For what values of a is the angle between the plane (below) and the xz plane 60 degrees?

Question

the equation of the plane is: ax+ay+z=a

Answer 1

The equation of the given plane must be put into a general form:
Ax + By + Cz + D = 0
in this case, A = 1/a, B = 1/a, C = 1, D = -a
The equation of the xz plane is
x + z = 0 or A = 1, B = 0, C = 1, D = 0
Now the cos of the angle between two planes is given by
cos(p1,p2) = |(1/a)(1) + 0 + (1/a)(1)|/{sqr[1/a^2 +1/a^2 + 1]sqr[1+0+1]}
= [2/a]/{sqr[2/a^2 + 1]sqr[2]}

Answer 2

The general equation of a plane is of the form:

ax + by + cz + d = 0

The xz plane is y = 0.
0x + y + 0z + 0 = 0

The other plane is
ax + ay + z - a = 0

The angle between the planes is the same as the angle between the normal vectors of the planes.

The normal vector N1 of the first plane is
N1 = <0, 1, 0>

Its magnitude is
|| N1 || = √(0² + 1² + 0²) = √1 = 1

The normal vector N2 of the second plane is
N2 =

Its magnitude is
|| N2 || = √(a² + a² + 1²) = √(2a² + 1)

Take the dot product of N1 and N2.
N1 • N2 = <0, 1, 1> • = 0 + a + 0 = a

The dot product of two vectors N1 and N2 is:

N1 • N2 = || N1 || || N2 || cosθ
where θ is the angle between them

cosθ = (N1 • N2) / (|| N1 || || N2 ||) = a / [1*√(2a² + 1)]
cosθ = a / √(2a² + 1)

But θ = 60°

cos60° = 1/2 = a / √(2a² + 1)
√(2a² + 1) = 2a

Square both sides
2a² + 1 = 4a²
2a² = 1
a² = 1/2
a = 1/√2

The negative solution must be rejected since the cosine is positive.
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