Question Video: Finding the Equation of the Tangent Plane to a Surface at a Given Point

Video Transcript

Find the equation of the tangent plane to the surface 𝑥 squared plus 𝑦 squared minus 𝑧 squared equals zero at the point three, four, five.

A plane in 3D space can be defined in a few different ways. One way is by finding a normal vector to the plane, 𝐧, and a known point in the plane, 𝑃. Given that we want to find the plane tangent to this surface at the point three, four, five, we already know that this point must lie on the plane. So, all we need is a normal vector to the plane. Let’s sketch our surface defined by this function, containing the point three, four, five.

We wish to find the tangent plane to the surface at this point, so this will look something like this. A normal vector to the surface at this point will also be normal to the plane. We can easily find a normal vector to any surface at any point by evaluating the gradient of the surface’s function at that point. In this case, the function 𝑓 of 𝑥, 𝑦, 𝑧 is given by 𝑥 squared plus 𝑦 squared minus 𝑧 squared.

By convention, when defining a function of a surface in 3D space, we set 𝑓 of 𝑥, 𝑦, 𝑧 equal to zero, which would involve taking any constants all over to the left-hand side. But in this case, we don’t have to since the right-hand side here is already zero. The gradient grad 𝑓 in Cartesian coordinates is given by the vector partial 𝑓 by partial 𝑥, partial 𝑓 by partial 𝑦, and partial 𝑓 by partial 𝑧. So, in this case, this gives us the vector two 𝑥, two 𝑦, negative two 𝑧. So, evaluating grad 𝑓 at the point three, four, five gives the vector six, eight, negative 10. This vector is normal to the surface at three, four, five, but so too is any scalar multiple of it.

For the sake of simplicity then, let’s multiply this vector by one-half to give our normal vector 𝐧 equals three, four, negative five. So, we now have our normal vector to the plane, 𝐧, and our known point on the plane, 𝑃. Let’s clear a little space first. So, how do we find the equation of the plane from these two pieces of information? Let’s consider a general plane in ℝ three. We have a known point on the plane, 𝑃, and we have a normal vector to the plane, 𝐧. Now, let’s consider another general point on the plane, 𝑄.

The position vector of 𝑃 is known. And in terms of the origin 𝑂, we can call this position vector 𝐎 to 𝐏. Likewise, we can call the unknown position vector of 𝑄 𝐎 to 𝐐. We can define a third vector 𝐏 to 𝐐, the vector from the point 𝑃 to the point 𝑄. We can express this vector 𝐏 to 𝐐 using the vector triangle in terms of 𝐎𝐏 and 𝐎𝐐. So, 𝐏 to 𝐐 is equal to negative 𝐎 to 𝐏 plus 𝐎 to 𝐐. We can rewrite this a little more nicely as 𝐎 to 𝐐 minus 𝐎 to 𝐏.

This vector 𝐏 to 𝐐 lies in the plane, so it must be perpendicular to the normal vector 𝐧. This means that the scalar product of 𝐏 to 𝐐 and 𝐧 must be equal to zero. From the expression for 𝐏 to 𝐐 above, this implies that 𝐎 to 𝐐 minus 𝐎 to 𝐏 dot 𝐧 is equal to zero. 𝐎 to 𝐐 is just the position vector of a general point on the plane 𝑥, 𝑦, 𝑧. And 𝐎 to 𝐏 is the position vector of the known point on the plane three, four, five.

Substituting in these expressions along with our normal vector 𝐧 gives us the dot product of 𝑥 minus three, 𝑦 minus four, 𝑧 minus five, and three, four, negative five. Expanding out this dot product gives us three times 𝑥 minus three plus four times 𝑦 minus four minus five times 𝑧 minus five equals zero. And expanding out these parentheses gives us the equation of the general point on the plane 𝑥, 𝑦, 𝑧, in other words, the equation of the plane three 𝑥 plus four 𝑦 minus five 𝑧 equals zero.