Finding the angle(In Radians) between 2 vectors?

Question

x=2i-j+2k,Y=5i+j-3k

(a)Find the angle in radians between them

(b) Find a vector of length 1 that is perpendicular to both x and y

Answer 1

Given vectors X and Y.

X = 2i - j + 2k

Y = 5i + j - 3k

(a) Find the angle θ, in radians between them.

Calculate the magnitue of the vectors.  We will use these later.

|| X || = √[2² + (-1)² + 2²) = √(4 + 1 + 4) = √9 = 3

|| Y || = √[5² + 1² + (-3)²] = √(25 + 1 + 9) = √35

Now calculate the dot product of X and Y.

X • Y = <2, -1, 2> • <5, 1, -3> = 10 - 1 - 6 = 3

The dot product can also be expressed in another way.

X • Y = || X || || Y || cosθ

3 = (3√35)cosθ

cosθ = 1/√35

θ = arccos(1/√35) ≈ 1.40095 radians

________

(b) Find a vector of length 1 that is perpendicular to both X and Y.

First take the cross product.

n = X x Y = <2, -1, 2> x <5, 1, -3> = <1, 16, 7>

Calculate the magnitude of vector n.

|| n || = √(1² + 16² + 7²) = √(1 + 256 + 49) = √306

Divide by the magnitude of n to get a unit vector perpendicular to both vectors X and Y.

n/√306 = <1/√306, 16/√306, 7/√306>

Answer 2

a) use the dot product

theta = 1.4010

x.y = |x| |y| cos(theta)  => cos(theta) = x.y / (|x| |y|)

x.y = 10-1-6 = 3

|x| = sqrt(4+1+4) = 3

|y| = sqrt(25+1+9) = sqrt(35)

cos(theta) = 1/sqrt(35)

acos(1/sqrt(35)) = 1.4010

b) use the cross product

You'll find (1/sqrt(306), 16/sqrt(306), 7/sqrt(306))

v = x cross y = (3-2, 10+6, 2+5) = (1, 16, 7)

All you have to do is divide by the norm to get a vector of length 1

|v| = sqrt(1 + 256 + 49) = sqrt(306)

u = (1/sqrt(306), 16/sqrt(306), 7/sqrt(306))