### Video Transcript

Find the length of the perpendicular drawn from point π΄ negative eight, one, 10 to the straight line π« equals negative one, two, negative seven plus π‘ times negative nine, negative nine, six, rounded to the nearest hundredth.

Alright, so here we have a line and we also have a point. And we want to solve for the distance, weβll call it π, of the perpendicular drawn from the point to the line. To do this, weβre going to make use of this expression. Here, the perpendicular distance between a straight line and a point depends on two vectors called ππ and π. In this equation, the vector π is parallel to the line involved. This means to solve for π in our scenario, weβll want to solve for a vector that is parallel to this line.

Beyond this, the vector ππ depends on two points, one of which is the point in space and one is a point along the given line. Looking again at our particular application, we know the coordinates of this point in space. Theyβre given to us. So the two bits of missing information we could say then are a point somewhere along this line and a vector parallel to it.

Note that weβre given the equation of this line in what is called vector form. Written this way, this vector in the equation represents a vector from an origin with coordinates zero, zero, zero to this point on our line, that is, a point with coordinates negative one, two, negative seven. We know then that this point is on our line, and weβll call this point π΅. The next thing we can say about this vector equation of our line is that this vector is parallel to the line. We can call it vector π, and it might look like this.

Now that we know a point in space, a point on our line, and a vector parallel to the line, we have everything we need to apply this relationship here. Our next step is to define a vector that goes from our point in space, point π΄, to the point on our line, point π΅. Weβll call this vector ππ, and itβs given by the vector form of the coordinates of point π΅ minus those of point π΄. These components work out to be seven, one, and negative 17.

We now know the two vectors that weβll use in our distance equation. Weβve called them ππ and π, whereas in the equation theyβre ππ and π. So long as we remember that ππ corresponds to ππ, weβll be alright. And now letβs move ahead by computing the cross product, ππ cross π. This is equal to the determinant of this matrix. Here, the top row are our three unit vectors, and the second and third rows are the corresponding components of vectors ππ and π. Calculating this cross product component by component, we find itβs equal to π’ hat times negative 147 minus π£ hat times negative 111 plus π€ hat times negative 54. In other words, itβs a vector with components negative 147, 111, negative 54.

Knowing what ππ cross π is, weβre now ready to calculate the magnitude of this cross product and divide it by the magnitude of our vector π. The magnitude of ππ cross π equals the square root of negative 147 squared plus 111 squared plus negative 54 squared. Dividing this by the magnitude of π, we see that this vector has a magnitude of the square root of negative nine squared plus negative nine squared plus six squared. If we enter this entire expression into our calculator, then rounding to the nearest hundredth, we get the result of 13.64.

Our final answer then is that the length of the perpendicular from the given point to the given straight line is 13.64 length units long.