Question

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

Answer 1

`:. theta=cos^-1(29/(sqrt3996))~~` `(62.69)^circ`

Explanation:

Here,

`A=<5,2,-5> and B=<6,5,3>`

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

`C= <-1,-3-8>`

Let , `veca=<5,2,-5> and vecc=<-1,-3,-8>` be

the two vectors.

`:.|veca|=sqrt((5)^2+(2)^2+(-5)^2)=sqrt(25+4+25)=sqrt54`

`|vec c|=sqrt((-1)^2+(-3)^2+(-8)^2)=sqrt(1+9+64)`=`sqrt74`

Dot product : `a*c= <5,2,-5> *<-1,-3,-8>`

`:.a*c=(5)xx(-1)+(2)xx(-3)+(-5)xx(-8)`

`:.a*c=-5-6+40=29`

Now the angle between `veca and vecc` is:

`theta=cos^-1((a*c)/(|veca||vecc|))=cos^-1((29)/(sqrt54sqrt74))`

`:. theta=cos^-1(29/(sqrt3996))~~` `(62.69)^circ`

Answer 2

Angle between `vec A and vec C` is `62.69^0`

Explanation:

`vec A =<5,2,-5> and vec B =<6,5,3>`

`vec C=vec A- vecB = (< 5,2,-5 >) - (< 6,5,3 >)`

`:. vec C=< 5-6,2-5,-5-3 > =< -1,-3,-8 >`

Let `theta` be the angle between `vec A and vec C` ; then we

know `cos theta= (vec A*vec C)/(||vec A||*||vec C||)` or

`cos theta=`

`((5*-1)+(2*-3)+(-5*-8))/(sqrt(5^2+2^2+(-5)^2)* sqrt((-1)^2+(-3)^2+(-8)^2))`

or `cos theta= 29/(sqrt 54*sqrt 74)=29/63.21 ~~0.4588`

`:. theta=cos^-1(0.4588)~~62.69^0`

The angle between `vec A and vec C` is `62.69^0` [Ans]