Video: MATH-ALG+GEO-2018-S1-Q11

Find the perpendicular distance from the point (2,3,1) to the plane 2π‘₯ βˆ’ 2𝑦 + 𝑧 = 5.

07:06

Video Transcript

Find the perpendicular distance from the point with coordinates two, three, one to the plane two π‘₯ minus two 𝑦 plus 𝑧 equals five.

So we have a plane with equation at two π‘₯ minus two 𝑦 plus 𝑧 equals five. And we have a point with coordinates two, three, one. We’re interested in the perpendicular distance of the point from the plane. And we do that by finding a point on the plane. It doesn’t matter where on the plane. And considering the vector from this point on the plane to the point two, three, one above it.

Now you might think that the distance we’re looking for is just the length of this vector. But this isn’t the perpendicular distance because there’s no reason why this vector should be perpendicular to the plane. But we can find a normal vector to the plane. We’ll call that the vector 𝑛. And we can put this normal vector wherever we want on the plane. And we can find the projection of our orange vector onto the normal vector, which as a multiple of the normal vector will be perpendicular to the plane. And so the perpendicular distance from the points to the plane will just be the length of this projection vector.

Okay, let’s give some of these points some names. Let 𝐴 be the point two, three, one whose distance from the plane we have to find. And let 𝐡 be the point we chose on the plane. Then our orange vector is the vector 𝐡𝐴. Then as we said before, the perpendicular distance that we’re looking for is the length of the projection vector. And using the notation we’ve laid down, the projection vector is the projection of the vector 𝐡𝐴 onto the vector 𝑛, the normal to the plane.

Now to find this length, we first need to find the vectors 𝐡𝐴 and 𝑛. Let’s start by finding the vector 𝐡𝐴. By definition, it is the position vector of the point 𝐴 minus the position vector of the point 𝐡. Now the position vector of the point 𝐴 is easy to find. The components of this vector are just the coordinates of the point 𝐴, which are two, three, one. But how about the point 𝐡?

Remember that 𝐡 is just some point on the plane. It doesn’t matter which one. We can pick any point on the plane we like. The plane has equation two π‘₯ minus two 𝑦 plus 𝑧 equals five. So the components π‘₯, 𝑦, and 𝑧 at any point on that plane will satisfy this equation. We can make things easy for ourselves by choosing to make the π‘₯- and 𝑦-coordinates of our point 𝐡 zero. And then we have no choice but to make the 𝑧-coordinate be five.

As two times zero minus two times zero plus five equals five, the coordinates of 𝐡 satisfy the equation. And so the point 𝐡 so defined is on the plane. So we continue to find the components of the vector 𝐡𝐴 using this choice for 𝐡. We could’ve chosen that the point 𝐡 to be some other point on the plane. But this point will do.

Now we just need to subtract the components of the position vectors 𝐴 and 𝐡. Two minus zero is two, three minus zero is three, and one minus five is negative four. Here are the components of 𝐡𝐴 then. Now we also need to find the components of the normal to the plane 𝑛. And the π‘₯-, 𝑦-, and 𝑧-components of 𝑛 are just the coefficients of π‘₯, 𝑦, and 𝑧 in the equation of the plane. The coefficient of π‘₯ in the equation is two, of 𝑦 is negative two, and of 𝑧 is one. So this is our normal vector.

You can see that the components of 𝑛 are just the coefficients of π‘₯, 𝑦, and 𝑧 by comparing the general vector equation of a plane to the equation we have. Now we have both of vectors 𝐡𝐴 and 𝑛. We’re ready to find the length of the projection of 𝐡𝐴 onto 𝑛, which is the perpendicular distance that we’re looking for. The projection of the vector 𝐡𝐴 onto the vector 𝑛 is 𝐡𝐴 dot 𝑛 over 𝑛 dot 𝑛 times the vector 𝑛. And we’re looking for the length or magnitude of this projection vector.

This fraction here is just a constant. So we can take it outside the magnitude. We have to make sure to take the absolute value of this constant as the magnitude must be positive. Now in the denominator, 𝑛 dot 𝑛 is just the magnitude of 𝑛 squared. And so bringing the magnitude of 𝑛 inside the absolute value, we can cancel it with one of the magnitudes of 𝑛 in the denominator. The length or magnitude of our projection vector is therefore the absolute value of the dot product of the vectors 𝐡𝐴 and 𝑛 over the magnitude of 𝑛. And remember that the magnitude of this projection vector is the length of the perpendicular distance from the point two, three, one to the plane that we’re looking for. So let’s get to calculating.

We know the components of the vector 𝐡𝐴. They are two, three, negative four. And we want to find the dot product of this with the normal vector 𝑛 whose components are two, negative two, one. So that’s the numerator. In the denominator, we have the magnitude of 𝑛. And of course the magnitude of 𝑛 is the square root of the sum of the squares of the components of 𝑛. So that’s the square root of two squared plus negative two squared plus one squared.

Now we can compute the dot product and the square root. The dot product is the product of the first components. That’s two times two. Plus the product of the second components, three times negative two, so negative six. Plus the product of the third components. That’s negative four times one, which is minus four. In the denominator, two squared is four, as is negative two squared. And one squared is one. So we have the square root of four plus four plus one.

Simplifying then in the numerator, four minus six is negative two. And subtracting four from that, we get negative six. And in the denominator, four plus four plus one is nine. And the square root of that nine is three. This fraction can be simplified. Negative six divided by three is negative two. And the absolute value of negative two is two. Hence, the perpendicular distance from the point two, three, one to the plane two π‘₯ minus two 𝑦 plus 𝑧 equals five is two.

We derived this by thinking about projections. But it may be worth remembering the formula perpendicular distance equals the absolute value of the dot product of 𝐡𝐴 and 𝑛 divided by the magnitude of the normal 𝑛. Whether vector 𝐡𝐴 is any vector from the plane to the point 𝐴 whose distance from the plane you want to find.

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