<Question>
The set of points $(x,y,z)$ that satisfy
\[2x = 3y = -z\]is a line.

The set of points $(x,y,z)$ that satisfy
\[6x = -y = -4z\]is another line.

Find the angle between these lines, in degrees.
</Question>

<think>
<Plan_1>
Okay, so I need to find the angle between two lines in 3D space. The lines are defined by the equations 2x = 3y = -z and 6x = -y = -4z. Hmm, I remember that to find the angle between two lines, we can use the dot product formula involving their direction vectors. But first, I need to figure out the direction vectors of each line.

Let me start with the first line: 2x = 3y = -z. I think when an equation is given in the form like this, it's a parametric equation where all the expressions are equal to some parameter. So maybe I can set 2x = 3y = -z = t, where t is a parameter. Then, I can express x, y, and z in terms of t.

<prompt_1.1> Find the direction vector for the line defined by 2x = 3y = -z by setting 2x = 3y = -z = t and solving for x, y, and z in terms of t. </prompt_1.1>

Now, the second line: 6x = -y = -4z. Similarly, I can set all these equal to another parameter, say s. So 6x = -y = -4z = s. Then, solving for x, y, z in terms of s.

<prompt_1.2> Find the direction vector for the line defined by 6x = -y = -4z by setting 6x = -y = -4z = s and solving for x, y, and z in terms of s. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's try that.

If 2x = t, then x = t/2. Similarly, 3y = t, so y = t/3. And -z = t, which means z = -t. So the parametric equations for the first line are x = t/2, y = t/3, z = -t. Therefore, the direction vector for this line can be found by looking at the coefficients of t in each equation. So when t increases by 1, x increases by 1/2, y by 1/3, and z decreases by 1. So the direction vector is (1/2, 1/3, -1). But direction vectors can be scaled by any non-zero scalar, so maybe it's easier to eliminate the fractions. If I multiply each component by 6 (the least common multiple of 2 and 3), the direction vector becomes (3, 2, -6). Let me check: if t is 6, then x = 3, y = 2, z = -6. That works because 2*3 = 6, 3*2 = 6, and -(-6) = 6. So yes, (3, 2, -6) is a direction vector for the first line.
</execution_1.1>

<execution_1.2>
Let me try that.

Starting with the first line: 2x = 3y = -z = t. Then, solving for each variable:

For x: 2x = t ⇒ x = t/2.

For y: 3y = t ⇒ y = t/3.

For z: -z = t ⇒ z = -t.

So, the parametric equations are x = t/2, y = t/3, z = -t. Therefore, the direction vector for this line can be found by looking at the coefficients of t in each equation. So, the direction vector v1 is (1/2, 1/3, -1). But direction vectors can be scaled by any non-zero scalar, so maybe it's easier to eliminate the fractions. If I multiply each component by 6 (the least common multiple of 2 and 3), I get (3, 2, -6). So, v1 = (3, 2, -6).

Now, moving on to the second line: 6x = -y = -4z = s. Let's do the same process.

For x: 6x = s ⇒ x = s/6.

For y: -y = s ⇒ y = -s.

For z: -4z = s ⇒ z = -s/4.

So, the parametric equations here are x = s/6, y = -s, z = -s/4. The direction vector v2 is (1/6, -1, -1/4). Again, to eliminate fractions, multiply each component by 12 (the least common multiple of 6 and 4). Let's see:

1/6 * 12 = 2, -1 * 12 = -12, -1/4 *12 = -3. So, the direction vector v2 becomes (2, -12, -3).
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Let me verify that.

<prompt_2.1> Verify that the direction vectors (3, 2, -6) and (2, -12, -3) are correct by checking the parametric equations. </prompt_2.1>

Based on execution_1.1:
Now, to find the angle between the two lines, we need the angle between their direction vectors v1 and v2. The formula for the angle θ between two vectors is:

cosθ = (v1 ⋅ v2) / (|v1| |v2|)

First, compute the dot product v1 ⋅ v2.

<prompt_2.2> Calculate the dot product of the direction vectors (3, 2, -6) and (2, -12, -3). </prompt_2.2>

Based on execution_1.1, execution_1.2:
Now, compute the magnitudes of v1 and v2.

<prompt_2.3> Calculate the magnitudes of the direction vectors (3, 2, -6) and (2, -12, -3). </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
For the first line, when s (the parameter) increases by 1, x increases by 1/6, y decreases by 1, and z decreases by 1/4. So scaling by 12: 1/6 *12 = 2, -1*12 = -12, -1/4*12 = -3. Yes, that's correct. So the direction vector for the second line is (2, -12, -3).
</execution_2.1>

<execution_2.2>
v1 ⋅ v2 = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = (6 + 18) - 24 = 24 - 24 = 0.
</execution_2.2>

<execution_2.3>
For v1 = (3, 2, -6):

|v1| = sqrt(3² + 2² + (-6)²) = sqrt(9 + 4 + 36) = sqrt(49) = 7.

For v2 = (2, -12, -3):

|v2| = sqrt(2² + (-12)² + (-3)²) = sqrt(4 + 144 + 9) = sqrt(157). Hmm, 157 is a prime number, so it doesn't simplify further. So |v2| = sqrt(157).
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Wait, the dot product is zero? That would mean the vectors are orthogonal, so the angle between them is 90 degrees.
But let me double-check my calculations because that seems surprising.

Let me recalculate the dot product:

<prompt_3.1> Recalculate the dot product of the direction vectors to verify the result. </prompt_3.1>

Based on execution_2.3:
Then, according to the formula:

cosθ = 0 / (7 * sqrt(157)) = 0.

Therefore, θ = arccos(0) = 90 degrees.

Based on execution_1.1, execution_1.2:
But let me confirm once more if the direction vectors are correct.

<prompt_3.2> Confirm the direction vectors and the final angle calculation. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
3*2 = 6

2*(-12) = -24

-6*(-3) = 18

Adding them up: 6 -24 +18 = (6 +18) -24 = 24 -24 = 0.
</execution_3.1>

<execution_3.2>
First line: 2x = 3y = -z. Let's pick a value for t. Let's say t = 6. Then x = 3, y = 2, z = -6. So direction vector is (3,2,-6). Correct.

Second line: 6x = -y = -4z. Let's pick s = 12. Then x = 12/6 = 2, y = -12, z = -12/4 = -3. So direction vector is (2, -12, -3). Correct.

So the direction vectors are indeed (3,2,-6) and (2,-12,-3). Then their dot product is 3*2 + 2*(-12) + (-6)*(-3) = 6 -24 +18 = 0. So yes, the dot product is zero. Therefore, the vectors are perpendicular, and the angle between the lines is 90 degrees.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
But just to be thorough, maybe I should check if the lines are actually skew or if they intersect? Wait, the problem states that each set of points defines a line, so they are straight lines in 3D space.
The angle between two lines is determined by their direction vectors regardless of their position. So even if they don't intersect, the angle is still calculated using the direction vectors.

Therefore, since the direction vectors are perpendicular, the angle between the lines is 90 degrees.
So the answer should be 90 degrees.

But let me just visualize this to make sure. If two lines in 3D have direction vectors that are orthogonal, then the angle between them is 90 degrees.

<prompt_4.1> Visualize and verify the orthogonality of the lines to ensure the angle is 90 degrees. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
That's correct. So unless there's a miscalculation in the direction vectors, which I double-checked, the angle is 90 degrees. 

Alternatively, maybe I made a mistake in setting up the parametric equations. Let me check again.

For the first line: 2x = 3y = -z. Let's set this equal to t. Then:

2x = t ⇒ x = t/2

3y = t ⇒ y = t/3

-z = t ⇒ z = -t

So when t = 6, x = 3, y = 2, z = -6. Plugging into 2x: 2*3=6, 3y=3*2=6, -z = -(-6)=6. So all equal to 6. Correct.

For the second line: 6x = -y = -4z. Let me set this equal to s. Then:

6x = s ⇒ x = s/6

-y = s ⇒ y = -s

-4z = s ⇒ z = -s/4

If s = 12, then x = 2, y = -12, z = -3. Check 6x = 6*2=12, -y = -(-12)=12, -4z = -4*(-3)=12. All equal to 12. Correct.

So the direction vectors are correctly found. Then their dot product is indeed zero. Therefore, the angle is 90 degrees.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
The angle between the lines is \boxed{90} degrees.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the angle between the two lines is $\boxed{90^\circ}$.
</Final_answer>
</think>