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:
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.
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.
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.
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.