Find the acute angle between the two lines... l1: r = (4, 28, 4) + λ(-1, -5, 1), l2: r = (5, 3, 1) + μ(3, 0, -4)

To do this, we will only consider the direction vectors of each line. That is to say, the vector 'attached' to our λ and μ terms (which are called scalar parameters).We will do this because the angle between the lines is only determined by the direction they are facing, the position vector of the line has nothing to do with it.
So, let d1 = (-1, -5, 1) ~ the direction vector of l1, and let d2 = (3, 0, -4) ~ the direction vector of l2.
Now, recall that d1.d2 = |d1||d2|cosθ, meaning cosθ = (d1.d2)/(|d1||d2|). This will be useful later. ()
To find d1.d2 (d1 'dot' d2), we will multiply each of their corresponding components, then sum them together.To put this simply, we multiply the 'top bit' of d1 by the 'top bit' of d2, the 'middle bit' of d1 by the 'middle bit' of d2, and so on... Then, add all these terms together will give d1.d2
So: d1.d2 = (-1)(3) + (-5)(0) + (1)(-4) = -7
To find |d1||d2|, i.e. the 'size' of each direction vector, we will just use Pythagoras' theorem: almost like each 'bit' of the vector is a part of a triangle (except in 3D).
So: |d1| = √ ((-1)² + (-5)² + (1)²) = 3√ 3|d2| = √ ((3)² + (0)² + (-4)²) = 5
Which gives |d1||d2| = 15√ 3.
Hence, we can sub these values into the equation I referenced earlier () to find:
cosθ = -7/(15√ 3).
Notice the minus sign. If we want to find the acute angle, we must ALWAYS remember that cosθ must be positive (to see these, draw out y = cosθ on a graph. You will see when 0 < θ < 90, cosθ > 0). So... We can just 'chop off' the minus sign as if it was never there in the first place.
Let α be the acute angle we want to find.
This gives: cosα = +7/(15√ 3), meaning α = cos^-1(+7/(15√ 3)) = 74.4° (1dp)
So the acute angle we have been asked to find is 74.4°