### Video Transcript

Find, to the nearest hundredth, the distance between the parallel lines π₯ equals one plus three π‘, π¦ equals seven plus two π‘, π§ equals four plus five π‘ and π₯ equals three minus three π‘, π¦ equals six minus two π‘, π§ equals four minus five π‘.

Okay, so here we have these two parallel lines, and weβll call the first line line one and the second line two. We want to solve for the perpendicular distance between these lines. Weβll call that π. To do this, there are three bits of information weβll need to find out. First, weβll need to know the coordinates of a point on the first line; weβll call that π one. Weβll also need to know a point on line two; weβll call that π two. And lastly, weβll need to know the components of a vector thatβs parallel to both these lines. Weβll call this vector π¬. Once we know all this, weβll be able to use this expression to compute the distance between our two parallel lines.

We see this equation involves a vector π¬, thatβs the vector thatβs parallel to our lines, and a second vector weβve called π one π two. On our sketch, thatβs a vector that looks like this. It goes from point one on line one to point two on line two. Letβs start by figuring out a point on line one, in other words, the coordinates of a point π one. Weβll do this by looking at the equation of line one, which we see is given to us in parametric form. Written this way, we have separate equations for the π₯-, π¦-, and π§-coordinates of every point on this line, and itβs possible to convert this lineβs equation from parametric form to whatβs called vector form. This involves combining all three equations into one, where π« is a vector with components π₯, π¦, and π§.

This form of the lineβs equation begins with the vector from the origin of our coordinate frame to the point one, seven, four. This point lies on line one, and then it moves out along this vector three, two, five multiplied by the scale factor π‘. We can say then that the point with coordinates one, seven, four lies along line πΏ one. So weβll call this point π one. Furthermore, this vector three, two, five is parallel to our line, and therefore we can call this the vector π¬.

Now that we know a point π one and a vector π¬, letβs clear a bit of space and start looking at the given equation for line two. Our goal in doing this is to discover a point π two that lies along this line. Just like for line one, line two is given to us in parametric form. That means we can write this line in vector form as a vector to the point three, six, four plus π‘ times another vector parallel to line two. Since the point three, six, four is on πΏ two, we can call this π two.

And so, now we have all the information we need to begin calculating the distance π. The first thing weβll do is solve for the components of this vector π one π two. We find these components by subtracting the coordinates of point π one from those of π two. With those values substituted in, we find that the vector π one π two has components three minus one or two, six minus seven or negative one, and four minus four, zero. Now that we know this vector, next, weβll take the cross product of this vector with π¬. That cross product is given by the determinant of this three-by-three matrix. In the top row, we have the π’, π£, and π€ unit vectors and then the π₯-, π¦-, and π§-components of π one, π two, and π¬, respectively.

The magnitude of the π’-component is given by the determinant of this two-by-two matrix. Negative one times five minus zero times two is negative five. And then the π£-component is negative the determinant of this matrix, which is two times five minus zero times three or 10. And lastly, thereβs the π€-component of this cross product equal to the determinant of this matrix. Two times two minus negative one times three is positive seven. This, then, is our overall cross product, which we can write in vector form with components negative five, negative 10, seven.

Okay, weβre ready now to calculate π by computing the magnitude of π one π two cross π¬ and dividing that by the magnitude of π¬. The magnitude of our cross product equals the square root of negative five squared plus negative 10 squared plus seven squared, while the magnitude of π¬ equals the square root of three squared plus two squared plus five squared. Entering this whole expression on our calculator, to the nearest hundredth, our answer is 2.14. This is the minimum distance between these two parallel lines.