Question Video: Finding the Distance between Three Parallel Lines Mathematics

Video Transcript

Find, to the nearest hundredth, the distance between the parallel lines 𝑥 equals eight plus three 𝑡, 𝑦 equals seven plus four 𝑡, 𝑧 equals six plus two 𝑡, and 𝐫 equals two, one, four plus 𝑡 times negative three, negative four, negative two.

So here we have these two parallel lines. Let’s say they’re 𝐿 one and 𝐿 two. The equation of the first line is given to us in what’s called parametric form. The second line, on the other hand, is written in what’s called vector form. When our problem statement talks about finding the distance between these two lines, it’s referring to the minimum possible distance between them, that is, their perpendicular distance, which we’ll call 𝑑. To solve for 𝑑, we can recall this relationship for the perpendicular distance between two parallel lines. Here, 𝐬 is a vector that’s parallel to both lines. On our sketch then, 𝐬 would look something like this. Also in our equation for 𝑑, we see there’s a second vector called 𝐏 one 𝐏 two.

The idea here is that we know a point on our first line. We’ll call that 𝐏 one. And we also know a point on our second line, 𝐏 two. And this vector 𝐏 one 𝐏 two looks like this. We see then that to calculate 𝑑 in our case, we’ll need to know a point on our first line, a point on our second line, and a vector parallel to them both. First, let’s look at a point on our first line, 𝐏 one. We mentioned earlier that this line’s equation is given to us in what’s called parametric form. That is, there’s a separate equation for the 𝑥-, 𝑦-, and 𝑧-coordinates of this line. Written this way, it doesn’t take much work to write the line in what’s called vector form.

What we do is combine the 𝑥-, 𝑦-, and 𝑧-coordinates so that they’re now the components of a vector. And our parametric form tells us that this line passes through the point with coordinates eight, seven, six and is parallel to a vector with components three, four, two. So then, we found a point on line one. And at the same time, we found the components of a vector parallel to line one, which must mean it’s parallel to line two as well. We can write then that the vector 𝐬 has components three, four, two.

Next, we want to solve for a point on line two. Because the equation of line two is given in vector form, we can do this fairly quickly. Written this way, we have a vector 𝐫, which starts at the origin of a coordinate frame and travels to the point two, one, four. This point is on our line 𝐿 two and then it moves according to this vector up and down that line. We have then the coordinates of a point on line two. And note that we’re also given a vector parallel to the line. Compared to our vector 𝐬, this vector would look like this. It’s antiparallel to 𝐬.

In any case, we’re now ready to compute the vector 𝐏 one 𝐏 two. This vector is equal to the vector form of the difference between the coordinates of point two and point one. The result is negative six, negative six, negative two. Now that we have a vector 𝐏 one 𝐏 two as well as a vector 𝐬 parallel to our lines, we’re ready to move ahead in calculating 𝑑.

Our next step will be to compute the cross product 𝐏 one 𝐏 two cross 𝐬. This equals the determinant of this three-by-three matrix, where the first row has our 𝐢, 𝐣, and 𝐤 unit vectors and the second and third rows have the corresponding components of 𝐏 one, 𝐏 two, and 𝐬. The 𝐢-component of this vector is equal to the determinant of this two-by-two matrix, that’s negative four, while the 𝐣-component is negative the determinant of this two-by-two matrix. Negative six times two minus negative two times three is negative six overall. And lastly, there’s the 𝐤-component equal to the determinant of this two-by-two matrix. That’s negative six. This then is our cross product. And we can write this as a vector with components negative four, six, negative six.

Knowing this, we’re finally ready to calculate the magnitude of this cross product and divide it by the magnitude of 𝐬. The magnitude of 𝐏 one 𝐏 two cross 𝐬 equals the square root of negative four squared plus six squared plus negative six squared, while the magnitude of 𝐬 equals the square root of three squared plus four squared plus two squared. Entering this whole expression in our calculator, to the nearest hundredth, we get 1.74. This is the distance and specifically the minimum distance between these two parallel lines.