Find the equation of the plane whose intersepts on x,y, z axes are 1, 2, 4 respectively

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    Find the equation of the plane whose intersepts on x,y, z axes are 1, 2, 4 respectively

    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 ๐ด(๐‘Ž,0,0), ๐ต(0,๐‘,0), and ๐ถ(0,0,๐‘). We say that the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts of the plane are ๐‘Ž, 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, โƒ‘๐‘ข=๏ƒ ๐ต๐ด=(๐‘Ž,โˆ’๐‘,0) and โƒ‘๐‘ฃ=๏ƒ ๐ถ๐ด=(๐‘Ž,0,โˆ’๐‘). A normal vector of the plane is then given by โƒ‘๐‘›=โƒ‘๐‘ขร—โƒ‘๐‘ฃ=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Žโˆ’๐‘0๐‘Ž0โˆ’๐‘|||||=๐‘๐‘โƒ‘๐‘–+๐‘Ž๐‘โƒ‘๐‘—+๐‘Ž๐‘โƒ‘๐‘˜.

    As 1๐‘Ž๐‘๐‘โƒ‘๐‘›=๏€ผ1๐‘Ž,1๐‘,1๐‘๏ˆ is also a normal vector of the plane, we can write a general equation of the form ๐‘ฅ๐‘Ž+๐‘ฆ๐‘+๐‘ง๐‘+๐‘‘=0.

    Substituting in the coordinates of one of three known points included in the plane, we find that ๐‘‘=โˆ’1.

    The intercept equation of a plane with ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts ๐‘Ž, ๐‘, and ๐‘, respectively, is given by ๐‘ฅ๐‘Ž+๐‘ฆ๐‘+๐‘ง๐‘=1.

    Let us look at the equation of planes that are parallel to one axis. Consider, for instance, a plane parallel to the ๐‘ง-axis that intersects the ๐‘ฅ-axis at ๐ด(๐‘Ž,0,0) and the ๐‘ฆ-axis at ๐ต(0,๐‘,0). Two vectors included in the plane are, for instance, โƒ‘๐‘ข=๏ƒ ๐ต๐ด=(๐‘Ž,โˆ’๐‘,0) and โƒ‘๐‘ฃ=(0,0,1), a vector parallel to the ๐‘ง-axis. A normal vector of the plane is then given by โƒ‘๐‘›=โƒ‘๐‘ขร—โƒ‘๐‘ฃ=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Žโˆ’๐‘0001|||||=โˆ’๐‘โƒ‘๐‘–โˆ’๐‘Žโƒ‘๐‘—.

    If we take, as a normal vector of the plane, โˆ’1๐‘Ž๐‘โƒ‘๐‘›=๏€ผ1๐‘Ž,1๐‘,0๏ˆ, we find the intercept equation to be ๐‘ฅ๐‘Ž+๐‘ฆ๐‘=1.

    It is equivalent to the equation of a line in the (๐‘ฅ,๐‘ฆ) plane, plus the fact that the ๐‘ง-coordinate can take up any value in โ„.

    Consider now a plane that is parallel to two axes, for example, the ๐‘ฅ- and ๐‘ง-axes. It intersects the third axis, here the ๐‘ฆ-axis, at a point with coordinates (0,๐‘,0). This plane contains all the points that have ๐‘ as the ๐‘ฆ-coordinate; hence, its equation is simply ๐‘ฆ=๐‘, which can be written in intercept form as ๐‘ฆ๐‘=1.

    Let us now find the intercept equation of a plane with the next example.

    Find the equation of the plane whose ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts are โˆ’7, 3, and โˆ’4 respectively.

    Answer

    The intercept equation of a plane with ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts ๐‘Ž, ๐‘, and ๐‘, respectively, is given by ๐‘ฅ๐‘Ž+๐‘ฆ๐‘+๐‘ง๐‘=1.

    Here, ๐‘Ž=โˆ’7, ๐‘=3, and ๐‘=โˆ’4. Hence, we find that the equation of the plane is โˆ’๐‘ฅ7+๐‘ฆ3โˆ’๐‘ง4=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 16๐‘ฅ+2๐‘ฆ+8๐‘งโˆ’16=0.

    Answer

    An intercept equation of a plane is of the form ๐‘ฅ๐‘Ž+๐‘ฆ๐‘+๐‘ง๐‘=1.

    Adding 16 to each side of the given equation, we find 16๐‘ฅ+2๐‘ฆ+8๐‘ง=16.

    As we want to have 1 on the right-hand side, we divide both sides by 16, which gives ๐‘ฅ1+๐‘ฆ8+๐‘ง2=1.

    To find this equation, we could also first find ๐‘Ž, ๐‘, and ๐‘, that is, the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts of the plane. We find the ๐‘ฅ-intercept by substituting 0 for ๐‘ฆ and ๐‘ง into our equation. We find 16๐‘ฅโˆ’16=0๐‘ฅ=1.

    Hence, ๐‘Ž=1. Similarly, we find that ๐‘=8 and ๐‘=2. Substituting these values into the intercept equation, we find ๐‘ฅ1+๐‘ฆ8+๐‘ง2=1.

    The intercept equation of the plane of general equation 16๐‘ฅ+2๐‘ฆ+8๐‘งโˆ’16=0 is ๐‘ฅ1+๐‘ฆ8+๐‘ง2=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‘€(๐‘ฅ,๐‘ฆ), its position vector is given by ๏ƒ ๐‘‚๐‘€=๐‘ฅโƒ‘๐‘–+๐‘ฆโƒ‘๐‘—, where ๐‘‚ is the origin of the coordinate system and โƒ‘๐‘– and โƒ‘๐‘— are the unit vectors along its two axes. We can write a similar equation with any two noncollinear vectors โƒ‘๐‘ข and โƒ‘๐‘ฃ in the plane: ๏ƒ ๐‘‚๐‘€=๐‘Žโƒ‘๐‘ข+๐‘โƒ‘๐‘ฃ, where ๐‘Ž 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‘ƒ(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ and two noncollinear vectors โƒ‘๐‘ข=๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and โƒ‘๐‘ฃ=๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™. For any point ๐‘€(๐‘ฅ,๐‘ฆ,๐‘ง) in the plane, we have ๏ƒ ๐‘ƒ๐‘€=๐‘กโƒ‘๐‘ข+๐‘กโƒ‘๐‘ฃ,๏Šง๏Šจ where ๐‘ก๏Šง and ๐‘ก๏Šจ are two real numbers.

    This gives three equations for the three components of the vector ๏ƒ ๐‘ƒ๐‘€: ๐‘ฅโˆ’๐‘ฅ=๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ฆโˆ’๐‘ฆ=๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘งโˆ’๐‘ง=๐‘ก๐‘ข+๐‘ก๐‘ฃ.๏Œฏ๏Šง๏—๏Šจ๏—๏Œฏ๏Šง๏˜๏Šจ๏˜๏Œฏ๏Šง๏™๏Šจ๏™

    By rearranging these equations, we get the parametric equations of a plane.

    The parametric equations of a plane in space that contains point ๐‘ƒ(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ and two noncollinear vectors โƒ‘๐‘ข=๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and โƒ‘๐‘ฃ=๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™ are a set of three equations of the form ๐‘ฅ=๐‘ฅ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ฆ=๐‘ฆ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ง=๐‘ง+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๏Œฏ๏Šง๏—๏Šจ๏—๏Œฏ๏Šง๏˜๏Šจ๏˜๏Œฏ๏Šง๏™๏Šจ๏™ where ๐‘ก๏Šง and ๐‘ก๏Šจ are two varying real numbers, called the parameters.

    By varying the parameters ๐‘ก๏Šง and ๐‘ก๏Šจ 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 ๐ด(1,2,1) and the two vectors โƒ‘๐‘‘=(1,โˆ’1,2)๏Šง and โƒ‘๐‘‘=(2,โˆ’1,1)๏Šจ.

    Answer

    The parametric equations of a plane that passes through the point ๐ด(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œ ๏Œ ๏Œ  and two vectors โƒ‘๐‘‘=๏€น๐‘‘,๐‘‘,๐‘‘๏…๏Šง๏Šง๏Ž•๏—๏Šง๏Ž•๏˜๏Šง๏Ž•๏™ and โƒ‘๐‘‘=๏€น๐‘‘,๐‘‘,๐‘‘๏…๏Šจ๏Šจ๏Ž•๏—๏Šจ๏Ž•๏˜๏Šจ๏Ž•๏™ are of the form ๐‘ฅ=๐‘ฅ+๐‘ก๐‘‘+๐‘ก๐‘‘,๐‘ฆ=๐‘ฆ+๐‘ก๐‘‘+๐‘ก๐‘‘,๐‘ง=๐‘ง+๐‘ก๐‘‘+๐‘ก๐‘‘,๏Œ ๏Šง๏Šง๏Ž•๏—๏Šจ๏Šจ๏Ž•๏—๏Œ ๏Šง๏Šง๏Ž•๏˜๏Šจ๏Šจ๏Ž•๏˜๏Œ ๏Šง๏Šง๏Ž•๏™๏Šจ๏Šจ๏Ž•๏™ where ๐‘ก๏Šง and ๐‘ก๏Šจ are two real numbers.

    Substituting in (๐‘ฅ,๐‘ฆ,๐‘ง)=(1,2,1)๏Œ ๏Œ ๏Œ , ๏€น๐‘‘,๐‘‘,๐‘‘๏…=(1,โˆ’1,2)๏Šง๏Ž•๏—๏Šง๏Ž•๏˜๏Šง๏Ž•๏™, and ๏€น๐‘‘,๐‘‘,๐‘‘๏…=(2,โˆ’1,1)๏Šจ๏Ž•๏—๏Šจ๏Ž•๏˜๏Šจ๏Ž•๏™, we find ๐‘ฅ=1+๐‘ก+2๐‘ก,๐‘ฆ=2โˆ’๐‘กโˆ’๐‘ก,๐‘ง=1+2๐‘ก+๐‘ก.๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ

    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 ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(2,3,4).

    1. ๐‘ฅ=1โˆ’2๐‘กโˆ’๐‘ก๏Šง๏Šจ, ๐‘ฆ=5+๐‘ก+2๐‘ก๏Šง๏Šจ, ๐‘ง=1+3๐‘ก+2๐‘ก๏Šง๏Šจ
    2. ๐‘ฅ=2+2๐‘ก+๐‘ก๏Šง๏Šจ, ๐‘ฆ=3โˆ’๐‘ก+๐‘ก๏Šง๏Šจ, ๐‘ง=4+2๐‘ก+๐‘ก๏Šง๏Šจ
    3. ๐‘ฅ=โˆ’1โˆ’๐‘ก+2๐‘ก๏Šง๏Šจ, ๐‘ฆ=โˆ’1+2๐‘ก+๐‘ก๏Šง๏Šจ, ๐‘ง=1+3๐‘ก+2๐‘ก๏Šง๏Šจ
    4. ๐‘ฅ=3โˆ’2๐‘กโˆ’๐‘ก๏Šง๏Šจ, ๐‘ฆ=4+๐‘กโˆ’๐‘ก๏Šง๏Šจ, ๐‘ง=3โˆ’2๐‘ก+๐‘ก๏Šง๏Šจ
    5. ๐‘ฅ=2โˆ’2๐‘กโˆ’๐‘ก๏Šง๏Šจ, ๐‘ฆ=3+4๐‘ก๏Šง, ๐‘ง=4+2๐‘ก+3๐‘ก๏Šง๏Šจ

    Answer

    We are given three points, ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(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 ๏ƒ ๐ด๐ต=(2,โˆ’1,2) and ๏ƒ ๐ด๐ถ=(1,โˆ’2,3), we find that ๏ƒ ๐ด๐ตร—๏ƒ ๐ด๐ถ=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜2โˆ’121โˆ’23||||=โƒ‘๐‘–โˆ’4โƒ‘๐‘—โˆ’3โƒ‘๐‘˜; 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‘ƒ(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ and two noncollinear vectors โƒ‘๐‘ข=๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and โƒ‘๐‘ฃ=๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™ are a set of three equations of the form ๐‘ฅ=๐‘ฅ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ฆ=๐‘ฆ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ง=๐‘ง+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๏Œฏ๏Šง๏—๏Šจ๏—๏Œฏ๏Šง๏˜๏Šจ๏˜๏Œฏ๏Šง๏™๏Šจ๏™ where ๐‘ก๏Šง and ๐‘ก๏Šจ are two real numbers.

    Let us now identify, for each option, the coordinates (๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ and the two sets of vector components ๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and ๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™. Our findings are given in the table below.

    Option(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™
    A(1,5,1)(โˆ’2,1,3)(โˆ’1,2,2)
    B(2,3,4)(2,โˆ’1,2)(1,1,1)
    C(โˆ’1,โˆ’1,1)(โˆ’1,2,3)(2,1,2)
    D(3,4,3)(โˆ’2,1,โˆ’2)(โˆ’1,โˆ’1,1)
    E(2,3,4)(โˆ’2,4,2)(โˆ’1,0,3)

    With the three points ๐ด(1,5,1), ๐ต(3,4,3), and ๐ถ(2,3,4), we can form the vectors ๏ƒ ๐ด๐ถ=(1,โˆ’2,3), ๏ƒ ๐ด๐ต=(2,โˆ’1,2), and ๏ƒŸ๐ต๐ถ=(โˆ’1,โˆ’1,1) - and of course the opposite vectors ๏ƒ ๐ถ๐ด=(โˆ’1,2,โˆ’3), ๏ƒ ๐ต๐ด=(โˆ’2,1,โˆ’2), and ๏ƒŸ๐ถ๐ต=(1,1,โˆ’1) 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 ๐ต(3,4,3) and the vectors ๏ƒ ๐ต๐ด=(โˆ’2,1,โˆ’2) and ๏ƒŸ๐ต๐ถ=(โˆ’1,โˆ’1,1).

    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‘›(๐‘ฅโˆ’๐‘ฅ)+๐‘›(๐‘ฆโˆ’๐‘ฆ)+๐‘›(๐‘งโˆ’๐‘ง)=0,๏—๏Œฏ๏˜๏Œฏ๏™๏Œฏ where โƒ‘๐‘›=๏€น๐‘›,๐‘›,๐‘›๏…๏—๏˜๏™ is a normal vector of the plane and ๐‘ƒ(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ is a point on the plane. The standard from can be easily rearranged to the general form ๐‘›๐‘ฅ+๐‘›๐‘ฆ+๐‘›๐‘ง+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 โƒ‘๐‘ข=๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and โƒ‘๐‘ฃ=๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™ from the parametric equations ๐‘ฅ=๐‘ฅ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ฆ=๐‘ฆ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ง=๐‘ง+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๏Œฏ๏Šง๏—๏Šจ๏—๏Œฏ๏Šง๏˜๏Šจ๏˜๏Œฏ๏Šง๏™๏Šจ๏™ we find โƒ‘๐‘›=๏€น๐‘›,๐‘›,๐‘›๏…=โƒ‘๐‘ขร—โƒ‘๐‘ฃ๏—๏˜๏™.

    The standard equation is then ๐‘›(๐‘ฅโˆ’๐‘ฅ)+๐‘›(๐‘ฆโˆ’๐‘ฆ)+๐‘›(๐‘งโˆ’๐‘ง)=0.๏—๏Œฏ๏˜๏Œฏ๏™๏Œฏ

    It can be rearranged to the general form ๐‘›๐‘ฅ+๐‘›๐‘ฆ+๐‘›๐‘ง+๐‘‘=0,๏—๏˜๏™ with ๐‘‘=โˆ’๏€น๐‘›๐‘ฅ+๐‘›๐‘ฆ+๐‘›๐‘ง๏….๏—๏Œฏ๏˜๏Œฏ๏™๏Œฏ

    Let us use this method in the next example.

    Find the general equation of the plane ๐‘ฅ=4+7๐‘ก+4๐‘ก๏Šง๏Šจ, ๐‘ฆ=โˆ’3โˆ’4๐‘ก๏Šจ, ๐‘ง=1+3๐‘ก๏Šง.

    Answer

    We are given the parametric equations of the plane. We know that the coefficients of ๐‘ก๏Šง on one side and those of ๐‘ก๏Šจ on the other side are the components of two vectors in the plane. Let us call these vectors โƒ‘๐‘ข and โƒ‘๐‘ฃ. We find โƒ‘๐‘ข=(7,0,3) and โƒ‘๐‘ฃ=(4,โˆ’4,0). One normal vector of the plane is thus โƒ‘๐‘›=โƒ‘๐‘ขร—โƒ‘๐‘ฃ=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜7034โˆ’40||||=โƒ‘๐‘–(0ร—0โˆ’(โˆ’4)ร—3)โˆ’โƒ‘๐‘—(7ร—0โˆ’3ร—4)+โƒ‘๐‘˜(7ร—(โˆ’4)โˆ’4ร—0)=12โƒ‘๐‘–+12โƒ‘๐‘—โˆ’28โƒ‘๐‘˜.

    We have found that โƒ‘๐‘›=(12,12,โˆ’28) is a normal vector of the plane. Normal vectors of a plane are all parallel, which means that they all can be written as ๐‘˜โƒ‘๐‘›, where โƒ‘๐‘› is one normal vector and ๐‘˜ is a real number. Here, all components of โƒ‘๐‘› are multiples of 4, so we can take 14โƒ‘๐‘›=(3,3,โˆ’7) as a normal vector to write our general equation.

    The general equation of a plane is of the form ๐‘›๐‘ฅ+๐‘›๐‘ฆ+๐‘›๐‘ง+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 3๐‘ฅ+3๐‘ฆโˆ’7๐‘ง+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,โˆ’3,1) belongs to the plane. We substitute these into the equation, which gives 3ร—4+3ร—(โˆ’3)โˆ’7ร—1+๐‘‘=0๐‘‘=4.

    The general equation of the plane is, therefore, 3๐‘ฅ+3๐‘ฆโˆ’7๐‘ง+4=0.

    It is worth noting that we could have kept our original normal vector to write the general equationโ€”we would have found the same final general equation after dividing both sides by 4.

    • The intercept equation of a plane with ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-intercepts ๐‘Ž, ๐‘, and ๐‘, respectively, is given by ๐‘ฅ๐‘Ž+๐‘ฆ๐‘+๐‘ง๐‘=1.
    • The parametric equations of a plane in space that contains point ๐‘ƒ(๐‘ฅ,๐‘ฆ,๐‘ง)๏Œฏ๏Œฏ๏Œฏ and two noncollinear vectors โƒ‘๐‘ข=๏€น๐‘ข,๐‘ข,๐‘ข๏…๏—๏˜๏™ and โƒ‘๐‘ฃ=๏€น๐‘ฃ,๐‘ฃ,๐‘ฃ๏…๏—๏˜๏™ are a set of three equations of the form ๐‘ฅ=๐‘ฅ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ฆ=๐‘ฆ+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๐‘ง=๐‘ง+๐‘ก๐‘ข+๐‘ก๐‘ฃ,๏Œฏ๏Šง๏—๏Šจ๏—๏Œฏ๏Šง๏˜๏Šจ๏˜๏Œฏ๏Šง๏™๏Šจ๏™ where ๐‘ก๏Šง and ๐‘ก๏Šจ are two varying real numbers, called the parameters.
    • A normal vector of a plane is given by โƒ‘๐‘›=โƒ‘๐‘ขร—โƒ‘๐‘ฃ.