vectors question?

Question

how do i find the angle of intersection between the two planes given by:

3x + 4y + 5z = 8
2x - 6y - 3z = 0

all i know how to do is find the dot product of vectors, how to find their modulus and how to find the angle between them from that but i dont see how it applies here. can anybody help out?
cheers.

its that bit with the 8 that confuses me

Answer 1

How do i find the angle of intersection between the two planes given by:

3x + 4y + 5z = 8
2x - 6y - 3z = 0
________________

The dihedral angle θ between the two intersecting planes is the same as the angle between the normal vectors n1 and n2 of the two planes.

n1 = <3, 4, 5>
n2 = <2, -6, -3>

Calculate the magnitudes of the normal vectors.

|| n1 || = √[3² + 4² + 5²] = √(9 + 16 + 25) = √50 = 5√2
|| n2 || = √[2² + (-6)² + (-3)²] = √(4 + 36 + 9) = √49 = 7

Take the dot product.

n1 • n2 = <3, 4, 5, > • <2, -6, -3> = 6 - 24 - 15 = -33

The dot product can also be calculated as:

n1 • n2 = || n1 || || n2 || cosθ

cosθ = (n1 • n2) / (|| n1 || || n2 ||) = -33 / [(5√2)*7]
cosθ = -33 / (35√2)

θ = arccos[-33 / (35√2)] ≈ 131.81293°

Answer 2

Write down the normal vector for each plane.
The angle between the planes is the angle between the normal vectors.

You know the modulus of each vector and you can find their dot product. So you therefore know the cosine of the angle between them, and hence the angle.


the 8 doesn't matter. If you change the 8 to any other number, you get a parallel plane. So changing the constant term does not affect the normal vector, and the angle with the other plane won't change