Question

If `A = <5 ,2 ,8 >`, `B = <6 ,5 ,3 >` and `C=A-B`, what is the angle between A and C?

Answer 1

Depending upon the sense of measurement (clockwise or opposite),

the angle is `59.45^o or 121.55^o`

Explanation:

The angle between the vectors A and C is

`theta="arcsin"(|A xx C|/(|A||B|))` or `theta = pi-"arcsin"(|A xx C|/(|A||B|))`

Here, `A=<5, 2, 8>`, and

`C = A - B =<5, 2, 8> - <6, 5, 3> = <-1, -3, 5>`.

`A xx C = <5, 2, 8> xx <-1, -3, 5>`

`<(2)(5)-(8)(-3), (8)(-1)-(5)(5), (5)(-3)-(2)(-1)>`

`=<34, -33, -13>`
So,

`|A|=sqrt(5^2+2^2+8^2)=sqrt 93`

`|C| = sqrt((-1)^2+(-3)^2+5^2)=sqrt 35`

`|A xx C|=sqrt((34)^2+(-33)^2+(-13)^2)=sqrt 2414`.

The angle

`theta = "arcsin" sqrt(2414/((93)(35)))=59.49^o`

Depending upon the sense of measurement (clockwise or opposite),

the angle is `59.49^o or 121.55^o`

Answer 2

I got `59.45^@` (using Cosine formula ...see A.S.Adikesavan's answer using Sine formula)

Explanation:

My understanding is that:

`color(blue)"-------------------------------------------------------"`
`color(blue)("| ")"Given vectors: "`
`color(blue)("| ")color(white)("XXX")color(blue)( vecu=< u_1,u_2,u_3 >" and"color(white)("XXXXX")color(blue)("|"`
`color(blue)("| ")color(white)("XXX")color(blue)(vecv = < v_1,v_2,v_3>color(white)("XXXXXxXXX")color(blue)("|")`

`color(blue)("| ")`For an angle `color(blue)(theta)` between `color(blue)(vecu)` and `color(blue)(vecv)` `color(white)("XXXXxX")color(blue)("|")`
`color(blue)("| ") color(white)("XXX")color(blue)(sin(theta)=(vecu * vecv)/(abs(vecu)*abs(vecv))color(white)("XXXXXXXxXX")color(blue)("|")`

`color(blue)("| ")`where`color(white)("XXXXXXXXXXXX")`
`color(blue)("| ")color(white)("XXX")color(blue)(vecu * vecv = u_1 * v_1+u_2 * v_2 + u_3 * v_3)color(white)("X")color(blue)("|")`
`color(blue)("| ")color(white)("XXX")color(blue)(abs(vecu)=sqrt(u_1^2+u_2^2+u_3^2))color(white)("XXXXXXxX")color(blue)("|")`
`color(blue)("| ")color(white)("XXX")color(blue)(abs(vecv)=sqrt(v_1^2+v_2^2+v_3^2))color(white)("XXXXXXXX")color(blue)("|")`
`color(blue)"-------------------------------------------------------"`

In this case, we are given
`color(white)("XXX")color(red)(vecA= < 5,2,8 >), color(red)(vecB= < 6,5,3>), and color(red)(vecC=vecA-vecB)`
`color(white)("XXX")rarr color(red)(vecC=< -1,-3,5 >)`

`color(red)(vecA) * color(red)(vecC)=(5)(-1)+(2)(-3)+(8)(5) color(red)(= 29)`

`color(red)(abs(vecA))=sqrt(5^2+2^2+8^2)color(red)(=sqrt(93))`
`color(red)(abs(vecC))=sqrt(1^1+3^2+5^2)color(red)(=sqrt(35))`

Therefore
`color(white)("XXX")color(green)(cos(theta))=color(red)(29)/(color(red)(sqrt(93) * sqrt(35)))`

Using a calculator:
`color(white)("XXX")color(green)(cos(theta))=0.508303`
and
`color(white)("XXX")color(green)(theta) = "arccos"(0.508303) = color(green)(59.45^@)`