Question Video: Finding the Distance between a Point and a Plane Mathematics

Video Transcript

Find the distance between the point two, negative one, three and the plane 𝐫 dot the vector negative two, two, one is equal to three.

In this question, we’re asked to find the distance between a given point and a plane. And the first question we can ask is, what does it mean by the distance between a point and a plane? Whenever we’re asked to find the distance between a point and a plane or a point and a line, this always means the perpendicular distance between these two things. This is because the perpendicular distance is also the shortest distance. And it’s very useful to be able to find the shortest distance between a point and a plane or a point and a line.

So to find this distance, we’re first going to need to recall our formula for finding this distance. We recall, the perpendicular distance from the point 𝑥 one, 𝑦 one, 𝑧 one to the plane 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 is equal to zero is given by capital 𝐷 is equal to the absolute value of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐𝑧 one plus 𝑑 all divided by the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared. So we’re gonna want to try and apply this formula to our question. First, we want to find the distance from the point two, negative one, three.

So we’re gonna want to set our value of 𝑥 one equal to two, our value of 𝑦 one equal to negative one, and our value of 𝑧 one equal to three. However, then we can see a problem. In the question, we’re given our plane in vector form. However, we want our plane written in Cartesian form. So we’re going to want to convert our plane into Cartesian form. And there’s several different ways of doing this. The easiest way is to rewrite our vector 𝐫 as the vector 𝑥, 𝑦, 𝑧. So by setting our vector 𝐫 to be the vector 𝑥, 𝑦, 𝑧, our plane is now given by 𝑥, 𝑦, 𝑧 dot negative two, two, one is equal to three.

Now we’re going to want to evaluate the dot product of these two vectors. And remember, to find the dot product of two vectors, we need to multiply the components together and then add the results of all three of these products. So we start by multiplying the first two components together. This gives us 𝑥 multiplied by negative two. We then want to add the second components multiplied together. So we add 𝑦 multiplied by two. Finally, we add the third components multiplied together. That’s 𝑧 multiplied by one. And of course, all of this is still equal to our value of three.

Finally, we just simplify and rearrange this equation to get the Cartesian form of our plane. It’s negative two 𝑥 plus two 𝑦 plus 𝑧 minus three is equal to zero. Now, we can use the Cartesian form of our plane to find the values of 𝑎, 𝑏, 𝑐, and lowercase 𝑑. Remember, 𝑎 is the coefficient of 𝑥 in the Cartesian form of our plane. That’s negative two. 𝑏 will be the coefficient of 𝑦, which is two. 𝑐 will be the coefficient of 𝑧, which is one. And 𝑑 is our constant, which is negative three.

And there is something worth pointing out here. We could just rewrite our formula for the distance by using the vector form instead. The normal vector in our vector equation gives us the values of 𝑎, 𝑏, and 𝑐. And our value of 𝑑 will be negative the value this is equal to. In either case, we’ve now found the values of 𝑎, 𝑏, 𝑐, and 𝑑 and 𝑥 one, 𝑦 one, and 𝑧 one. So we can now find the perpendicular distance between the point and the plane. We just substitute these values into our formula.

We get the distance, capital 𝐷, is equal to the absolute value of negative two times two plus two times negative one plus one times three minus three all divided by the square root of negative two squared plus two squared plus one squared. Now all that’s left to do is evaluate this expression. Simplifying the expression in our numerator, we get the absolute value of negative six. And calculating the expression in our denominator, we get the square root of nine.

Of course, the absolute value of negative six is six, and the square root of nine is equal to three. So we get that the distance, capital 𝐷, is equal to six divided by three. And we can simplify this to give us two. And remember, this represents a distance. So we can say that this is two length units. Therefore, we were able to show that there were two-length-units distance between the point two, negative one, three and the plane 𝐫 dot the vector negative two, two, one is equal to three.