### Video Transcript

Find, to the nearest hundredth, the distance between the parallel lines ๐ฅ equals eight plus three ๐ก, ๐ฆ equals seven plus four ๐ก, ๐ง equals six plus two ๐ก, and ๐ซ equals two, one, four plus ๐ก times negative three, negative four, negative two.

So here we have these two parallel lines. Letโs say theyโre ๐ฟ one and ๐ฟ two. The equation of the first line is given to us in whatโs called parametric form. The second line, on the other hand, is written in whatโs called vector form. When our problem statement talks about finding the distance between these two lines, itโs referring to the minimum possible distance between them, that is, their perpendicular distance, which weโll call ๐. To solve for ๐, we can recall this relationship for the perpendicular distance between two parallel lines. Here, ๐ฌ is a vector thatโs parallel to both lines. On our sketch then, ๐ฌ would look something like this. Also in our equation for ๐, we see thereโs a second vector called ๐ one ๐ two.

The idea here is that we know a point on our first line. Weโll call that ๐ one. And we also know a point on our second line, ๐ two. And this vector ๐ one ๐ two looks like this. We see then that to calculate ๐ in our case, weโll need to know a point on our first line, a point on our second line, and a vector parallel to them both. First, letโs look at a point on our first line, ๐ one. We mentioned earlier that this lineโs equation is given to us in whatโs called parametric form. That is, thereโs a separate equation for the ๐ฅ-, ๐ฆ-, and ๐ง-coordinates of this line. Written this way, it doesnโt take much work to write the line in whatโs called vector form.

What we do is combine the ๐ฅ-, ๐ฆ-, and ๐ง-coordinates so that theyโre 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 itโs parallel to line two as well. We can write then that the vector ๐ฌ 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 ๐ซ, which starts at the origin of a coordinate frame and travels to the point two, one, four. This point is on our line ๐ฟ 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 weโre also given a vector parallel to the line. Compared to our vector ๐ฌ, this vector would look like this. Itโs antiparallel to ๐ฌ.

In any case, weโre now ready to compute the vector ๐ one ๐ 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 ๐ one ๐ two as well as a vector ๐ฌ parallel to our lines, weโre ready to move ahead in calculating ๐.

Our next step will be to compute the cross product ๐ one ๐ two cross ๐ฌ. This equals the determinant of this three-by-three matrix, where the first row has our ๐ข, ๐ฃ, and ๐ค unit vectors and the second and third rows have the corresponding components of ๐ one, ๐ two, and ๐ฌ. The ๐ข-component of this vector is equal to the determinant of this two-by-two matrix, thatโs negative four, while the ๐ฃ-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, thereโs the ๐ค-component equal to the determinant of this two-by-two matrix. Thatโs 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, weโre finally ready to calculate the magnitude of this cross product and divide it by the magnitude of ๐ฌ. The magnitude of ๐ one ๐ two cross ๐ฌ equals the square root of negative four squared plus six squared plus negative six squared, while the magnitude of ๐ฌ 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.