Find the angle in radians between the planes 1x + z = 1 and -3y + z = 1.?

Answer 1

The angle between the two planes is the same as the angle between their normal vectors.

The given planes are:

x + z = 1
-3y + z = 1

Their normal vectors n1 and n2, are taken from the coefficients of x, y, and z. They are:

n1 = <1, 0, 1>
n2 = <0, -3, 1>

First find the magnitude of the normal vectors.

|| n1 || = √[1² + 0² + 1²) = √(1 + 0 + 1) = √2
|| n2 || = √[0² + (-3)² + 1²) = √(0 + 9 + 1) = √10

Now calculate the dot product of the normal vectors.

n1 • n2 = <1, 0, 1> • <0, -3, 1> = 0 + 0 + 1 = 1

The dot product can also be expressed in another way.

n1 • n2 = || n1 || || n2 || cosθ
where θ is the angle between the two vectors

cosθ = (n1 • n2) / (|| n1 || || n2 ||) = 1 / [(√2)(√10)]
cosθ = 1/√20

θ = arccos(1/√20) ≈ 1.3452829 radians

Answer 2

Find the angle in radians between the planes 1x + z = 1 and -3y + z = 1

general equation of a plane is :
ax +by +cz +d =0

normal to the plane is the vector
1x+z =1 ----> Normal N1= <1, 0, 1>
-3y +z =1 ---> Normal N2 = <0, -3 , 1>

using the dot product of the 2 normals:
let's find |N1 . N2| = | (1)(0) + (0)(-3) +(1)(1)| = 1

let's find |N1||N2|
|N1| = sqrt(1^2 +0^2 +1^2) = sqrt(2)
|N2| = sqrt( 0^2 +(-3)^2 +1^2) = sqrt(10)
|N1||N2| = sqrt(20) = 2 sqrt(5)

The angle between the planes is given by
arcos( 1/(2sqrt(5)) = arcos(0.2236) = 1.3453 rads.

Answer 3

The angle between the planes is the same as the angle between the two normal vectors

The two normal vectors are <1, 0, 1> and <0, -3, 1>

The dot product of these two vectors is

adotb = 1(0) + 0(-3) + 1(1) = 1

The magnitudes of the two vectors are

sqrt(2) and sqrt(10) respectively

sqrt(20) = 2*sqrt(5)

Then the formula for the two angles becomes

arccos [(adotb)/(|a||b|)]

That will be your final answer.