INSTRUCTOR: Find the distance between the point 0 comma 3 comma 6 and the line of parametric equations x equals 1 minus t, y equals 1 plus 2t, and z equals 5 plus 3t. I'm going to refer to this theorem. Let L be a line through point P with direction vector v. If M is any point not on L, then the distance from M to L is d equals the magnitude of PM cross v divided by the magnitude of v.

Now, in order to use this definition, we need to identify what things we have here. So first off, the point M is the 0.036. That's a point that is not on the line. However, it's a little less obvious about the point that is on the line.

Well, we should be able to use any point here. But if we set the value of t in each of these parametric equations equal to 0, then we get the point P equals the x value of 1, a y value of 1, and a z value of 5. That is definitely on the line. And also, the direction vector is the vector v equal to angle bracket negative 1 comma 2 comma 3 close angle bracket.

Because if you look at each of these equations, because these are all linear parametric equations, the slope, negative 1, 2t, so slope of 2, and 3t, so have a slope of 3, those are the direction. Those are actually the derivatives of each of those with respect to t since we're going to get there eventually.

And that should be all we need so that we can find what the vector PM is. The vector PM, to find that, we're going to subtract the terminal minus the initial. That's what we're going to do, the points for these. So we'll take the M values and subtract the P values from that coordinate.

So that'll be 0 minus 1. And then we'll have 3 minus 1. Then we'll have 6 minus 5. So this vector should be negative 1 comma 2 comma 1.

All right. So next, to fit this formula for distance that we need, we need to find PM cross v. So just kind of building all of this up. Our first step really was to identify our information. Our second step was finding PM. The following step is finding PM cross v.

I'm going to use the determinant notation for this. So I'm going to have i, j, k. PM is negative 1, 2, 1. So that's my second row. My third row is v. That is negative 1, 2, 3. So this will be determinant of the matrix 2, 1, 2, 3 i minus the determinant of the matrix negative 1, 1, negative 1, 3 for the j plus determinant of the matrix negative 1, 2, negative 1, 2 k.

So PM cross v. Double-checking all of what I've written here. OK. So this should come out to be a 4i plus 2j plus 0k. And I'm going to write this in component form. That's 4 comma 2 comma 0, that vector.

Our next step is going to have to be to find the magnitude of this vector along with the magnitude of v because we'll then divide them. So the magnitude of PM cross v is the square root of 4 squared plus 2 squared plus 0 squared, which is the square root of 20, which we could also write as-- that'd be 2 square roots of 5. Yes. OK.

Let me add on here the magnitude of v alone. Magnitude v alone is the square root of negative 1 squared plus 2 squared plus 3 squared, which is square root of 14. So putting this all together, d equals 2 square roots of 5 divided by the square root of 14.

I'm going to go ahead and rationalize this. Multiply by the square root of 14, numerator and denominator. So that this would be 2 square roots of 70 over 14, which will reduce to be square root of 70 over 7. That is the distance from that point to line defined by those parametric equations.