Video: Equation of a Plane: Vector, Scalar, and General Forms

In this video, we will learn how to find the vector, scalar (standard or component), and general (Cartesian or normal) forms of the equation of a plane given the normal vector and a point on it.

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Video Transcript

In this video, we’re talking about the equation of a plane: vector, scalar, and general forms. We see then that there are several different ways to represent a plane, which is a two-dimensional surface. In this lesson, we’ll learn these different forms, and we’ll see how they’re related to one another.

Let’s get started by considering this plane. Even though we’ve drawn in all four sides, we know that, really, this plane goes infinitely far in these directions as a two-dimensional surface. And in order to define a plane, we need to have a frame of reference. With that frame of reference in place, our objective is to write the equation of a plane by knowing two things about it. First, let’s assume we have a vector, we’ve called it 𝐧, that is normal or perpendicular to the plane. And second, let’s assume we also know a point on the plane. We’ve called it 𝑃 zero. With these two ingredients, we could call them, we can use vectors to develop a vector equation for this plane.

Since 𝑃 zero is a known point, we know the three coordinates of that point. We’ll call them 𝑥 zero, 𝑦 zero, and 𝑧 zero. And that means if we draw a vector from the origin of our coordinate frame to 𝑃 zero, then that vector, we can call it 𝐫 zero, has components 𝑥 zero, 𝑦 zero, and 𝑧 zero. So far, then, we have these two vectors, our normal vector and 𝐫 zero. But we can see that there’s no particular spatial relation between them. In general, they won’t be, for example, parallel or perpendicular.

But now let’s say that we pick an arbitrary point elsewhere in our plane. And we’ll call this point 𝑃. Since 𝑃 is a general unknown point, we’ll give it general coordinates 𝑥, 𝑦, and 𝑧. And to this point also, we can draw a vector from our origin. This vector we’ll call 𝐫 with components 𝑥, 𝑦, 𝑧. And notice what we can do now with our two vectors 𝐫 and 𝐫 zero. If we subtract them, specifically if we take 𝐫 zero and subtract it from 𝐫, then we’ll get a resultant vector that looks like this. And the important thing is that this resultant, 𝐫 minus 𝐫 zero, is in the plane. That’s because it travels in a straight line from one point in the plane to another.

Recalling that our normal vector 𝐧 is perpendicular to the plane, it must therefore also be true that 𝐧 is perpendicular to 𝐫 minus 𝐫 zero. And that means we can write this. The vector 𝐧 dotted with the vector 𝐫 minus 𝐫 zero is equal to zero. This is so because the dot product of two perpendicular vectors is zero; the vectors don’t overlap at all.

By writing this equation, we have defined this plane. And notice that another way of writing this equation is 𝐧 dot 𝐫 is equal to 𝐧 dot 𝐫 zero. Written this way, we have what’s often called the vector form of a plane equation. And so we’ve reached our goal. Given a normal vector 𝐧 and a point 𝑃 zero on a plane, we were able to define a specific vector 𝐫 zero based on 𝑃 zero and a general vector 𝐫 to a general point on the plane 𝑃 and combine these vectors in a way that uniquely describes this plane.

So mission accomplished. But let’s recall that in addition to the vector form of our plane, we also want to solve for the scalar and general forms. We’ll work on those in order. And we can start on our scalar form by recognizing that we know the components of the three vectors in this expression. The normal vector 𝐧 has components 𝑎, 𝑏, 𝑐, our vector 𝐫 has components 𝑥, 𝑦, 𝑧, and 𝐫 zero has components 𝑥 zero, 𝑦 zero, 𝑧 zero. And now if we compute these dot products multiplying together the respective components of each vector and then adding those products, we find that 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 equals 𝑎 times 𝑥 zero plus 𝑏 times 𝑦 zero plus 𝑐 times 𝑧 zero.

And then if we take the whole right-hand side of this expression and subtract it from both sides of the equation and also group together terms by 𝑎, 𝑏, and 𝑐, we’ll find that 𝑎 times the quantity 𝑥 minus 𝑥 zero plus 𝑏 times the quantity 𝑦 minus 𝑦 zero plus 𝑐 times the quantity 𝑧 minus 𝑧 zero all is equal to zero. Written this way, we have what’s called the scalar form of a plane equation. And note that to get this form of our equation, we plugged in for our vectors in the vector form and then rearranged until only scalar quantities remain. The scalar form of a plane’s equation is sometimes also called standard or component form.

Lastly, let’s look at what’s called the general form of a plane’s equation. To solve for that, we can actually return to this step right here, after we took the dot product on either side of our equation. Say we were to take the entire right-hand side of this equation and we call it 𝑑. If we do that and then subtract 𝑑 from both sides, we can write that 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 minus 𝑑 equals zero. And arranged this way, we have what’s called the general form of a plane’s equation.

So now we have a vector form, a scalar form, and a general form for the equation of our plane. And note that the general and scalar forms both came from the vector form. This means if we can remember this mathematical relationship and what it means, then we’ll stand a good chance of being able to generate all three equations. And we can also note that the basic fact, which undergirds the vector form of our equation, is that if we take the dot product of two vectors that are perpendicular to one another, that product is zero.

Knowing all this, let’s get some practice with these ideas now through an example.

Find the vector form of the equation of the plane that has normal vector 𝐧 equals 𝐢 hat plus 𝐣 hat plus 𝐤 hat and contains the point two, six, six.

Okay, so in this example, we have a plane. We’ll say this is it. And we’re told that relative to some set of coordinate axes, there’s a vector normal or perpendicular to the plane. And if we write that vector in vector form, we see that it has components of one, one, and one in the 𝑥-, 𝑦-, and 𝑧-directions, respectively. Along with this, we’re told that the plane contains a point, we’ll call it 𝑃 zero, with coordinates two, six, six. And knowing all this, we want to solve for the vector form of the plane’s equation.

To write the plane’s equation that way, we’ll want to define a vector that lies in the plane so that if we take the dot product of that vector and the normal vector 𝐧, we get zero. Here’s how we can go about doing that. First, let’s define a vector from our origin to the point 𝑃 zero. We’ll call the vector 𝐫 zero. And since it comes from the origin, it must have components two, six, six. And next, let’s do this. Let’s pick a point at random in our plane, we’ll call it 𝑃, and we’ll say this point has coordinates 𝑥, 𝑦, 𝑧.

We don’t specify what these values are, but nonetheless this point will help us because now we can draw a vector from our origin to this point 𝑃, call that vector 𝐫, which we see has components 𝑥, 𝑦, 𝑧. And then if we subtract the vector 𝐫 zero from 𝐫, and this was the whole purpose of defining 𝐫 in the first place, then we get a vector shown here, which lies in the plane, which means that this vector is indeed perpendicular to the normal vector 𝐧. And that means if we take the dot product of 𝐧 and 𝐫 minus 𝐫 zero, the result we’ll get is zero.

Another way to write this is 𝐧 dot 𝐫 is equal to 𝐧 dot 𝐫 zero. And we can now substitute in the known values for the normal vector 𝐧 and the vector 𝐫 zero. We’ll do that down here, where we see that the vector one, one, one dotted with 𝐫 is equal to one, one, one dotted with two, six, six. We see the vector 𝐫 is made up of the components 𝑥, 𝑦, 𝑧. But since we don’t know those more specifically, we’ll leave this left-hand side as it is. On the right-hand side, though, we can compute this dot product by multiplying the respective components together, giving us two plus six plus six.

And since these numbers add up to 14, we can write a simplified form of our equation like this. This is the vector form of the equation of our plane.

Let’s now look at another example, where we give the equation of a plane in a different form.

Give the equation of the plane with normal vector 10, eight, three that contains the point 10, five, five.

Okay, let’s say that this here is our plane. And we’re told that it has a normal vector, a vector perpendicular to the plane with components 10, eight, and three. And along with this, the plane contains the point, we’ve called 𝑃 zero, which has coordinates 10, five, and five. We want to solve for the equation of this plane. And we’ll solve for what’s called the general or normal form of that equation.

Given a vector normal to our plane with components 𝑎, 𝑏, and 𝑐, the general form of the plane’s equation is given by this expression, where 𝑑 is equal to the dot product of our normal vector and a vector called 𝐫 zero. In the case of our plane, we don’t have an 𝐫 zero vector, but we do have a point 𝑃 zero. So if we draw a vector from the origin of our coordinate frame to 𝑃 zero, we can call that 𝐫 zero, where its components are the same as the coordinates of 𝑃 zero.

Returning to our general form of the equation of a plane, we can see that to write this expression out for a particular instance, we’ll need to know the components of a normal vector as well as the components of this vector 𝐫 zero. We now have that information for our plane, with 𝐧 having components 10, eight, three and 𝐫 zero having components 10, five, five. So applying this equation to our scenario, here is what we can write. 10𝑥 plus eight 𝑦 plus three 𝑧, where 10, eight, and three are the respective components of 𝐧, minus that normal vector 𝐧 dotted with the vector to our point 𝑃 zero, that vector is 𝐫 zero, is all equal to zero.

To simplify this further, let’s take the dot product of these two vectors. In doing this, we multiply 10 by 10, eight by five, and three by five so that this dot product is equal to 100 plus 40 plus 15. And adding those values up, we get 155. This, then, is the expression that gives us the general form for the equation of our plane.

Let’s look now at another example, where we solve for the normal form of a plane’s equation.

Write, in normal form, the equation of the plane one, zero, three; one, two, negative one; and six, one, six.

Okay, so let’s say that this here is our plane defined with respect to these coordinate axes. And we’re told that our plane has these three points, we’ve called them 𝑃 one, 𝑃 two, and 𝑃 three, in it. We’re given the coordinates of these three points. And based on this, we want to solve for the equation of the plane in normal form.

In order to do that, we need two things: first, a normal vector, we can call it 𝐧, perpendicular or normal to our plane and also a point that’s located on our plane. And we see that we actually have three of those. Since we already have that second bit of information, let’s look into how we would solve for a vector that’s normal to our plane.

To start, note that we could draw vectors from the origin of our coordinate frame to these three points. And just like the points are 𝑃 one, 𝑃 two, and 𝑃 three, we can call these vectors 𝐫 one, 𝐫 two, and 𝐫 three, respectively. Say next that we take our vector 𝐫 two and we subtract it from the vector 𝐫 one. Graphically, that would make a vector that lies in our plane that looks like this. And since the vector 𝐫 one has components which are equal to the coordinates of point 𝑃 one and the vector 𝐫 two has components that are equal to the coordinates of point 𝑃 two, we can actually compute the components of this resultant vector.

We see it’s equal to zero, negative two, four. So that’s one vector that lies entirely in our plane. And we can then go about identifying another by considering the vectors 𝐫 two and 𝐫 three this time. If we subtract 𝐫 three from 𝐫 two on our plane, that would give us a vector that looks this way. And once again, we can use the coordinates of our points translated to components of vectors to solve for this vector. One minus six is negative five. Two minus one is one. Negative one minus six is negative seven. And so we now have the components of a second vector that also lies in our plane.

Knowing this, we can say that if we were to take the cross product of these two vectors, then the result would be perpendicular to them both. And since both these vectors lie in the plane, that resultant would be normal to it. In other words, we would have a normal vector 𝐧. And that’s good news because if we had a normal vector, we would have all the information we need to write the equation of our plane in normal form.

Let’s recall, then, that if we were to take the cross product of two three-dimensional vectors, we’ve called them 𝐀 and 𝐁, then that would be equal to the determinant of this three-by-three matrix. Note that the first row of this matrix is the 𝐢-, 𝐣-, and 𝐤-unit vectors and the next two rows are the corresponding components of our two vectors 𝐀 and 𝐁. The cross product of two vectors yields a vector. And as we mentioned, this resultant is perpendicular to both the vectors we used to calculate it.

In our situation, our vectors are 𝐫 one minus 𝐫 two and 𝐫 two minus 𝐫 three. And this cross product is equal to the determinant of this matrix, where we still have to fill in the second and third rows. Recall that these rows are completed by the components of the two vectors we’re crossing together. Looking at our first vector, 𝐫 one minus 𝐫 two, we see from its components that its 𝑥-component is zero, its 𝑦-component negative two, and its 𝑧-component four.

Considering next our second vector, 𝐫 two minus 𝐫 three, this has an 𝑥-component of negative five, a 𝑦-component of one, and a 𝑧-component of negative seven. We’re now ready to calculate this cross product. And we’ll start with the 𝐢-component. Striking through the row and column containing this element, we know the 𝐢-component is equal to the determinant of this two-by-two matrix. Negative two times negative seven is positive 14. And from that, we subtract four times one, or four. Moving on to the 𝐣-component, that’s equal to negative the determinant of this two-by-two matrix, where zero times negative seven is zero. And from that, we subtract four times negative five.

Lastly, with our 𝐤-component, this equals the determinant of this matrix, which we see is zero times one, or zero, minus negative two times negative five. That’s positive 10. If we add all these components together, the result is the cross product of 𝐫 two minus 𝐫 one and 𝐫 two minus 𝐫 three. This is equal to 10𝐢 minus 20𝐣 minus 10𝐤 or, written in vector form, 10, negative 20, negative 10. So now finally, we have 𝐧, a vector that’s normal to our plane. And as we said, this is because the two vectors we crossed together are both in the plane. So by the property of the cross product, their resultant must be perpendicular to them both.

Knowing 𝐧, let’s clear some space and now start writing out the equation of our plane in normal form. We mentioned earlier that to do this, we need a normal vector, which we now have, as well as a point in the plane. We’ve got three of those. Let’s choose to work with point 𝑃 one. We can recall that in general, the vector form of the equation of a plane is given by this expression. It tells us that a vector normal to the plane dotted with a vector to a generic or general point in the plane is equal to the dot product of that normal vector and a vector to a known point.

As we apply this relationship, we’ll use our normal vector 𝐧 and 𝐫, we’ll say, has components 𝑥, 𝑦, and 𝑧. For the vector to a known point in our plane, we’ll use the vector we defined, 𝐫 one. Substituting in for 𝐧 and 𝐫 one, we get this expression. And we can now take these two dot products. Multiplying the respective components of these vectors together, on the left we get 10𝑥 minus 20𝑦 minus 10𝑧 and on the right we get 10 times one minus 20 times zero minus 10 times three. That adds up to negative 20.

We’re almost there. But to get the equation for our plane in normal form, we’ll want a zero to appear on the right-hand side. So if we add positive 20 to this side and this side of the equation, we’ll find that 10𝑥 minus 20𝑦 minus 10𝑧 plus 20 equals zero. And note that if we divide both sides of this equation by 10, then that yields this expression. 𝑥 minus two 𝑦 minus 𝑧 plus two equals zero. And this is our final answer for the normal form of the equation of our plane.

Let’s now finish up our lesson by reviewing some key points. In this lesson, we’ve seen that a plane can be uniquely defined by a normal vector and a point on the plane. Given a normal vector with components 𝑎, 𝑏, and 𝑐 and a point 𝑃 zero in the plane with coordinates 𝑥 zero, 𝑦 zero, and 𝑧 zero, we can define a vector 𝐫 zero from the origin of our coordinate frame to 𝑃 zero. And choosing another point at random on our plane and calling it 𝑃, we can define a vector to that point, calling it 𝐫. The vector 𝐫 minus 𝐫 zero is in the plane and perpendicular to the normal vector 𝐧 so that the equation of the plane can be written in vector form like this, scalar form like this, or in general form like this. These are three equivalent ways of writing the equation of a plane.

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