Question

What is the angle between `<1,3,9 >` and `< 4,9,2 >`?

Answer 1

The angle between the two vectors `=59.2622^0`

Explanation:

In order to find the angle between the two vectors, it helps to take the vectors in standard form (their tails are at the origin).Then take them as two sides of a triangle, and the vector of their difference would be the third side of that triangle.

So the triangle will have:

1. Side `vecA=<1,3,9>`, the length of which is the magnitude of the vector`=|vecA|=sqrt((sqrt(1^2+3^2))^2+9^2)`
   `|vecA|=sqrt(91)`

2. Side `vecB=<4,9,2>`, the length of which is the magnitude of the vector`=|vecB|=sqrt((sqrt(4^2+9^2))^2+2^2)`
   `|vecB|=sqrt(101)`

3. Side `vecC=vecB-vecA=<3,6,-7>`, the length of which is the magnitude of the vector`=|vecC|=sqrt((sqrt(3^2+6^2))^2+(-7)^2)`
   `|vecC|=sqrt(94)`

4. Three angles, one of which (the angle between `vecA`and`vecB` `=theta`) can be calculated using the law of cosines:
   `(cos(theta)=( |vecA|^2 +| vecB|^2 -|vecC|^2)/(2|vecA|*|vecB|))`(MathIsFun, 2014)
   `cos(theta)=(91+101-94)/(2*sqrt(91)*sqrt(101))`
   `cos(theta)=0.5111`
   `:.theta=cos^(-1)0.5111`
   `theta=59.2622^0`

References:
-MathIsFun, 2014. The Law of Cosines . [Online]. Available from:https://www.mathsisfun.com/algebra/trig-cosine-law.html [Accessed:13th Jan 2016].