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5.7: Planes in Space

Difficulty Level: At Grade Created by: CK-12
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Orlando is building a coffee table for his mother. He has the round surface cut and sanded and is ready to attach the legs. As he tries a few different heights, he notices that unless the spot on the floor is really flat, and he is extremely careful to make the legs exactly the same length, the table is unsteady and wobbles annoyingly.

How can Orlando ensure that the table remains stable, even if the floor is not entirely flat where his mom decides to put it?

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- James Sousa: Determining the Equation af a Plane

Guidance

Vectors can be used to identify directions in space and also the orientation of a plane by identifying the direction perpendicular to that plane. In this section we will look at that calculation in reverse. Rather than determining the normal vector to a plane using two vectors which lie in that plane, we will be using the normal vector to determine the equation for the plane itself. We will also use the normal vectors to determine the intersection angle between any pair of planes.

Intercept Form

The diagram below shows a plane which crosses all three coordinate axes. Points A, B, and C are the locations where the plane crosses each of the coordinate axes, called intercepts. Their locations are given by A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c). The line segments AB, BC, and CA all lie in the plane. Furthermore, segment AB is a portion of the line of intersection between this plane and the x-y axis; segment BC is a portion of the line of intersection between this plane and the y-z axis; and segment CA is a portion of the line of intersection between this plane and the z-x axis.

The intercept form of the equation for a plane is given by

\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}

Using a Normal Vector

Another way to specify a plane is to know two vectors within the plane. Recall that if two vectors lie in the same plane, the normal to that plane can be found using the cross product of the two vectors. Sometimes we are given the equations of the vectors themselves. Sometimes, however, we are only given a set of points that lie on the plane. If we know three points that lie on the plane, we can use the method developed when looking at cross products to find equations for the vectors between those points and then use those vectors to identify the plane. We only need three points to accomplish this task.

Since the normal to the plane is, by definition, perpendicular to all possible vectors within a plane and since the dot product of two vectors is equal to zero for any two perpendicular vectors, we can define a plane in terms of the dot product of the normal vector with any vector, \begin{align*}\overrightarrow{v}\end{align*}, within the plane:

\begin{align*}\overrightarrow{n} \times \overrightarrow{v} = 0\end{align*}

Which we can also write as

\begin{align*}\left \langle n_x, n_y, n_z \right \rangle \times \left \langle (x - x_0), (y - y_0), (z - z_0) \right \rangle = 0\end{align*}

If we compute the dot product, we obtain another equation which specifies the plane in terms of the normal vector and two points on the plane, (x, y, z) and (x0, y0, z0).

nx (x - x0) + ny (y - y0) + nz (z - z0) = 0

This equation is frequently written as

nxx + nyy + nzz + d = 0

Where

d = -nxxo - nyyo - nzzo

and the intercepts of the plane with the x, y, and z axes are given by:

\begin{align*}a = -\frac{d}{n_x}, \ b = -\frac{d}{n_y},\end{align*} and \begin{align*}c = -\frac{d}{n_z}\end{align*}

If you only have the normal vector and one point on the plane, first determine the vector projection of the position vector of that point onto the normal vector using the dot product. Then you will have the locations of two points on the plane and can use the normal and two points method described above.

Example A

Find the intercepts of the plane given by the equation 3x + 5y - 2z - 4 = 0.

Solution

Rewrite the equation of the plane in the format of the intercept form of the plane equation.

3x + 5y - 2z = 4

\begin{align*}1 = \frac{3}{4} x + \frac{5}{4} y + \frac{-2}{4}z\end{align*}

\begin{align*}a = \frac{4}{3}, \ b = \frac{4}{5}\end{align*}, and \begin{align*}c = \frac{4}{-2} = -2\end{align*}. Thus the intercepts of this plane are

\begin{align*}\left (\frac{4}{3}, 0, 0 \right ), \ \left (0, \frac {4}{5}, 0 \right )\end{align*}, and \begin{align*}(0, 0, -2)\end{align*}.

Example B

A plane has intersections at (12, 0, 0), (0, 6, 0), and (0, 0, 4). Write the equation of the plane.

Solution

\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{x}{12} + \frac{y}{6} + \frac{z}{4}\end{align*} or \begin{align*}12 = x + 2y + 3z\end{align*}

Example C

Use the equation 3x + 5y - 2z - 4 = 0 to determine the normal unit-vector to that plane.

Solution

Comparing this equation to nxx + nyy + nzz + d = 0, we can see that \begin{align*}\overrightarrow{n} = \left \langle 3, 5, -2\right \rangle\end{align*}. To find the unit normal vector, find the magnitude of this normal vector and divide each component by the magnitude.

\begin{align*}|\overrightarrow{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} = \sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{38}\end{align*}

\begin{align*}\hat{n} = \left \langle \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}, \frac{-2}{\sqrt{38}} \right \rangle\end{align*}

Example D

The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the equation of the plane.

Solution

First find the vectors between two pairs of the points.

\begin{align*}\overrightarrow{PQ} = \left \langle (Q_x - P_x), (Q_y - P_y), (Q_z - P_z) \right \rangle = \left \langle (1 - 3), (4 - 7), (3 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -2, -3, 1 \right \rangle\end{align*}

\begin{align*}\overrightarrow{PR} = \left \langle (R_x - P_x), (R_y - P_y), (R_z - P_z) \right \rangle = \left \langle (2 - 3), (3 - 7), (4 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -1, -4, 2 \right \rangle\end{align*}

The cross product of these two vectors is normal to the plane.

\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle (PQ_yPR_z - PQ_zPR_y), (PQ_zPR_x - PQ_xPR_z), (PQ_xPR_y - PQ_yPR_x)\right \rangle\end{align*}

\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-3 \cdot 2) - (1 \cdot -4)], [(1 \cdot -1) - (-2 \cdot 2)], [(-2 \cdot -4) - (-1 \cdot -3)] \right \rangle\end{align*}

\begin{align*}\overrightarrow{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-6) - (-4)], [(-1) - (-4)], [(8) - (3)] \right \rangle = \left \langle -2, 3, 5 \right \rangle\end{align*}

This normal vector and one of the points will give an equation for the plane.

nx (x - Px) + ny (y - Py) + nz (z - Pz) = 0

-2(x - 3) + 3(y - 7) + 5(z - 2) = 0

-2x + 3y + 5z + (6 - 21 - 10) = 0

-2x + 3y + 5z - 25 = 0

Concept question wrap-upHow can Orlando ensure that the table he is making for his mother remains stable, even if the floor is not entirely flat where his mom decides to put it?

Recall from the lesson that any plane can be defined by three points. If Orlando makes the table with only three legs, then the endpoints of each leg will define a stable plane for the table to sit on. Even if the plane defined by the leg endpoints is not exactly parallel to the average plane of the floor across the entire room, the table itself will remain stable.

Vocabulary

A plane is the 3 dimensional equivalent of a line on a standard rectangular graph. It can be conceptualized as a sheet of paper of infinite area.

The intercept form of the equation for a plane is given by: \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}

The normal vector to a plane is perpendicular to all possible vectors within that plane

Guided Practice

Questions

1) Rewrite the equation of the plane 7x + 3y + z + 12 = 0 in intercept form.

2) Determine the equation for the unit vector which is perpendicular to the plane, 7x + 3y + z + 12 = 0.

3) Find the intercepts of the plane described by the equation 2.4x + 3.6y - 4.8z - 5.9 = 0.

4) A plane is defined by three points having position vectors \begin{align*}\overrightarrow{r_1} = \left \langle 1, 0, -1 \right \rangle, \ \overrightarrow{r_2} = \left \langle 2, 4, 6 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{r_3} = \left \langle -3, 7, 5 \right \rangle\end{align*}. Determine the components of the unit vector which is perpendicular to the plane passing through those points.

5) Determine the components of the unit vector which is perpendicular to the plane 12x + 23y + 14z - 5 = 0.

Solutions

1) The equation \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*} must be true for all points on a plane. Therefore, we should first rearrange 7x + 3y + z + 12 = 0 into the form \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.

7x + 3y + z = -12
\begin{align*}\frac{7}{-12}x + \frac{3}{-12}y + \frac{1}{-12}z = 1\end{align*}
Therefore, \begin{align*}a = \frac{-12}{7}, \ b = \frac{-12}{3} = -4\end{align*}, and \begin{align*}c = \frac{-12}{1} = -12\end{align*} and the position vectors of the three intercepts are \begin{align*}\overrightarrow{A} = \left \langle -1.714, 0, 0 \right \rangle, \ \overrightarrow{B} = \left \langle 0, -4, 0 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{C} = \left \langle 0, 0, -12 \right \rangle\end{align*}

2) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 7, 3, 1 \right \rangle\end{align*}

\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{(7)^2 + (3)^2 + (1)^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{49 + 9 + 1}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{59}} = \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle\end{align*}

3) First write the equation of the plane in intercept form, \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.

2.4x + 3.6y - 4.8z = 5.9
\begin{align*}\frac{2.4}{5.9}x + \frac{3.6}{5.9}y - \frac{4.8}{5.9}z = 1\end{align*}
Therefore the x-intercept is \begin{align*}\left \langle \frac{5.9}{2.4}, 0, 0 \right \rangle\end{align*}, the y-intercept is \begin{align*}\left \langle 0, \frac{5.9}{3.6}, 0 \right \rangle\end{align*}, and the z-intercept is \begin{align*}\left \langle 0, 0, \frac{5.9}{4.8} \right \rangle\end{align*}.

4) The cross-product determines the direction perpendicular to a pair of vectors. Therefore we can use these three points to define two vectors in the same plane. The vector from point 1 to point 2 is given by subtracting vector 2 from vector 1:

\begin{align*}\overrightarrow{r_{1-2}} = \overrightarrow{r_1} - \overrightarrow{r_2} = \left \langle 1, 0, -1\right \rangle - \left \langle 2, 4, 6 \right \rangle = \left \langle 1 - 2, 0 - 4, -1 - 6 \right \rangle = \left \langle -1, -4, -7 \right \rangle\end{align*}
Likewise, the vector from point 1 to point 3 is given by subtracting vector 3 from vector 1:
\begin{align*}\overrightarrow{r_{1-3}} = \overrightarrow{r_1} - \overrightarrow{r_3} = \left \langle 1, 0, -1\right \rangle - \left \langle -3, 7, 5 \right \rangle = \left \langle 1 - (-3), 0 - 7, -1 - 5 \right \rangle =\end{align*} \begin{align*}\left \langle 4, -7, -6 \right \rangle\end{align*}
Now we can use the cross-product of the two vectors in the plane to determine a vector which is perpendicular to that plane,
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle (r_{1-2y}r_{1-3z} - r_{1-2z}r_{1-3y}), (r_{1-2z}r_{1-3x} - r_{1-2x}r_{1-3z}), (r_{1-2x}r_{1-3y} - r_{1-2y}r_{1-3x}) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((-4)(-6) - (-7)(-7)), ((-7)(4) - (-1)(-6)), ((-1)(-7) - (-4)(4)) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((24) - (49), ((-28) - (6)), ((7) - (-16)) \right \rangle = \left \langle -25, -34, 23 \right \rangle\end{align*}
Now use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{(-25)^2 + (-34)^2 + (23)^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{625 + 1156 + 529}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{2310}} \end{align*}
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle {-25, -34, 23}\right \rangle}{48.06} = \left \langle -0.5202, -0.7074, 0.4785 \right \rangle\end{align*}

5) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 12, 23, 14 \right \rangle\end{align*}.

Now we can use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{12^2 + 23^2 + 14^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{869}} = \frac{\left \langle 12, 23, 14 \right \rangle}{29.5} = \left \langle \frac{12}{29.5}, \frac{23}{29.5}, \frac{14}{29.5} \right \rangle\end{align*}

Practice

Given the following intersections, write the equation of the plane.

  1. \begin{align*}(7, 0, 0), (0, 3, 0)\end{align*} and \begin{align*}(0, 0, 19)\end{align*}
  2. \begin{align*}(2, 0, 0), (0, 8, 0)\end{align*} and \begin{align*}(0, 0, 5)\end{align*}
  3. \begin{align*}(13, 0, 0), (0, 21, 0)\end{align*} and \begin{align*}(0, 0, 17)\end{align*}
  4. \begin{align*}(5, 0, 0), (0, 1, 0)\end{align*} and \begin{align*}(0, 0, 2)\end{align*}
  5. \begin{align*}(27, 0, 0), (0, 12, 0)\end{align*} and \begin{align*}(0, 0, 18)\end{align*}

Find the intercepts of the plane given the following equations:

  1. \begin{align*}3x + 2y + z - 6 = 0\end{align*}
  2. \begin{align*}1x - 7y - z + 10 = 0\end{align*}
  3. \begin{align*}-2x + 9y + 4z - 1 = 0\end{align*}
  4. \begin{align*}6x - 11y + 2z + 3 = 0\end{align*}
  5. \begin{align*}-2x + 5y + 5z + 6 = 0\end{align*}

Use the given equations to determine the normal unit-vector to that plane

  1. \begin{align*}7x + 5y - 1z - 10 = 0\end{align*}
  2. \begin{align*}4x - 13y + 5z - 3 = 0\end{align*}
  3. \begin{align*}-8x + 7y + 2z + 5 = 0\end{align*}
  4. \begin{align*}10x + 3y - z - 2 = 0\end{align*}
  5. \begin{align*}-1x - 2y + 7z + 16 = 0\end{align*}

Determine the equation of the planes below using the three points given:

  1. \begin{align*}P = (3, 6, 9), Q = (9, 6, 3)\end{align*} and \begin{align*}R = (6, -9, 9)\end{align*}
  2. \begin{align*}P = (1, -7, 2), Q = (4, 2, 9)\end{align*} and \begin{align*}R = (3, -5, 1)\end{align*}
  3. \begin{align*}P = (3, 8, 10), Q = (-2, 5, 8)\end{align*} and \begin{align*}R = (7, 4, 8)\end{align*}
  4. \begin{align*}P = (9, -1, 4), Q = (6, 2, -8)\end{align*} and \begin{align*}R = (12 , 9, 10)\end{align*}
  5. \begin{align*}P = (5, 8,-9), Q = ( -5, 3, 9)\end{align*} and \begin{align*}R = (10, 4, -6)\end{align*}

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Vocabulary

Cross product

The cross product of two vectors is a third vector that is perpendicular to both of the original vectors.

dot product

The dot product is also known as inner product or scalar product. The two forms of the dot product are \vec{a} \cdot \vec{b} = \Big \| \vec{a}\Big \| \ \Big \| \vec{b}\Big \| \cos \theta and \vec{a} \cdot \vec{b} = x_a x_b + y_a y_b.

intercept form

The intercept form of the equation for a plane is given by 1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}.

Intercepts

The intercepts of a curve are the locations where the curve intersects the x and y axes. An x intercept is a point at which the curve intersects the x-axis. A y intercept is a point at which the curve intersects the y-axis.

normal vector

A normal vector is a vector that is perpendicular to a given surface or plane. A unit normal vector is a normal vector with a magnitude of one.

plane

A plane is a flat, two-dimensional surface. It can be conceptualized as a sheet of paper of infinite area.

unit vector

A unit vector is a vector with a magnitude of one.

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