/sqrt10)=pi-arc cos(3/sqrt10).`