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In my personal opinion the lecturers's attempt to "make things look easy" by showing "magic tricks" ("let us now divide by the norm as if we were dealing with one-dimensional vectors AND VOILA") is rather unhelpful as it does not really explain neither the methodology nor the math nor the intuition.

There are some basic, but important notions hidden behind that "voila" step, let me help you by writing them out explicitly.

One way to find the distance between two parallel lines is to:

Pick any point $x_1$ on the first line.

Pick any point $x_2$ on the second line.

Project the vector $x_1 - x_2$ on the direction, perpendicular to two lines and measure the length of the projection.

Here's an illustration, just in case: you have two lines, one point taken on each line, the vector $x_2-x_1$ connects the two points. The dotted line is a perpendicular to the two lines. The vector $v$ points in the direction of the perpendicular, and the red segment shows the distance you want to measure.

This red distance is exactly the length of the projection of $x_2 - x_1$ onto the dotted line.

At this point we shall need three facts:

(a) For any unit vector $v$ the value $x^T v$ is equal to $|x|\cos \alpha$, where $\alpha$ is the angle between the vectors. Coincidentally, this is exactly the (signed) length of the projection of $x$ on the line with direction defined by $v$.

vector $v$ the value $x^T v$ is equal to $|x|\cos \alpha$, where $\alpha$ is the angle between the vectors. Coincidentally, this is exactly the (signed) length of the projection of $x$ on the line with direction defined by $v$. (b) Any vector $v$ can be made a unit vector (i.e. vector of length 1) by dividing it with its actual length: $\frac{v}{|v|}$.

(c) The equation $w^T x + b = 0$ defines a line with the normal $w$. I.e. $w$ is a vector, perpendicular to the line.

We know from (c) that $w$ is perpendicular to the line, we know from (b) that $\frac{w}{|w|}$ is a unit vector, so we plug it into (a) to get that the (signed) distance between the lines must be equal to $$\frac{w^T(x_2-x_1)}{|w|}.$$

If we know that $w^T x_2 + b = -1$ and $w^Tx_1 + b = 1$ we can play a bit with the equations to get $w^T(x_2 - x_1) = -2$ and hence the (signed) distance must be $\frac{-2}{|w|}$. We can drop the sign easily. ... and voila!

This is certainly not the only way to solve the problem, so for completeness sake let me show you a similar, yet still useful alternative. Here, you need to know just one fact (which I suggest you memorize just like the facts above):

(d) (Signed) distance from point $p$ to a line defined implicitly by the equation $w^T x + b = 0$ (note the zero on the right) is equal to $\frac{w^T p + b}{|w|}$.

Now, we can compute the distance between two lines by taking a point on one of them and substituting it in (d).

Let $x_1$ be a point on the line $w^T x+ b = 1$. The second line has the form $w^T x + b = -1$ which is the same as $w^T x + b + 1 = 0$. The distance between the lines is therefore:

$$\frac{w^T x_1 + b + 1}{|w|} = \frac{1 + 1}{|w|} = \frac{2}{|w|}.$$

Yet another way to derive the necessary distance would be to note that the line $x = sw$ is perpendicular to the two lines of interest. If $x_1$ and $x_2$ are the intersection points of this perpendicular with the first and the second parallel line correspondingly, the necessary distance is just the Euclidean distance between the two points: $$ d = |x_1 - x_2| = \sqrt{(x_1 - x_2)^T(x_1 - x_2)}. $$ I'll leave solving this to you as an exercise.