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# 5.5: Planes in Space

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Determine the equation of a plane given three points on the plane.
• Determine the x, y, and z intercepts of a plane.
• Determine the equation of a plane given the normal vector and a point on the plane.
• Determine the dihedral angle between two planes.

## Introduction

As we have already discussed, vectors are used to identify directions in space. We also discussed how vectors can be used to identify 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.

## The Equation of a Plane, 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

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

This equation must hold true for all points (x, y, z) on the plane. To check this equation, insert the coordinates of point A, B, or C into the equation. For points A, B, and C respectively

1=aa+0b+0c=aa1=0a+bb+0c=bb1=0a+0b+cc=cc\begin{align*}1 = \frac{a}{a} + \frac{0}{b} + \frac{0}{c} = \frac{a}{a} \qquad \qquad 1 = \frac{0}{a} + \frac{b}{b} + \frac{0}{c} = \frac{b}{b} \qquad \qquad 1 = \frac{0}{a} + \frac{0}{b} + \frac{c}{c} = \frac{c}{c}\end{align*}

Example: 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

1=34x+54y+24z\begin{align*}1 = \frac{3}{4} x + \frac{5}{4} y + \frac{-2}{4}z\end{align*}

a=43, b=45\begin{align*}a = \frac{4}{3}, \ b = \frac{4}{5}\end{align*}, and c=42=2\begin{align*}c = \frac{4}{-2} = -2\end{align*}. Thus the intercepts of this plane are

(43,0,0), (0,45,0)\begin{align*}\left (\frac{4}{3}, 0, 0 \right ), \ \left (0, \frac {4}{5}, 0 \right )\end{align*}, and (0,0,2)\begin{align*}(0, 0, -2)\end{align*}.

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

Solution: 1=xa+yb+zc=x12+y6+z4\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{x}{12} + \frac{y}{6} + \frac{z}{4}\end{align*} or 12=x+2y+3z\begin{align*}12 = x + 2y + 3z\end{align*}

## Identification of a Plane Using a Normal Vector

Another way to specify a plane is to know two vectors within the plane. As we showed before 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.

Look again at the plane we showed above. Now we will use the intercepts to determine two vectors, AB and AC, which lie within the plane.

Remember that the location of points A, B, and C are given by A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c). The vectors AB and AC are then given by

AB=(BxAx),(ByAy),(BzAz)=(0a),(b0),(00)=\begin{align*}\overrightarrow{AB} = \left \langle (B_x - A_x), (B_y - A_y), (B_z - A_z) \right \rangle = \left \langle (0 - a), (b - 0), (0 - 0) \right \rangle =\end{align*} a,b,0\begin{align*}\left \langle -a, b, 0 \right \rangle\end{align*}

AC=(CxAx),(CyAy),(CzAz)=(0a),(00),(c0)=\begin{align*}\overrightarrow{AC} = \left \langle (C_x - A_x), (C_y - A_y), (C_z - A_z) \right \rangle = \left \langle (0 - a), (0 - 0), (c - 0) \right \rangle =\end{align*} a,0,c\begin{align*}\left \langle -a, 0, c \right \rangle\end{align*}

Now that we know two vectors in the plane, we can use the cross product of those two vectors to determine the normal to the plane.

AB×AC=(AByACzABzACy),(ABzACxABxACz),(ABxACyAByACx)\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} = \left \langle (AB_y AC_z - AB_z AC_y), (AB_z AC_x - AB_x AC_z), (AB_x AC_y - AB_y AC_x) \right \rangle\end{align*}

AB×AC=(bc0),(0(ac)),(0(ab))=bc,ac,ab\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} = \left \langle (bc - 0), (0-(-ac)), (0-(-ab)) \right \rangle = \left \langle bc, ac, ab \right \rangle\end{align*}

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, v\begin{align*}\overrightarrow{v}\end{align*}, within the plane:

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

Which we can also write as

nx,ny,nz×(xx0),(yy0),(zz0)=0\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:

a=dnx, b=dny,\begin{align*}a = -\frac{d}{n_x}, \ b = -\frac{d}{n_y},\end{align*} and c=dnz\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: Use the equation of the plane given in the example above to determine the normal unit-vector to that plane.

Solution: The equation of the plane in the first example was 3x + 5y - 2z - 4 = 0. Comparing this equation to nxx + nyy + nzz + d = 0, we can see that n=3,5,2\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.

|n|=n2x+n2y+n2z=32+52+(2)2=38\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*}

n^=338,538,238\begin{align*}\hat{n} = \left \langle \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}, \frac{-2}{\sqrt{38}} \right \rangle\end{align*}

Example: 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.

PQ=(QxPx),(QyPy),(QzPz)=(13),(47),(32)=\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*} 2,3,1\begin{align*}\left \langle -2, -3, 1 \right \rangle\end{align*}

PR=(RxPx),(RyPy),(RzPz)=(23),(37),(42)=\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*} 1,4,2\begin{align*}\left \langle -1, -4, 2 \right \rangle\end{align*}

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

PQ×PR=(PQyPRzPQzPRy),(PQzPRxPQxPRz),(PQxPRyPQyPRx)\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*}

PQ×PR=[(32)(14)],[(11)(22)],[(24)(13)]\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*}

n=PQ×PR=[(6)(4)],[(1)(4)],[(8)(3)]=2,3,5\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

## Determining the Distance from a Plane to the Origin

One point on the plane is unique from all the others. No matter what the orientation of the plane, this point is located closer to the origin than any other point on the plane. This means that the position vector for this point is shorter than any other point in the plane. The diagram below shows a two-dimensional projection of a plane, in grey, near a point not on the plane, in black. The position vectors to a variety of point are shown in the diagram. The position vector marked in blue is shorter than the position vectors for the other points. This shortest vector is perpendicular to the plane. You can also see that the blue line is the vector projection of any orange vector onto the perpendicular direction.

This orthagonality (i.e. being perpendicular) is useful for us because it means that the position vector for this special point is parallel to the normal vector. Therefore, if we know the equation for a normal vector and the position vector for any point on the plane, we can determine the location of the point on the plane closest to the origin by finding the projection of the given point’s position vector onto the normal direction.

Example: The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the point on the plane which is closest to the origin.

Solution:

In the previous example, we determined the normal vector for the plane defined by points P, Q, and R: n=PQ×PR=[(6)(4)],[(1)(4)],[(8)(3)]=2,3,5\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*}.

The point on the plane which is nearest to the origin can be found by determining the projection of the position vector of either of these three points onto the normal vector. Remember that the vector projection of one vector onto the direction of another, is given by the dot-product of the first vector onto the unit vector defining the direction of the second vector: (A×B)B\begin{align*}\left (\overrightarrow{A} \times \overrightarrow{B} \right ) \overrightarrow{B}\end{align*}.

Since we know three points on the plane, we can use one of them to solve the problem. Let’s start with point P. The vector projection of P\begin{align*}\overrightarrow{P}\end{align*} onto n^\begin{align*}\hat{n}\end{align*} is given by (P×n^)n^\begin{align*}\left (\overrightarrow{P} \times \hat{n} \right ) \hat{n}\end{align*}, so first we need to determine the unit vector n^\begin{align*}\hat{n}\end{align*} which is given by n^=n^|n|=nx,ny,nzn2x+n2y+n2z=2,3,5(2)2+32+(5)2=2,3,538=0.32,0.49,0.81\begin{align*}\hat{n} = \frac{\hat{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 -2, 3, 5 \right \rangle}{\sqrt{(-2)^2 + 3^2 + (5)^2}} = \frac{\left \langle -2, 3, 5 \right \rangle}{\sqrt{38}} = \left \langle -0.32, 0.49, 0.81 \right \rangle\end{align*}

P×n^=Pxnx^+Pyny^+Pznz^=(3)(0.32)+(7)(0.49)+(2)(0.81)=\begin{align*}\overrightarrow{P} \times \hat{n} = P_x\hat{n_x} + P_y\hat{n_y} + P_z\hat{n_z} = (3)(-0.32) + (7)(0.49) + (2)(0.81) =\end{align*} 0.96+3.43+1.62=4.09\begin{align*}-0.96 + 3.43 + 1.62 = 4.09\end{align*}

(P×n^)n^=(4.09)0.32,0.49,0.81=1.3088,2.0041,3.3129\begin{align*}\left (\overrightarrow{P} \times \hat{n} \right )\hat{n} = (4.09) \left \langle -0.32, 0.49, 0.81 \right \rangle = \left \langle -1.3088, 2.0041, 3.3129 \right \rangle\end{align*}

Therefore, the point on the plane closest to the origin is (-1.3088, 2.0041, 3.3129).

## The Dihedral Angle

The angle between two planes is called the dihedral angle. The angle between two planes is the same as the angle between their normal vectors. If we want to determine the dihedral angle between two planes, we identify normal vectors to the two planes, then we can use the dot-product of the two normal vectors to determine the angle between the two normals which is also the two planes. Recall A×B=|A||B| cos θ\begin{align*}\overrightarrow{A} \times \overrightarrow{B} =|\overrightarrow{A}||\overrightarrow{B}| \ \mbox{cos} \ \theta\end{align*}

Example: The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the dihedral angle between this plane and the x-y plane.

Solution: As we saw in the example above, these three points define a plane which has a normal vector

n=2,3,5\begin{align*}\overrightarrow{n} = \left \langle -2, 3, 5 \right \rangle\end{align*}

The normal to the x-y plane is the unit vector z^=0,0,1\begin{align*}\hat{z} = \left \langle 0, 0, 1 \right \rangle\end{align*}. To find the angle between these two vectors we use the fact that A×B=|A||B| cos θ\begin{align*}\overrightarrow{A} \times \overrightarrow{B} =|\overrightarrow{A}||\overrightarrow{B}| \ \mbox{cos} \ \theta\end{align*} and that A×B=AxBx+AyBy+AzBz\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z\end{align*}

First find a numerical value for the dot product:

n×z^=nxzx+nyzy+nzzz=(20)+(30)+(51)=5\begin{align*}\overrightarrow{n} \times \hat{z} = n_xz_x + n_yz_y + n_zz_z = (-2 \cdot 0) + (3 \cdot 0) + (5 \cdot 1) = 5\end{align*}

|n|=n2x+n2y+n2z=(2)2+(3)2+(5)2=4+9+25=38\begin{align*}|\overrightarrow{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} = \sqrt{(-2)^2 + (3)^2 + (5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}\end{align*}

|z^|=z2x+z2y+z2z=02+02+12=1\begin{align*}|\hat{z}| = \sqrt{z_x^2 + z_y^2 + z_z^2} = \sqrt{0^2 + 0^2 + 1^2} = 1\end{align*}

Then find the cosine version of the dot product:

n×z^=38 cos θ\begin{align*}\overrightarrow{n} \times \hat{z} = \sqrt{38} \ \mbox{cos} \ \theta\end{align*}

Now equate the two and solve for the angle, θ

n×z^=5=38 cos θ\begin{align*}\overrightarrow{n} \times \hat{z} = 5 = \sqrt{38} \ \mbox{cos} \ \theta\end{align*}

θ=cos1(538)=62.5\begin{align*}\theta = \mbox{cos}^{-1} \left (\frac{5}{\sqrt{38}} \right ) = 62.5^\circ\end{align*}

## Lesson Summary

Planes in space are defined by an equation which must hold true for every point on the plane. One common form of this equation is the intercept form 1=xa+yb+zc\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*} where the intercepts are given by x=a,0,0,y=0,b,0\begin{align*}\overrightarrow{x} = \left \langle a, 0, 0 \right \rangle, \overrightarrow{y} = \left \langle 0, b, 0 \right \rangle\end{align*}, and z=0,0,c\begin{align*}\overrightarrow{z} = \left \langle 0, 0, c \right \rangle\end{align*}. Another common form of this equation can be used to determine the direction perpendicular to the plane, nxx+nyy+nzz+d=0\begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*} where the normal vector n=nx,ny,nz\begin{align*}\overrightarrow{n} = \left \langle n_x, n_y, n_z \right \rangle\end{align*}, and the intercepts of the plane with the x, y, and z axes are given by: n1×n2= cos θ\begin{align*}\overrightarrow{n_1} \times \overrightarrow{n_2} = \ \mbox{cos} \ \theta\end{align*}.

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

Knowing the normal vector for a plane is useful in that it allows us to determine the orientation of this plane. The dihedral angle is the angle between two planes and can be calculated using the dot product of the two normal unit-vectors. The normal vector also allows us to determine the distance from a plane to any given point in space using the scalar projection of the position vector for the given point onto the normal unit-vector defining the plane.

## Practice Problems

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 r1=1,0,1, r2=2,4,6\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 r3=3,7,5\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.
6. Determine the dihedral angle between the two planes 12x + 23y + 14z - 5 = 0 and 7x + 3y + z + 12 = 0.
7. Determine the angle between the plane 2x - 5y + 8 - 10 = 0 and the y-z plane.
8. The three points P=2,3,4, Q=5,6,7\begin{align*}\overrightarrow{P} = \left \langle -2, 3, 4 \right \rangle, \ \overrightarrow{Q} = \left \langle 5, -6, 7 \right \rangle\end{align*}, and R=8,9,1\begin{align*}\overrightarrow{R} = \left \langle 8, 9, -1 \right \rangle\end{align*}. Determine the point on the plane which is closest to the origin.
9. Determine the point on the plane 7x + 3y + z + 12 = 0 which is closest to the origin.

## Date Created:

Feb 23, 2012

Jun 08, 2015
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