Determine a plane that is parallel with line B and with an angle of 30 with the line A.
#Equations of line and planes
young shoal
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Determine a plane that is parallel with line B and with an angle of 30 with the ...
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young shoal
A plane that is parallel with the line, means that a•b = 0 right so <x,y,z> • <b> = 0 and the angle between the two should be 30
so do I just use the scalar product? However you do have 3 unknowns
This is line A
quick bane
Hm...
This plane needs to have a normal vector perpendicular to the first line. Maybe we can try parametrizing a family of unit vectors perpendicular to that line, and then see which ones (really, just two) make a plane with the required angle to the other line?
young shoal
Wdym by the last bit
Not sure I understood the thought process
So the line is perpendicular to the plane, and the plane has an angle of 30 with the line
this seems quite hard
quick bane
Wdym by the last bit
Let's make up an example.
Suppose we have two lines with direction vectors s1 and s2. We need to find a plane (with normal vector n) that is parallel to line 1 and makes an angle θ with line 2. Note that this is only possible for nonzero θ if s1 and s2 are not parallel.
First, let's take a parametric vector u = {cos(t), sin(t), 0} and find its cross product with s (and call it n):
n = s1⨯u = {s11, s12, s13}⨯{cos(t), sin(t), 0}
We don't care about the value for now, but this does necessarily generate a family of vectors that are perpendicular to s, which is what we need.
Next, we want our plane to make an angle θ with the second line. This means we must have:
n·s2 = |n||s2|sin(θ)
So, substituting n, we get:
(s1⨯u)·s2 = |s1⨯u||s2|sin(θ)
We can square both parts to get rid of the square roots in the magnitudes:
|(s1⨯u)·s2|^2 = |s1⨯u|^2 |s2|^2 sin(θ)^2
This will be a trigonometric equation in t. After finding t you can find n.
young shoal
I don’t really see it at all with this generalized method
quick bane
Well, start by finding s1 and s2 from your case.
young shoal
Ok the direction of both vector is pretty easy, for line A I just take the cross product to find the direction of my line, for B it is simply given
quick bane
Ok the direction of both vector is pretty easy, for line A I just take the cross...
Right. So, in our case (if I didn't mess up):
s1 = {1, -7, 14}
s2 = {0, -1, 2}
Now, we find n in parametric form:
n = s1⨯u = {1, -7, 14}⨯{cos(t), sin(t), 0} = {-14sin(t), 14cos(t), 7cos(t) + sin(t)}
So, now try finding its dot product with s2, then find its square, as well as squared magnitudes of s2 and s1⨯u.
I'll also try doing it so we can check the answer.
n·s2 = {-14sin(t), 14cos(t), 7cos(t) + sin(t)}·{0, -1, 2} = 2sin(t)
(n·s2)^2 = 4sin(t)^2
|n|^2 = 196sin(t)^2 + 196cos(t)^2 + (7cos(t) + sin(t))^2 = 196 + 49cos(t)^2 + 14sin(t)cos(t) + sin(t)^2 = 197 + 48cos(t)^2 + 14sin(t)cos(t)
|s2|^2 = 0 + 1 + 4 = 5
In our case θ = π/6, so sin(θ)^2 = 1/4. So, the equation becomes:
(n·s2)^2 = |n|^2 |s2|^2 sin(θ)^2
4sin(t)^2 = (5/4)(197 + 48cos(t)^2 + 14sin(t)cos(t))
Which doesn't seem to have solutions. Hm...
Actually, is this situation even possible?
These lines seem to be nearly parallel.
young shoal
n·s2 = {-14sin(t), 14cos(t), 7cos(t) + sin(t)}·{0, -1, 2} = 2sin(t)
(n·s2)^2 = 4...
Sorry, I havent really seen any of this so it’s quite hard for me to understand what you’re doing
quick bane
Sorry, I havent really seen any of this so it’s quite hard for me to understand ...
Oh, it's an approach I made up a long time ago for problems like this.
Maybe there are others.
quick bane
Sorry, I havent really seen any of this so it’s quite hard for me to understand ...
Ah, actually, I just noticed that we need to be parallel to the second line, rather. It will still be impossible, but let's see what we have.
s1 = {0, -1, 2}
s2 = {1, -7, 14}
n = s1⨯u = {0, -1, 2}⨯{cos(t), sin(t), 0} = {-2sin(t), 2cos(t), cos(t)}
n·s2 = {-2sin(t), 2cos(t), cos(t)}·{1, -7, 14} = -2sin(t)
|s2|^2 = 1 + 49 + 196 = 246
|n|^2 = 4 + cos(t)^2
(n·s2)^2 = 4sin(t)^2
sin(θ)^2 = 1/4
So, we get:
(n·s2)^2 = |n|^2 |s2|^2 sin(θ)^2
4sin(t)^2 = (123/2)(4 + cos(t)^2)
Still no solution. However, we can instead try to see what angles we can get.
4sin(t)^2 = 246sin(θ)^2 (4 + cos(t)^2)
2sin(t)^2 = 123sin(θ)^2 (5 - sin(t)^2)
(123sin(θ)^2 + 2)sin(t)^2 = 615sin(θ)^2
sin(t)^2 = 615sin(θ)^2/(123sin(θ)^2 + 2)
This needs to be between 0 and 1.
Let u = sin(θ). Then we have a system:
615u^2/(123u^2 + 2) ≥ 0
615u^2/(123u^2 + 2) ≤ 1
The first inequality is unneeded, since this is obviously always nonnegative. The second one we can multiply by (123u^2 + 2), since that's always positive.
615u^2 ≤ 123u^2 + 2
492u^2 ≤ 2
u^2 ≤ 1/246
|u| ≤ 1/√(246)
So, |sin(θ)| ≤ 1/√(246). But since the angle between a line and a plane is always between 0 and π/2, that means θ is nonnegative, so we can find the maximum angle from the equation:
sin(θ) = 1/√(246), 0 ≤ θ ≤ π/2
θ = arcsin(1/√(246))
So, this is the maximum possible angle. Let's work with it.
sin(θ)^2 = 1/246
(n·s2)^2 = |n|^2 |s2|^2 sin(θ)^2
4sin(t)^2 = 4 + cos(t)^2
5sin(t)^2 = 5
sin(t)^2 = 1
sin(t) = ±1
Doesn't matter which value we take, so let's take sin(t) = 1, cos(t) = 0. Then:
n = {-2, 0, 0}
So, our plane is just x = 0. This will make the angle θ = arcsin(1/√(246)) with the line with line A, which is the maximum possible.
young shoal
Ah, actually, I just noticed that we need to be parallel to the second line, rat...
Hmm okay, thanks for helping me!
quick bane
You're welcome!
young shoal
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