Lesson Video: Equation of a Plane: Intercept and Parametric Forms | Nagwa Lesson Video: Equation of a Plane: Intercept and Parametric Forms | Nagwa

# Lesson Video: Equation of a Plane: Intercept and Parametric Forms Mathematics

In this video, we will learn how to find the equation of a plane in different forms, such as intercept and parametric forms.

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### Video Transcript

In this video, our topic is the equation of a plane intercept and parametric forms. Weβll learn here how to write the equation of a plane in these two forms. And weβll also see how they relate to other ways of expressing a planeβs equation.

As we get started, letβs remind ourselves of other ways weβve learned of writing out the equation of a plane. If we have a plane existing in three-dimensional space and we know a vector thatβs normal to that plane as well as a point that lies in it, then we can define a vector β here weβve called it π« zero β traveling from the origin of our coordinate frame to our known point. And we can define a vector β we called it π« β going from the origin of our frame to an arbitrary point π in our plane so that if we subtract π« zero from π«, we get a vector thatβs entirely in this plane. Therefore, the dot product of this vector and our normal vector π§ must be zero.

Equivalently, we can write that π§ dot π« is equal to π§ dot π« zero. And this is what is known as the vector form of the equation of a plane. Now, if we say that our known point π zero has coordinates π₯ zero, π¦ zero, and π§ zero, that means we can write the vector π« zero as a vector with those components. And so carrying out the two dot products in the vector form of a planeβs equation, we get this result. And if we group together the terms that have π, π, and π in common, then we arrive at what is known as the scalar form of a planeβs equation.

And then to generate the general form of the equation of a plane, we can come back to this line in our calculation. And what weβll do is group together everything on the right-hand side, and weβll call that negative π. If we then add π to both sides, we arrive at this expression, which is the general form of a planeβs equation.

Knowing all this, weβre now ready to look into two more ways of writing out the equation of a plane. As weβve seen, one is called the intercept form and the other the parametric form. The idea behind the intercept form is this. In general, a two-dimensional plane in three-dimensional space will intersect the π₯-, π¦-, and π§-axes of our coordinate frame. After all, even though weβve represented our plane here as a four-sided object, we know that really all of these sides extend infinitely far outward. Except in very exceptional cases then, this plane will at some point intersect all three of our axes.

To see that more clearly, say that we change our perspective so that now our plane is oriented to our axes like this. If we were to mark in the points where our plane intersects the π₯-, π¦-, and π§-axes, they might look like this. Say further that we know these points of intersection. Along the π₯-axis, for example, our point has coordinates capital π΄, zero, zero. Going further, on the π¦-axis, weβll say this intersection occurs a distance capital π΅ from the origin and then a distance of capital πΆ along the π§-axis.

Now, knowing these three points, we can insert them one by one into the general form of the equation of our plane. For example, inserting the intersection point along the π₯-axis into our equation, we would get π times capital π΄ plus π times zero plus π times zero plus π equals zero. We see that this simplifies to π times capital π΄ plus π equals zero, which then tells us if we solve for lowercase π that it equals negative π over capital π΄.

What we can do is store that result off to the side and next use the intersection point along the π¦-axis in the same way. Substituting this point into our general form, we get π times zero plus π times capital π΅ plus π times zero plus π equals zero. This simplifies to π times capital π΅ plus π equals zero, which tells us that π is equal to negative π over capital π΅. Weβll record this result too. And finally, weβll substitute in the intersection point along the π§-axis. π times zero plus π times zero plus π times capital πΆ plus π equals zero. So π times capital πΆ plus π is zero or π equals negative π over capital πΆ.

Now that we have expressions for π, π, and π in terms of the points of intersection of our plane, we can return once more to our general form and substitute these values for π, π, and π into this equation. That gives us this equation. And as we look at the left-hand side, we notice that our factor of π is common to all four terms. So hereβs what we can do. If we subtract π from both sides and then divide both sides by negative π, notice that on the right-hand side, weβll get positive one. And on the left-hand side, all instances of the variable π cancel out.

Since we were dividing by negative π, we also lose our minus signs. And so we end up with this expression. This is the intercept form of the equation of our plane, where the plane intersects the π₯-, π¦-, and π§-axes at the points weβve identified.

This brings us to the last form of the equation of a plane weβll talk about. As our start point, letβs go back to our plane and letβs say that we know some point on it, not the points weβve identified, but some other one somewhere on the planeβs surface. To completely define this plane, to be able to access any point in the plane, we want to be able to start at this point we know to be in the plane and then move out any distance in any direction thatβs also in the plane. Letβs say that this known point is π zero.

And in order to access all points on the plane from π zero, what we need are two vectors β weβll call them π― one and π― two β that lie in the plane and are not parallel. The idea here is that if each one of these two vectors is multiplied by a scale factor β weβll call them π‘ one and π‘ two, respectively β then by letting π‘ one and π‘ two range over all possible negative and positive values, we allow this expression to sweep through all the points in our plane.

To say this another way, letβs return to our sketch where we have π zero and these three points of intersection. Since our intersection points all lie in the plane, by definition, any vector going from one of these points to another will also be in the plane. What we do then is move out on a vector starting at our origin until we reach the known point in the plane π zero. Mathematically, we can represent that as a vector π« zero. If we then add to that value in the plane all possible multiples of our nonparallel vectors π― one and π― two, then we will have comprehensively covered all points in our plane. Weβll have defined it.

In this equation, the values π‘ one and π‘ two are known as our parameters. They are the values we can vary in order to sweep through or scan over all the points in our plane. Now, because weβre looking at an equation for a three-dimensional vector, really there are three separate equations involved here. If we write π« as a vector with components π₯, π¦, and π§; our zero as a vector with components π₯ zero, π¦ zero, and π§ zero; and likewise π― one and π― two in terms of their components, then we can write out separate equations for π₯, π¦, and π§ like this. The parameters π‘ one and π‘ two appear in all three equations. And this then is the parametric form of the equation of a plane.

So we now know five different ways to write out a planeβs equation. Letβs now get some practice with these forms through a few example exercises.

Find, in parametric form, the equation of the plane that passes through the point π΄ one, two, one and the two vectors π one equals one, negative one, two and π two equals two, negative one, one.

Okay, we have here all these possible answers for the parametric form of the equation of our plane. And as weβve seen, this plane passes through this point π΄ and contains these two vectors π one and π two. Visually then, our plane might look like this. In writing the equation of this plane, the idea is that we start at our known point and then we move out from that point in multiples of the vectors π one and π two. We could say then that a vector describing our planeβs entire surface β weβll call it π« β is equal to a vector to our known point on the plane plus one parameter that varies across all possible numbers multiplied by our one vector π one added to another parameter that also varies across all possible numbers multiplied by the vector π two.

Even though it may seem weβre looking at just one equation here, actually, there are three involved: one for the π₯-dimension, one for the π¦, and one for the π§. To see that, we can replace this vector π« with its components. And now we see that, for example, π₯ is equal to one plus π‘ one times one plus π‘ two times two. Then, likewise, π¦ is equal to two plus π‘ one times negative one plus π‘ two times negative one. And then, lastly, we also have an equation for the π§-component of our vector. π§ equals one plus two times π‘ one plus π‘ two. This is the parametric form of the equation of our plane. And if we look through our answer options, we see it matches up with option (A). The equation of our plane in parametric form is π₯ equals one plus π‘ one plus two times π‘ two, π¦ equals two minus π‘ one minus π‘ two, π§ equals one plus two π‘ one plus π‘ two.

Now letβs look at an example where we solve for the parametric form of a planeβs equation using three points.

Find the parametric form of the equation of the plane that passes through the points π΄ one, five, one; π΅ three, four, three; and πΆ two, three, four.

Okay, so we have this plane that contains these three points π΄, π΅, and πΆ. And we want to figure out which of these five options gives us the parametric form of the equation of the plane. We can recall that to define the parametric form of a planeβs equation, we need to know one point on the plane and two vectors that lie in it. Here we have three points. And we can actually use these points to define two coplanar vectors. For example, if we subtract point π΄ from point π΅, we get this vector in pink. And likewise, if we subtract π΄ from πΆ, we get this vector.

Substituting in the coordinates of these points, for π΅ minus π΄, three minus one is two, four minus five is negative one, and three minus one is two. And then, for πΆ minus π΄, two minus one is one, three minus five is negative two, and four minus one is three. And what weβre saying is that these coordinates are actually the components of two vectors in our plane. Weβll call them π― one and π― two.

And now note that we have a point β weβve picked point π΄ β as well as two vectors that lie in this plane. We can now recall the most general way to write the parametric form of a planeβs equation. Itβs in terms of a point in the plane that we have a vector going to and two vectors, π― one and π― two, that lie in the plane, each of which is multiplied by its own parameter.

Applying this equation to our scenario, we can write that the vector π« with components π₯, π¦, π§ is equal to a vector to our known point, one, five, one, plus a parameter π‘ one times our first vector π― one plus a second parameter π‘ two times π― two. Note that from this expression, we can get equations for π₯ and π¦ and π§. For example, π₯ is equal to one plus two times π‘ one plus one times π‘ two. π¦ is equal to five minus one times π‘ one minus two times π‘ two. And then π§ equals one plus two times π‘ one plus three times π‘ two. And these equations altogether are the parametric form of the equation of our plane. Looking over our answer options, we see a match with choice (E).

Now, itβs true that there are many equivalent ways to express the parametric form of a planeβs equation. For example, we mightβve picked a different point, say π΅ or πΆ, rather than point π΄. And we mightβve solved for different coplanar vectors than these two here. Nonetheless, we still wouldnβt have chosen options (A), (B), (C), or (D) because none of those options use point π΄, π΅, or πΆ as the point that lies in the plane.

Our final answer then is choice (E). π₯ equals one plus two times π‘ one plus π‘ two, π¦ equals five minus π‘ one minus two π‘ two, and π§ equals one plus two π‘ one plus three π‘ two.

Letβs now look at an example where we start with the parametric form of a planeβs equation and convert to another form.

Find the general equation of the plane π₯ equals four plus seven π‘ one plus four π‘ two, π¦ equals negative three minus four π‘ two, π§ equals one plus three π‘ one.

Alright, so in this exercise, weβre given the equation of a plane in parametric form. That means we have three equations for π₯, π¦, and π§. And theyβre written in terms of two parameters, π‘ one and π‘ two. We want to convert from the parametric form to the general form of the equation of this plane. And we can remember that the general or Cartesian form of a planeβs equation is given by ππ₯ plus ππ¦ plus ππ§ plus π equaling zero. So the question is, how do we express these equations this way?

We can start by remembering that the parametric form of a planeβs equation involves one point in the plane and two vectors that also lie in it. We can write that mathematically like this. This equation means that we can get to any point in our plane by starting at a known point in the plane and then moving out from that in these two different directions defined by these vectors π― one and π― two, where these vectors critically are varied by these parameters π‘ one and π‘ two.

Now, these three given equations for π₯, π¦, and π§ fit into this general parametric form. What we mean is that, for example, π₯ is equal to four β that would be π₯ zero in this form β plus seven times π‘ one β so π― one π₯ would be seven β plus four times π‘ two β so π― two π₯ would be four. Carrying this out for π₯, π¦, and π§ altogether, we get this result, which tells us that according to these parametric equations, our plane contains the point four, negative three, one and the vectors seven, zero, three and four, negative four, zero.

Now, at this point, letβs recall that the general form of a plane is based on defining a plane in terms of a vector normal to it and one point on it, while on the other hand the parametric form comes from defining a plane in terms of a point on it and two vectors that lie in it. To go from parametric to general form then, we need to take these two coplanar vectors and combine them in some way to get a vector normal to the plane.

We can do exactly this by taking the cross product of our two coplanar vectors. Weβve called them π― one and π― two. Note that the components of π― one are seven, zero, three and those of π― two are four, negative four, zero. And so those are the values we use in the second two rows of our three-by-three matrix. Calculating this determinant, we get π’ times zero minus negative 12 minus π£ times zero minus 12 plus π€ times negative 28 minus zero. Thatβs equal to 12π’ plus 12π£ minus 28π€ or, written another way, 12, 12, negative 28.

Notice that if we divide all the components by four in this vector, then we will get a vector that is reduced but that is still normal to the plane of interest. For simplicityβs sake then, weβll say that this is our normal vector π§.

Okay, so now weβve got a vector thatβs normal to our plane. And we also have a point that lies in it: four, negative three, one. We can use this information to write the equation of our plane as our normal vector dotted with a vector to a general point in the plane being equal to our normal vector dotted with a vector to a point we know to lie in the plane. Carrying out those dot products gives us three π₯ plus three π¦ minus seven π§ being equal to 12 minus nine minus seven, which is equal to negative four.

With our equation written this way, we can see weβre very close to the general form of a plane. As a last step, letβs add positive four to both sides of this equation, leaving us with this result. So the general form of the equation of our plane is three π₯ plus three π¦ minus seven π§ plus four equals zero.

Letβs finish up this lesson now by summarizing a few key points. We started this lesson reminding ourselves that the equation of a plane can be written in vector, scalar, and general forms. We then saw that we can also use whatβs called intercept form, where we use the π₯-, π¦-, and π§-coordinates where our plane intersects the π₯-, π¦-, and π§-axes to define the plane. And lastly, we learned itβs possible to write the equation of a plane in whatβs called parametric form. This involves using two vectors that lie in the plane as well as a vector to a point in the plane to generate a set of equations describing the π₯-, π¦-, and π§-coordinates of all the points in the plane.