Shortest Distance Between Two Lines

Transcript

0:00Introduction

Hello everyone! Picture a flyover, with a road running underneath it in a different direction. They never meet, and they are not parallel either. So how far apart are they, and which gap would you even measure? Interesting, right? Lines like these are called skew lines, and today we find the shortest distance between them. It comes up often in JEE, and one neat formula does the job. So, let's begin.

0:29Skew lines and the common perpendicular

So, here is our picture. The first line, L one, is vector r equals vector a one, plus lambda times vector b one. The little arrows on top mean vectors, the small number written below a letter is a subscript for the first line, and lambda, a Greek letter, slides us along it. The thick black segment P Q is labelled shortest distance d, and the formula at the bottom of the figure is where we are heading today. Okay? The second line, L two, is vector r equals vector a two, plus mu times vector b two, with mu, another Greek letter, as its own slider. And remember, skew lines are neither parallel nor intersecting. With me? Now, join any point of L one to any point of L two. Of all such segments, which one is the shortest? Yes, exactly, the one perpendicular to both lines! In the figure, that is P Q, and the small squares at its ends mark the right angles. We call it the common perpendicular. So vector P Q dotted with vector b one is zero, and dotted with vector b two is zero too; the small dot is the dot product. These two equations fix lambda and mu, and so they give the points of closest approach. Clear? Now, the direction of this segment. Which vector is perpendicular to both vector b one and vector b two at the same time? The cross product, of course! Vector b one cross vector b two, with the small cross sign, is perpendicular to both. So P Q runs along it, as the label in the figure says; the two vertical lines there mean parallel. Got it?

2:08The formula: project onto the common perpendicular

Now, the length of this segment. As the caption under the figure says, vector a two minus vector a one joins a point of L one to a point of L two, and the length of its projection on b one cross b two is exactly the distance d. So d equals the modulus of, vector a two minus vector a one, dot, b one cross b two, over the modulus of b one cross b two. The vertical bars mean the modulus. Okay? The numerator is one vector dotted with the cross product of two others, so it is a scalar triple product. Right? In Cartesian form, it is the same three by three determinant that we used to test whether two lines are coplanar, now inside a modulus. Clear? So the distance is zero exactly when this numerator is zero, and for non parallel lines, that means they intersect. If they meet, there is no gap. With me?

2:57Parallel lines: a different formula

Now, parallel lines. Both have the same direction, vector b, with no subscript. So vector r equals vector a one plus lambda times vector b, and vector r equals vector a two plus mu times vector b. Okay? Can we use the same formula here? Think about it: what is b one cross b two when the directions are parallel? It is the zero vector, the zero with an arrow on top that you see on screen! So the skew formula gives zero over zero, and it simply fails. For parallel lines, d equals the modulus of, vector a two minus vector a one, cross vector b, over the modulus of vector b. Notice, a cross here, not a dot. Right? And it makes sense: this is just the distance of the point vector a two from the first line, a formula we already know. Got it? So here is the JEE trap: rushing into the skew formula without checking. Always test the numerator first. If it is zero, the lines are coplanar: the distance is zero if they intersect, and you need the parallel formula if they are parallel. Clear?

4:06Example 1: skew lines in vector form, d = 3 over root 2

Now, Example one. Find the shortest distance between vector r equals, i hat plus two j hat plus k hat, plus lambda times, i hat minus j hat plus k hat, and vector r equals, two i hat minus j hat minus k hat, plus mu times, two i hat plus j hat plus two k hat. The little caps on i, j and k mean unit vectors along the axes. Pause the video and try it first. Done? Okay, let's check. Vector a one is one, two, one, and vector b one is one, minus one, one. Vector a two is two, minus one, minus one, and vector b two is two, one, two. With me? Now b one cross b two. The components are minus two minus one, then two minus two, then one plus two. So we get minus three, zero, three. Okay? Next, vector a two minus vector a one. Two minus one, minus one minus two, and minus one minus one give one, minus three, minus two. Right? Now dot these two vectors. Minus three, plus zero, minus six, which gives minus nine. Clear? The modulus of the cross product is the root of nine plus zero plus nine; the tick shaped sign is the square root. Root eighteen is three root two. Got it? So now the distance is minus nine over three root two, right? No, no, wait, a distance can never be negative! That is exactly why the formula has a modulus on top. The modulus of minus nine is nine, and nine over three root two gives three over root two. And that is our answer, perfect!

5:49Example 2: Cartesian skew lines, d = 2 root 29

Now a harder one, Example two, in Cartesian form. Find the shortest distance between x plus one over seven equals y plus one over minus six equals z plus one over one, and x minus three over one equals y minus five over minus two equals z minus seven over one. Pause the video and try it first. Done? Okay, let's check. Many students read the first point straight off as one, one, one. Is that right? No, careful, that is the sign trap! x plus one means x minus minus one, so vector a one is minus one, minus one, minus one. Vector b one, from the denominators, is seven, minus six, one. Got it? The second line is friendlier: vector a two is three, five, seven, and vector b two is one, minus two, one. Okay? Now the cross product: minus six plus two, then one minus seven, then minus fourteen plus six. So we get minus four, minus six, minus eight. With me? These are all multiples of minus two. Can we use the simpler vector two, three, four instead? Yes, absolutely, and that is a topper's trick! We divide by the modulus anyway, so any non zero multiple of the direction works. The modulus of two, three, four is the root of four plus nine plus sixteen, which is root twenty nine. Much smaller numbers, right? Next, vector a two minus vector a one is three plus one, five plus one, seven plus one, so four, six, eight. Clear? Dot it with two, three, four: eight plus eighteen plus thirty two, which is fifty eight. So d equals fifty eight over root twenty nine, and that simplifies to two root twenty nine. That is our answer, and it was really simple, right?

7:44Quick revision: skew and parallel formulas, coplanar test

So, quick revision. For skew lines, take the modulus of the scalar triple product, over the modulus of b one cross b two. For parallel lines, take the modulus of, vector a two minus vector a one, cross vector b, over the modulus of vector b. And always test the numerator first, since zero means the lines are coplanar. Now go and try the two practice problems on this page. All the best!