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In this explainer, we will learn how to find the equation of a plane in different forms, such as intercept and parametric forms. Consider a plane that does not pass through the origin, is not parallel to any of the axes, and thus intersects the three axes at three points with coordinates ๐ด(๐,0,0), ๐ต(0,๐,0), and ๐ถ(0,0,๐). We say that the ๐ฅ, ๐ฆ, and ๐งintercepts of the plane are ๐, ๐, and ๐. Having the coordinates of three points in the plane, we can easily find two noncollinear vectors included in the plane, for instance, โ๐ข=๏ ๐ต๐ด=(๐,โ๐,0) and โ๐ฃ=๏ ๐ถ๐ด=(๐,0,โ๐). A normal vector of the plane is then given by โ๐=โ๐ขรโ๐ฃ=โ๐โ๐โ๐๐โ๐0๐0โ๐=๐๐โ๐+๐๐โ๐+๐๐โ๐. As 1๐๐๐โ๐=๏ผ1๐,1๐,1๐๏ is also a normal vector of the plane, we can write a general equation of the form ๐ฅ๐+๐ฆ๐+๐ง๐+๐=0. Substituting in the coordinates of one of three known points included in the plane, we find that ๐=โ1. The intercept equation of a plane with ๐ฅ, ๐ฆ, and ๐งintercepts ๐, ๐, and ๐, respectively, is given by ๐ฅ๐+๐ฆ๐+๐ง๐=1. Let us look at the equation of planes that are parallel to one axis. Consider, for instance, a plane parallel to the ๐งaxis that intersects the ๐ฅaxis at ๐ด(๐,0,0) and the ๐ฆaxis at ๐ต(0,๐,0). Two vectors included in the plane are, for instance, โ๐ข=๏ ๐ต๐ด=(๐,โ๐,0) and โ๐ฃ=(0,0,1), a vector parallel to the ๐งaxis. A normal vector of the plane is then given by โ๐=โ๐ขรโ๐ฃ=โ๐โ๐โ๐๐โ๐0001=โ๐โ๐โ๐โ๐. If we take, as a normal vector of the plane, โ1๐๐โ๐=๏ผ1๐,1๐,0๏, we find the intercept equation to be ๐ฅ๐+๐ฆ๐=1. It is equivalent to the equation of a line in the (๐ฅ,๐ฆ) plane, plus the fact that the ๐งcoordinate can take up any value in โ. Consider now a plane that is parallel to two axes, for example, the ๐ฅ and ๐งaxes. It intersects the third axis, here the ๐ฆaxis, at a point with coordinates (0,๐,0). This plane contains all the points that have ๐ as the ๐ฆcoordinate; hence, its equation is simply ๐ฆ=๐, which can be written in intercept form as ๐ฆ๐=1. Let us now find the intercept equation of a plane with the next example. Find the equation of the plane whose ๐ฅ, ๐ฆ, and ๐งintercepts are โ7, 3, and โ4 respectively. AnswerThe intercept equation of a plane with ๐ฅ, ๐ฆ, and ๐งintercepts ๐, ๐, and ๐, respectively, is given by ๐ฅ๐+๐ฆ๐+๐ง๐=1. Here, ๐=โ7, ๐=3, and ๐=โ4. Hence, we find that the equation of the plane is โ๐ฅ7+๐ฆ3โ๐ง4=1. Finally, let us convert a general equation of a plane into an intercept equation in the next example. Write, in intercept form, the equation of the plane 16๐ฅ+2๐ฆ+8๐งโ16=0. AnswerAn intercept equation of a plane is of the form ๐ฅ๐+๐ฆ๐+๐ง๐=1. Adding 16 to each side of the given equation, we find 16๐ฅ+2๐ฆ+8๐ง=16. As we want to have 1 on the righthand side, we divide both sides by 16, which gives ๐ฅ1+๐ฆ8+๐ง2=1. To find this equation, we could also first find ๐, ๐, and ๐, that is, the ๐ฅ, ๐ฆ, and ๐งintercepts of the plane. We find the ๐ฅintercept by substituting 0 for ๐ฆ and ๐ง into our equation. We find 16๐ฅโ16=0๐ฅ=1. Hence, ๐=1. Similarly, we find that ๐=8 and ๐=2. Substituting these values into the intercept equation, we find ๐ฅ1+๐ฆ8+๐ง2=1. The intercept equation of the plane of general equation 16๐ฅ+2๐ฆ+8๐งโ16=0 is ๐ฅ1+๐ฆ8+๐ง2=1. Let us now look at another form of equation of a plane, namely, the parametric form. Any point in the coordinate plane is uniquely defined by its two coordinates. In other words, for any point ๐(๐ฅ,๐ฆ), its position vector is given by ๏ ๐๐=๐ฅโ๐+๐ฆโ๐, where ๐ is the origin of the coordinate system and โ๐ and โ๐ are the unit vectors along its two axes. We can write a similar equation with any two noncollinear vectors โ๐ข and โ๐ฃ in the plane: ๏ ๐๐=๐โ๐ข+๐โ๐ฃ, where ๐ and ๐ are two real numbers. This means that any vector in the plane can be written as a linear combination of two noncollinear vectors. It follows that two noncollinear vectors allow us to define any point in the plane. We use this property to write parametric equations of a plane in space. Let us consider a plane in space that contains point ๐(๐ฅ,๐ฆ,๐ง)๏ฏ๏ฏ๏ฏ and two noncollinear vectors โ๐ข=๏น๐ข,๐ข,๐ข๏ ๏๏๏ and โ๐ฃ=๏น๐ฃ,๐ฃ,๐ฃ๏ ๏๏๏. For any point ๐(๐ฅ,๐ฆ,๐ง) in the plane, we have ๏ ๐๐=๐กโ๐ข+๐กโ๐ฃ,๏ง๏จ where ๐ก๏ง and ๐ก๏จ are two real numbers. This gives three equations for the three components of the vector ๏ ๐๐: ๐ฅโ๐ฅ=๐ก๐ข+๐ก๐ฃ,๐ฆโ๐ฆ=๐ก๐ข+๐ก๐ฃ,๐งโ๐ง=๐ก๐ข+๐ก๐ฃ.๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏ By rearranging these equations, we get the parametric equations of a plane. The parametric equations of a plane in space that contains point ๐(๐ฅ,๐ฆ,๐ง)๏ฏ๏ฏ๏ฏ and two noncollinear vectors โ๐ข=๏น๐ข,๐ข,๐ข๏ ๏๏๏ and โ๐ฃ=๏น๐ฃ,๐ฃ,๐ฃ๏ ๏๏๏ are a set of three equations of the form ๐ฅ=๐ฅ+๐ก๐ข+๐ก๐ฃ,๐ฆ=๐ฆ+๐ก๐ข+๐ก๐ฃ,๐ง=๐ง+๐ก๐ข+๐ก๐ฃ,๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏ where ๐ก๏ง and ๐ก๏จ are two varying real numbers, called the parameters. By varying the parameters ๐ก๏ง and ๐ก๏จ across โ, the three equations describe the coordinates of all points in the plane. Let us look at a first example. Find, in parametric form, the equation of the plane that passes through the point ๐ด(1,2,1) and the two vectors โ๐=(1,โ1,2)๏ง and โ๐=(2,โ1,1)๏จ. AnswerThe parametric equations of a plane that passes through the point ๐ด(๐ฅ,๐ฆ,๐ง)๏ ๏ ๏ and two vectors โ๐=๏น๐,๐,๐๏ ๏ง๏ง๏๏๏ง๏๏๏ง๏๏ and โ๐=๏น๐,๐,๐๏ ๏จ๏จ๏๏๏จ๏๏๏จ๏๏ are of the form ๐ฅ=๐ฅ+๐ก๐+๐ก๐,๐ฆ=๐ฆ+๐ก๐+๐ก๐,๐ง=๐ง+๐ก๐+๐ก๐,๏ ๏ง๏ง๏๏๏จ๏จ๏๏๏ ๏ง๏ง๏๏๏จ๏จ๏๏๏ ๏ง๏ง๏๏๏จ๏จ๏๏ where ๐ก๏ง and ๐ก๏จ are two real numbers. Substituting in (๐ฅ,๐ฆ,๐ง)=(1,2,1)๏ ๏ ๏ , ๏น๐,๐,๐๏ =(1,โ1,2)๏ง๏๏๏ง๏๏๏ง๏๏, and ๏น๐,๐,๐๏ =(2,โ1,1)๏จ๏๏๏จ๏๏๏จ๏๏, we find ๐ฅ=1+๐ก+2๐ก,๐ฆ=2โ๐กโ๐ก,๐ง=1+2๐ก+๐ก.๏ง๏จ๏ง๏จ๏ง๏จ Let us now find the parametric equations of a plane that passes through three given points. Find the parametric form of the equation of the plane that passes through the points ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(2,3,4).
AnswerWe are given three points, ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(2,3,4), that are in the plane. Let us first check that these three points are noncollinear by using the cross product, since if two vectors are collinear, then their cross product is zero. With ๏ ๐ด๐ต=(2,โ1,2) and ๏ ๐ด๐ถ=(1,โ2,3), we find that ๏ ๐ด๐ตร๏ ๐ด๐ถ=โ๐โ๐โ๐2โ121โ23=โ๐โ4โ๐โ3โ๐; hence, the points are not collinear, which means that they define a plane indeed. We know that the parametric equations of a plane in space that contains point ๐(๐ฅ,๐ฆ,๐ง)๏ฏ๏ฏ๏ฏ and two noncollinear vectors โ๐ข=๏น๐ข,๐ข,๐ข๏ ๏๏๏ and โ๐ฃ=๏น๐ฃ,๐ฃ,๐ฃ๏ ๏๏๏ are a set of three equations of the form ๐ฅ=๐ฅ+๐ก๐ข+๐ก๐ฃ,๐ฆ=๐ฆ+๐ก๐ข+๐ก๐ฃ,๐ง=๐ง+๐ก๐ข+๐ก๐ฃ,๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏ where ๐ก๏ง and ๐ก๏จ are two real numbers. Let us now identify, for each option, the coordinates (๐ฅ,๐ฆ,๐ง)๏ฏ๏ฏ๏ฏ and the two sets of vector components ๏น๐ข,๐ข,๐ข๏ ๏๏๏ and ๏น๐ฃ,๐ฃ,๐ฃ๏ ๏๏๏. Our findings are given in the table below.
With the three points ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(2,3,4), we can form the vectors ๏ ๐ด๐ถ=(1,โ2,3), ๏ ๐ด๐ต=(2,โ1,2), and ๏๐ต๐ถ=(โ1,โ1,1)  and of course the opposite vectors ๏ ๐ถ๐ด=(โ1,2,โ3), ๏ ๐ต๐ด=(โ2,1,โ2), and ๏๐ถ๐ต=(1,1,โ1) as well. Let us now analyze for each option what point and vectors have been used. In the table below, the cell has been shaded when the point coordinates or vector components do not correspond to any of the information given in the question. We find that option D is correct. The parametric equations correspond to a plane that contains point ๐ต(3,4,3) and the vectors ๏ ๐ต๐ด=(โ2,1,โ2) and ๏๐ต๐ถ=(โ1,โ1,1). Let us look now at how to write the equation of a plane in general form from its parametric equations. Recall that the standard form of the equation of a plane is ๐(๐ฅโ๐ฅ)+๐(๐ฆโ๐ฆ)+๐(๐งโ๐ง)=0,๏๏ฏ๏๏ฏ๏๏ฏ where โ๐=๏น๐,๐,๐๏ ๏๏๏ is a normal vector of the plane and ๐(๐ฅ,๐ฆ,๐ง)๏ฏ๏ฏ๏ฏ is a point on the plane. The standard from can be easily rearranged to the general form ๐๐ฅ+๐๐ฆ+๐๐ง+๐=0.๏๏๏ As the normal vector of a plane is given by the cross product of any two noncollinear vectors lying in the plane, it is possible to write the general form of the equation of a plane from its parametric equations. The normal vector of the plane is given by the cross product of any two noncollinear vectors of the plane. Therefore, taking โ๐ข=๏น๐ข,๐ข,๐ข๏ ๏๏๏ and โ๐ฃ=๏น๐ฃ,๐ฃ,๐ฃ๏ ๏๏๏ from the parametric equations ๐ฅ=๐ฅ+๐ก๐ข+๐ก๐ฃ,๐ฆ=๐ฆ+๐ก๐ข+๐ก๐ฃ,๐ง=๐ง+๐ก๐ข+๐ก๐ฃ,๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏๏ฏ๏ง๏๏จ๏ we find โ๐=๏น๐,๐,๐๏ =โ๐ขรโ๐ฃ๏๏๏. The standard equation is then ๐(๐ฅโ๐ฅ)+๐(๐ฆโ๐ฆ)+๐(๐งโ๐ง)=0.๏๏ฏ๏๏ฏ๏๏ฏ It can be rearranged to the general form ๐๐ฅ+๐๐ฆ+๐๐ง+๐=0,๏๏๏ with ๐=โ๏น๐๐ฅ+๐๐ฆ+๐๐ง๏ .๏๏ฏ๏๏ฏ๏๏ฏ Let us use this method in the next example. Find the general equation of the plane ๐ฅ=4+7๐ก+4๐ก๏ง๏จ, ๐ฆ=โ3โ4๐ก๏จ, ๐ง=1+3๐ก๏ง. AnswerWe are given the parametric equations of the plane. We know that the coefficients of ๐ก๏ง on one side and those of ๐ก๏จ on the other side are the components of two vectors in the plane. Let us call these vectors โ๐ข and โ๐ฃ. We find โ๐ข=(7,0,3) and โ๐ฃ=(4,โ4,0). One normal vector of the plane is thus โ๐=โ๐ขรโ๐ฃ=โ๐โ๐โ๐7034โ40=โ๐(0ร0โ(โ4)ร3)โโ๐(7ร0โ3ร4)+โ๐(7ร(โ4)โ4ร0)=12โ๐+12โ๐โ28โ๐. We have found that โ๐=(12,12,โ28) is a normal vector of the plane. Normal vectors of a plane are all parallel, which means that they all can be written as ๐โ๐, where โ๐ is one normal vector and ๐ is a real number. Here, all components of โ๐ are multiples of 4, so we can take 14โ๐=(3,3,โ7) as a normal vector to write our general equation. The general equation of a plane is of the form ๐๐ฅ+๐๐ฆ+๐๐ง+๐=0,๏๏๏ where ๐๏, ๐๏, and ๐๏ are the components of a normal vector of the plane. Substituting these components into the general equation of the plane, we find 3๐ฅ+3๐ฆโ7๐ง+๐=0. To find the constant ๐, we need to know the coordinates of a point that belongs to a plane. We can find the coordinates of a point on the plane using the parametric equations: the constants in each equation correspond to the coordinates of a point on the plane. We find that the point with coordinates (4,โ3,1) belongs to the plane. We substitute these into the equation, which gives 3ร4+3ร(โ3)โ7ร1+๐=0๐=4. The general equation of the plane is, therefore, 3๐ฅ+3๐ฆโ7๐ง+4=0. It is worth noting that we could have kept our original normal vector to write the general equationโwe would have found the same final general equation after dividing both sides by 4.
