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, let me recall how to approach this. 

First, from what I remember, the angle between two lines in three dimensions is determined by the angle between their direction vectors. So, if I can find the direction vectors of each line, then I can use the dot product formula to find the angle between them. The formula for the angle θ between two vectors u and v is given by cosθ = (u • v) / (|u||v|). Then, θ = arccos[(u • v)/(|u||v|)]. That's the plan.

So, the first step is to find the direction vectors of each line. Let's start with the first line: 2x = 3y = -z. I think this represents a line in 3D space, and the coefficients here can be used to find the direction ratios. 

Wait, when a line is given in the form a = b = c, then the direction ratios are the coefficients of the variables in the equations. So, if 2x = 3y = -z, let me set each of these equal to a parameter t. So, 2x = t, 3y = t, and -z = t. Then, solving for x, y, z in terms of t: x = t/2, y = t/3, z = -t. Therefore, the parametric equations of the line are x = (1/2)t, y = (1/3)t, z = -t. 

So, the direction vector of the first line is the coefficients of t in the parametric equations. That would be (1/2, 1/3, -1). But maybe it's better to represent it without fractions. To eliminate the denominators, I can multiply each component by 6, the least common multiple of 2 and 3. Multiplying by 6 gives (3, 2, -6). So, the direction vector can be (3, 2, -6). That seems right. Let me check: if t is 6, then x = 3, y = 2, z = -6, which satisfies 2x = 6, 3y = 6, -z = 6. Yep, that works. So direction vector u = (3, 2, -6).

Now, moving on to the second line: 6x = -y = -4z. Similarly, let's set each of these equal to a parameter s. So, 6x = s, -y = s, -4z = s. Then, solving for x, y, z: x = s/6, y = -s, z = -s/4. Therefore, the parametric equations are x = (1/6)s, y = -s, z = (-1/4)s. 

So, the direction vector is (1/6, -1, -1/4). Again, to eliminate fractions, multiply each component by 12. That gives (2, -12, -3). Let me check: if s = 12, then x = 2, y = -12, z = -3. Then 6x = 12, -y = 12, -4z = 12. All equal, so that's correct. Therefore, direction vector v = (2, -12, -3).

Now, I need to compute the dot product of u and v. The dot product is u • v = (3)(2) + (2)(-12) + (-6)(-3). Let's calculate each term: 3*2 = 6, 2*(-12) = -24, (-6)*(-3) = 18. Adding them up: 6 -24 +18 = (6 +18) -24 = 24 -24 = 0. Wait, the dot product is zero? That would mean the vectors are perpendicular, so the angle between them is 90 degrees. But that seems too straightforward. Let me verify.

Wait, maybe I made a mistake in calculating the direction vectors. Let me double-check. For the first line, 2x = 3y = -z. Let me set 2x = 3y = -z = k, where k is a parameter. Then x = k/2, y = k/3, z = -k. So direction vector is (1/2, 1/3, -1). When scaled by 6, it's (3, 2, -6). Correct.

For the second line: 6x = -y = -4z. Let me set each equal to m. So 6x = m, -y = m, -4z = m. Then x = m/6, y = -m, z = -m/4. So direction vector is (1/6, -1, -1/4). Scaling by 12: (2, -12, -3). Correct. Then the dot product: (3)(2) + (2)(-12) + (-6)(-3). 6 -24 +18 = 0. So yes, the dot product is zero. Therefore, the angle is 90 degrees. 

But let me check once more. Maybe the direction vectors are correct. For the first line: 2x = 3y = -z. Let's pick a point on the line. If t = 6, then x = 3, y = 2, z = -6. So direction vector is from the origin to (3,2,-6), which is correct. For the second line, if s = 12, x = 2, y = -12, z = -3. So direction vector is (2, -12, -3). Correct. Then their dot product is indeed 0. 

Wait, but if the dot product is zero, then the angle is 90 degrees. So the answer is 90 degrees. But maybe I did something wrong in the direction vectors? Let me think. 

Alternatively, direction vectors can be found by taking the ratios from the equations. For the first line, 2x = 3y = -z. Let me write this as x/(1/2) = y/(1/3) = z/(-1). So direction ratios are 1/2 : 1/3 : -1. To make them integers, multiply by 6: 3 : 2 : -6. So direction vector (3,2,-6). Correct. 

For the second line, 6x = -y = -4z. Let's write x/(1/6) = y/(-1) = z/(1/4). So direction ratios are 1/6 : -1 : 1/4. Multiply by 12: 2 : -12 : 3. Wait, hold on. Wait, in the parametrization earlier, z was -1/4, so direction vector was (1/6, -1, -1/4). Scaling by 12 gives (2, -12, -3). But according to the ratios, it's 1/6 : -1 : 1/4. So when scaling by 12, 1/6 *12 = 2, -1*12 = -12, 1/4 *12=3. But in the parametrization, z was -1/4, so the direction vector was (1/6, -1, -1/4). So when scaling by 12, (2, -12, -3). So the z-component is -3, not 3. So the direction vector is (2, -12, -3). So the dot product would be 3*2 + 2*(-12) + (-6)*(-3) = 6 -24 +18 = 0. So yes, correct. 

Therefore, the angle is 90 degrees. But let me check another way. Maybe using another parametrization. 

Alternatively, for the first line, direction vector can be found by solving the equations. Suppose we set 2x = 3y = -z = k. Then, x = k/2, y = k/3, z = -k. So direction vector is (1/2, 1/3, -1). For the second line, 6x = -y = -4z = m. So x = m/6, y = -m, z = -m/4. Direction vector is (1/6, -1, -1/4). Now, the dot product is (1/2)(1/6) + (1/3)(-1) + (-1)(-1/4). Let's compute this: (1/12) + (-1/3) + (1/4). Convert to twelfths: 1/12 - 4/12 + 3/12 = (1 -4 +3)/12 = 0/12 = 0. So again, the dot product is zero. 

So regardless of scaling, the dot product is zero. Therefore, the vectors are perpendicular. Therefore, the angle between the lines is 90 degrees. 

But wait, just to make sure, maybe there's a miscalculation here. Let's recompute the dot product with the original direction vectors (1/2, 1/3, -1) and (1/6, -1, -1/4). 

Dot product = (1/2)(1/6) + (1/3)(-1) + (-1)(-1/4) = (1/12) + (-1/3) + (1/4). 

Convert to twelfths: 1/12 - 4/12 + 3/12 = (1 -4 +3)/12 = 0/12 = 0. 

Yes, that's correct. So indeed, the dot product is zero. Therefore, the angle is 90 degrees. 

Therefore, the answer is 90 degrees. 

But just to make sure, sometimes when dealing with lines in 3D, the angle between them is defined as the angle between their direction vectors, and if the dot product is zero, they are perpendicular. So yes, 90 degrees is the correct answer. 

I think that's solid. I can't see any mistakes here. The direction vectors are correctly derived from the parametric equations, and the dot product being zero confirms perpendicularity. So the angle is 90 degrees.
Therefore, the angle between the two lines is $\boxed{90}$ degrees.
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