What is the angle between <1,3,9 ><1,3,9> and < 4,9,2 ><4,9,2>?
1 Answer
The angle between the two vectors
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:
-
Side
vecA=<1,3,9>→A=<1,3,9> , the length of which is the magnitude of the vector=|vecA|=sqrt((sqrt(1^2+3^2))^2+9^2)=∣∣∣→A∣∣∣=√(√12+32)2+92
|vecA|=sqrt(91)∣∣∣→A∣∣∣=√91 -
Side
vecB=<4,9,2>→B=<4,9,2> , the length of which is the magnitude of the vector=|vecB|=sqrt((sqrt(4^2+9^2))^2+2^2)=∣∣∣→B∣∣∣=√(√42+92)2+22
|vecB|=sqrt(101)∣∣∣→B∣∣∣=√101 -
Side
vecC=vecB-vecA=<3,6,-7>→C=→B−→A=<3,6,−7> , the length of which is the magnitude of the vector=|vecC|=sqrt((sqrt(3^2+6^2))^2+(-7)^2)=∣∣∣→C∣∣∣=√(√32+62)2+(−7)2
|vecC|=sqrt(94)∣∣∣→C∣∣∣=√94 -
Three angles, one of which (the angle between
vecA→A andvecB→B =theta=θ ) can be calculated using the law of cosines:
(cos(theta)=( |vecA|^2 +| vecB|^2 -|vecC|^2)/(2|vecA|*|vecB|))⎛⎜ ⎜ ⎜ ⎜⎝cos(θ)=∣∣∣→A∣∣∣2+∣∣∣→B∣∣∣2−∣∣∣→C∣∣∣22∣∣∣→A∣∣∣⋅∣∣∣→B∣∣∣⎞⎟ ⎟ ⎟ ⎟⎠ (MathIsFun, 2014)
cos(theta)=(91+101-94)/(2*sqrt(91)*sqrt(101))cos(θ)=91+101−942⋅√91⋅√101
cos(theta)=0.5111cos(θ)=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].
