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

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. 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 𝐴(𝑎,0,0), 𝐵(0,𝑏,0), and 𝐶(0,0,𝑐). We say that the 𝑥-, 𝑦-, and 𝑧-intercepts of the plane are 𝑎, 𝑏, and 𝑐.

Having the coordinates of three points in the plane, we can easily find two noncollinear vectors included in the plane, for instance, ⃑𝑢=𝐵𝐴=(𝑎,−𝑏,0) and ⃑𝑣=𝐶𝐴=(𝑎,0,−𝑐). A normal vector of the plane is then given by ⃑𝑛=⃑𝑢×⃑𝑣=|||||⃑𝑖⃑𝑗⃑𝑘𝑎−𝑏0𝑎0−𝑐|||||=𝑏𝑐⃑𝑖+𝑎𝑐⃑𝑗+𝑎𝑏⃑𝑘.

As 1𝑎𝑏𝑐⃑𝑛=1𝑎,1𝑏,1𝑐 is also a normal vector of the plane, we can write a general equation of the form 𝑥𝑎+𝑦𝑏+𝑧𝑐+𝑑=0.

Substituting in the coordinates of one of three known points included in the plane, we find that 𝑑=−1.

The intercept equation of a plane with 𝑥-, 𝑦-, and 𝑧-intercepts 𝑎, 𝑏, and 𝑐, respectively, is given by 𝑥𝑎+𝑦𝑏+𝑧𝑐=1.

Let us look at the equation of planes that are parallel to one axis. Consider, for instance, a plane parallel to the 𝑧-axis that intersects the 𝑥-axis at 𝐴(𝑎,0,0) and the 𝑦-axis at 𝐵(0,𝑏,0). Two vectors included in the plane are, for instance, ⃑𝑢=𝐵𝐴=(𝑎,−𝑏,0) and ⃑𝑣=(0,0,1), a vector parallel to the 𝑧-axis. A normal vector of the plane is then given by ⃑𝑛=⃑𝑢×⃑𝑣=|||||⃑𝑖⃑𝑗⃑𝑘𝑎−𝑏0001|||||=−𝑏⃑𝑖−𝑎⃑𝑗.

If we take, as a normal vector of the plane, −1𝑎𝑏⃑𝑛=1𝑎,1𝑏,0, we find the intercept equation to be 𝑥𝑎+𝑦𝑏=1.

It is equivalent to the equation of a line in the (𝑥,𝑦) plane, plus the fact that the 𝑧-coordinate can take up any value in ℝ.

Consider now a plane that is parallel to two axes, for example, the 𝑥- and 𝑧-axes. It intersects the third axis, here the 𝑦-axis, at a point with coordinates (0,𝑏,0). This plane contains all the points that have 𝑏 as the 𝑦-coordinate; hence, its equation is simply 𝑦=𝑏, which can be written in intercept form as 𝑦𝑏=1.

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

Find the equation of the plane whose 𝑥-, 𝑦-, and 𝑧-intercepts are −7, 3, and −4 respectively.

The intercept equation of a plane with 𝑥-, 𝑦-, and 𝑧-intercepts 𝑎, 𝑏, and 𝑐, respectively, is given by 𝑥𝑎+𝑦𝑏+𝑧𝑐=1.

Here, 𝑎=−7, 𝑏=3, and 𝑐=−4. Hence, we find that the equation of the plane is −𝑥7+𝑦3−𝑧4=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 16𝑥+2𝑦+8𝑧−16=0.

An intercept equation of a plane is of the form 𝑥𝑎+𝑦𝑏+𝑧𝑐=1.

Adding 16 to each side of the given equation, we find 16𝑥+2𝑦+8𝑧=16.

As we want to have 1 on the right-hand side, we divide both sides by 16, which gives 𝑥1+𝑦8+𝑧2=1.

To find this equation, we could also first find 𝑎, 𝑏, and 𝑐, that is, the 𝑥-, 𝑦-, and 𝑧-intercepts of the plane. We find the 𝑥-intercept by substituting 0 for 𝑦 and 𝑧 into our equation. We find 16𝑥−16=0𝑥=1.

Hence, 𝑎=1. Similarly, we find that 𝑏=8 and 𝑐=2. Substituting these values into the intercept equation, we find 𝑥1+𝑦8+𝑧2=1.

The intercept equation of the plane of general equation 16𝑥+2𝑦+8𝑧−16=0 is 𝑥1+𝑦8+𝑧2=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 𝑀(𝑥,𝑦), its position vector is given by 𝑂𝑀=𝑥⃑𝑖+𝑦⃑𝑗, where 𝑂 is the origin of the coordinate system and ⃑𝑖 and ⃑𝑗 are the unit vectors along its two axes. We can write a similar equation with any two noncollinear vectors ⃑𝑢 and ⃑𝑣 in the plane: 𝑂𝑀=𝑎⃑𝑢+𝑏⃑𝑣, where 𝑎 and 𝑏 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 𝑃(𝑥,𝑦,𝑧) and two noncollinear vectors ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣. For any point 𝑀(𝑥,𝑦,𝑧) in the plane, we have 𝑃𝑀=𝑡⃑𝑢+𝑡⃑𝑣, where 𝑡 and 𝑡 are two real numbers.

This gives three equations for the three components of the vector 𝑃𝑀: 𝑥−𝑥=𝑡𝑢+𝑡𝑣,𝑦−𝑦=𝑡𝑢+𝑡𝑣,𝑧−𝑧=𝑡𝑢+𝑡𝑣.

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

The parametric equations of a plane in space that contains point 𝑃(𝑥,𝑦,𝑧) and two noncollinear vectors ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣 are a set of three equations of the form 𝑥=𝑥+𝑡𝑢+𝑡𝑣,𝑦=𝑦+𝑡𝑢+𝑡𝑣,𝑧=𝑧+𝑡𝑢+𝑡𝑣, where 𝑡 and 𝑡 are two varying real numbers, called the parameters.

By varying the parameters 𝑡 and 𝑡 across ℝ, 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 𝐴(1,2,1) and the two vectors ⃑𝑑=(1,−1,2) and ⃑𝑑=(2,−1,1).

The parametric equations of a plane that passes through the point 𝐴(𝑥,𝑦,𝑧) and two vectors ⃑𝑑=𝑑,𝑑,𝑑 and ⃑𝑑=𝑑,𝑑,𝑑 are of the form 𝑥=𝑥+𝑡𝑑+𝑡𝑑,𝑦=𝑦+𝑡𝑑+𝑡𝑑,𝑧=𝑧+𝑡𝑑+𝑡𝑑, where 𝑡 and 𝑡 are two real numbers.

Substituting in (𝑥,𝑦,𝑧)=(1,2,1), 𝑑,𝑑,𝑑=(1,−1,2), and 𝑑,𝑑,𝑑=(2,−1,1), we find 𝑥=1+𝑡+2𝑡,𝑦=2−𝑡−𝑡,𝑧=1+2𝑡+𝑡.

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 𝐴(1,5,1), 𝐵(3,4,3), and 𝐶(2,3,4).

1. 𝑥=1−2𝑡−𝑡, 𝑦=5+𝑡+2𝑡, 𝑧=1+3𝑡+2𝑡
2. 𝑥=2+2𝑡+𝑡, 𝑦=3−𝑡+𝑡, 𝑧=4+2𝑡+𝑡
3. 𝑥=−1−𝑡+2𝑡, 𝑦=−1+2𝑡+𝑡, 𝑧=1+3𝑡+2𝑡
4. 𝑥=3−2𝑡−𝑡, 𝑦=4+𝑡−𝑡, 𝑧=3−2𝑡+𝑡
5. 𝑥=2−2𝑡−𝑡, 𝑦=3+4𝑡, 𝑧=4+2𝑡+3𝑡

We are given three points, 𝐴(1,5,1), 𝐵(3,4,3), and 𝐶(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 𝐴𝐵=(2,−1,2) and 𝐴𝐶=(1,−2,3), we find that 𝐴𝐵×𝐴𝐶=||||⃑𝑖⃑𝑗⃑𝑘2−121−23||||=⃑𝑖−4⃑𝑗−3⃑𝑘; 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 𝑃(𝑥,𝑦,𝑧) and two noncollinear vectors ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣 are a set of three equations of the form 𝑥=𝑥+𝑡𝑢+𝑡𝑣,𝑦=𝑦+𝑡𝑢+𝑡𝑣,𝑧=𝑧+𝑡𝑢+𝑡𝑣, where 𝑡 and 𝑡 are two real numbers.

Let us now identify, for each option, the coordinates (𝑥,𝑦,𝑧) and the two sets of vector components 𝑢,𝑢,𝑢 and 𝑣,𝑣,𝑣. Our findings are given in the table below.

Option(𝑥,𝑦,𝑧)𝑢,𝑢,𝑢𝑣,𝑣,𝑣
A(1,5,1)(−2,1,3)(−1,2,2)
B(2,3,4)(2,−1,2)(1,1,1)
C(−1,−1,1)(−1,2,3)(2,1,2)
D(3,4,3)(−2,1,−2)(−1,−1,1)
E(2,3,4)(−2,4,2)(−1,0,3)

With the three points 𝐴(1,5,1), 𝐵(3,4,3), and 𝐶(2,3,4), we can form the vectors 𝐴𝐶=(1,−2,3), 𝐴𝐵=(2,−1,2), and 𝐵𝐶=(−1,−1,1) - and of course the opposite vectors 𝐶𝐴=(−1,2,−3), 𝐵𝐴=(−2,1,−2), and 𝐶𝐵=(1,1,−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 𝐵(3,4,3) and the vectors 𝐵𝐴=(−2,1,−2) and 𝐵𝐶=(−1,−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 𝑛(𝑥−𝑥)+𝑛(𝑦−𝑦)+𝑛(𝑧−𝑧)=0, where ⃑𝑛=𝑛,𝑛,𝑛 is a normal vector of the plane and 𝑃(𝑥,𝑦,𝑧) is a point on the plane. The standard from can be easily rearranged to the general form 𝑛𝑥+𝑛𝑦+𝑛𝑧+𝑑=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 ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣 from the parametric equations 𝑥=𝑥+𝑡𝑢+𝑡𝑣,𝑦=𝑦+𝑡𝑢+𝑡𝑣,𝑧=𝑧+𝑡𝑢+𝑡𝑣, we find ⃑𝑛=𝑛,𝑛,𝑛=⃑𝑢×⃑𝑣.

The standard equation is then 𝑛(𝑥−𝑥)+𝑛(𝑦−𝑦)+𝑛(𝑧−𝑧)=0.

It can be rearranged to the general form 𝑛𝑥+𝑛𝑦+𝑛𝑧+𝑑=0, with 𝑑=−𝑛𝑥+𝑛𝑦+𝑛𝑧.

Let us use this method in the next example.

Find the general equation of the plane 𝑥=4+7𝑡+4𝑡, 𝑦=−3−4𝑡, 𝑧=1+3𝑡.

We are given the parametric equations of the plane. We know that the coefficients of 𝑡 on one side and those of 𝑡 on the other side are the components of two vectors in the plane. Let us call these vectors ⃑𝑢 and ⃑𝑣. We find ⃑𝑢=(7,0,3) and ⃑𝑣=(4,−4,0). One normal vector of the plane is thus ⃑𝑛=⃑𝑢×⃑𝑣=||||⃑𝑖⃑𝑗⃑𝑘7034−40||||=⃑𝑖(0×0−(−4)×3)−⃑𝑗(7×0−3×4)+⃑𝑘(7×(−4)−4×0)=12⃑𝑖+12⃑𝑗−28⃑𝑘.

We have found that ⃑𝑛=(12,12,−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 𝑘⃑𝑛, where ⃑𝑛 is one normal vector and 𝑘 is a real number. Here, all components of ⃑𝑛 are multiples of 4, so we can take 14⃑𝑛=(3,3,−7) as a normal vector to write our general equation.

The general equation of a plane is of the form 𝑛𝑥+𝑛𝑦+𝑛𝑧+𝑑=0, where 𝑛, 𝑛, and 𝑛 are the components of a normal vector of the plane. Substituting these components into the general equation of the plane, we find 3𝑥+3𝑦−7𝑧+𝑑=0.

To find the constant 𝑑, 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,−3,1) belongs to the plane. We substitute these into the equation, which gives 3×4+3×(−3)−7×1+𝑑=0𝑑=4.

The general equation of the plane is, therefore, 3𝑥+3𝑦−7𝑧+4=0.

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

• The intercept equation of a plane with 𝑥-, 𝑦-, and 𝑧-intercepts 𝑎, 𝑏, and 𝑐, respectively, is given by 𝑥𝑎+𝑦𝑏+𝑧𝑐=1.
• The parametric equations of a plane in space that contains point 𝑃(𝑥,𝑦,𝑧) and two noncollinear vectors ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣 are a set of three equations of the form 𝑥=𝑥+𝑡𝑢+𝑡𝑣,𝑦=𝑦+𝑡𝑢+𝑡𝑣,𝑧=𝑧+𝑡𝑢+𝑡𝑣, where 𝑡 and 𝑡 are two varying real numbers, called the parameters.
• A normal vector of a plane is given by ⃑𝑛=⃑𝑢×⃑𝑣. 