This part picks up from Part 1 and looks at the ‘shortest distance’ scenarios in FP3 Vectors. I’m always stunned by the textbooks and formula books that promote such formulae as:

and

The former of these is for the distance between two skew lines, and the latter for the perpendicular distance from a point to a plane. Not only are these a hideous mass of letters but, even more concerning to me, they are both disguising *the same mathematical principle*. Let’s take a clearer look at things.

## Understanding Components

Here’s my starter question before we consider the ‘shortest distance’ scenarios:

This can take a little time but because the given statement is true, students can invest some time and thought into *why it is so*. After some short discussion, we are happy that dot and cross product can be used to find components. Moreover, it is only the ‘vertical’ component that we will need: sometimes found with dot product, sometimes with cross.

## Similar Situations

Mathematics to me is all about identifying similarities. The formulae quoted at the start of this post are totally anathema to me. Consider these (beautiful…) diagrams:

In every one of these situations, the ‘formula’ for finding the shortest distance is simply

Much less to remember, much more understanding developed.

## Two Other Cases

In the case of parallel lines or perpendicular distance from a point to a line, this approach breaks down: we don’t have a convenient unit normal vector. This is where the cross product alternative for finding a component comes in.

## Many Ways

Of course there are many ways to solve these problems. I simply like the above approach because it follows from a clear understanding of components and the geometry of the objects involved.

I wouldn’t often openly criticise a textbook but, for comparison, here is a worked example from the Edexcel/Pearson book [FP3, p128]:

Of course it works. But. Yuck.

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