The material in this section will rely on your knowledge of these topics:

Vector dot product

Vector cross product

Matrix determinant

Solving systems of equations

Before we get to planes, let's remind ourselves that the equation of a line in a plane (ℝ^{2}), which represents the collection of all (x, y) points that lie on that line, is entirely specified by knowing a single point on the line, and the direction of the line, which we commonly know as slope.

$$y = mx + b,$$

where $y = b$ is the point (0, b) – the y-intercept – which specifies the location of the line, and * m* is the slope, which specifies its direction. Here's a refresher of the general idea of a line in the Cartesian plane.

We can generalize that equation to a line in 3-D space (ℝ^{3}), using a point for location and a **vector** for the direction or slope. Consider this diagram:

We're trying to come up with an equation for the line L that depends on the three coordinates of our ℝ^{3} system. To do that, we'll form the vector v, which is parallel to L, but can be located anywhere. At the origin is a convenient point. We'll specify the location of our line at the point $(x_0, y_0, z_0),$ to which we draw vector $\vec{r}_0.$

The vector r is going to specify any point (x, y, z) on our line; it's our target. Notice that by vector addition, $\vec{r} = \vec{r}_0 + \vec{a}.$ Now $\vec{a}$ is just $\vec{v}$ translated to where we need it. If we adjust the length of $\vec{v},$ we have the equation of our line:

$$\vec{r} = \vec{r}_0 + t \vec{v},$$ where *t* is a parameter that changes the length of $\vec{v}$ to access any point on the line. If $\vec{v} \lt 0,$ it points to the left of $(x_0, y_0, z_0).$

Now if $\vec{v}$ has coordinates $(v_1, v_2, v_3),$ any point on our line is given by

$$(x, y, z) = (x_0 + v_1 t, y_0 + v_2 t, z_0 + v_3 t),$$

where $x = x_0 + v_1 t, \; y = y_0 + v_2 t,$ and $z = z_0 + v_3 t.$ These are parametric equations of our line, one for each coordinate in ℝ^{3}. We can solve all of these for the parameter t and use the resulting equations to eliminate t to give

$$\frac{x - x_0}{v_1} = \frac{y - y_0}{v_2} = \frac{z - z_0}{v_3}$$

These are sometimes called the **symmetric equations** of the line.

Find an equation of the line passing through the point $(1, 2, -2),$ and parallel to the line $\hat{i} + 2 \hat{j} - \hat{k}.$

**Solution**

$$ \begin{align} (x, y, z) &= (\hat{i} + 2\hat{j} - 2\hat{k}) + t(\hat{i} + 2 \hat{j} - \hat{k}) \\[5pt] &= \hat{i}(1 + t) + 2\hat{j}(1 + t) - \hat{k}(2 + t) \end{align}

So our three parametric equations are

$$x = 1 + t, \; \; \; y = 2 + 2t, \; \; \text{and} \; z = 2 + t$$

Or, in terms of symmetric equations, solving each of these for t and setting the results equal to one another gives

$$x - t \; = \; \frac{y - 2}{2} \; = \; z - 2$$

Notice that the formula for a line is more awkward in 3-D space than it is in the 2-D plane. Lines are "native" to the plane, and easy to represent and manipulate. In what follows, you'll see that planes are native to the 3-D axis system, and will be relatively easy to represent and manipulate there.

Now let's consider the equation of a plane in ℝ^{3}. We'll look at a plane passing through the origin (0, 0, 0) with normal vector $\vec{N} = (1, 3, 5).$ There's nothing special about that normal vector, only that we know the normal to our plane. We'll use it to discover a few things about planes.

The question we'll ask here is, does point P lie in the plane with normal $\vec{N} = (1, 3, 5)?$ If we establish the condition for P = (x, y, z) to be in the plane, then we'll have an equation for the plane as a function of x, y and z. If P is in the plane, then vector $\vec{OP}$ must be perpendicular to $\vec{N},$ which means that the dot product between them has to be zero:

$$\vec{OP} \cdot \vec{N} = 0.$$

Let's write that dot product out and simplify it:

$$ \begin{align} (x, y, z) \cdot (1, 3, 5) &= 0 \\[5pt] x + 3y + 5z &= 0 \end{align}$$

So $x + 3y + 5z = 0$ is an equation for our plane. Notice that the vector of coefficients (1, 3, 5) is just our normal vector. Let's see if this holds if we move the plane around a bit, but keep the same normal vector.

This plane also has normal vector $\vec{N} = (1, 3, 5),$ but it passes through $P = (2, -1, 2).$ We set up the same condition on whether P lies in the plane, except this time, the vector from the point we know in the plane to P is $(x + 2, y + 1, z - 1):$

$$ \begin{align} (x + 2, y + 1, z - 2) \cdot (1, 3, 5) &= 0 \\[5pt] x + 2 + 3y + 3 + 5z - 10 &= 0 \\[5pt] x + 3y + 5z &= -5 \end{align}$$

Notice that the coefficients of this equation of our plane are still the coordinates of our normal vector, and the -5 on the right indicates it shift from the origin. So if we know a vector normal to a plane, we know an equation of the plan. Note that because there are an infinite number of normal vectors to any plane, there are also an infinite number of equations specifying that plane. We need only one.

Here is a Mathematica plot of our plane so you can see what it actually looks like on a 3D axis system.

We also know that three points describe a plane, so we can determine the equation of a plane from those points. Here is such a plane:

We can calculate two vectors, $\vec{P_oP_1}$ and $\vec{P_oP_2}$:

$$ \begin{align} \vec{P_oP_1} = (6, -1, 1) - (-2, -1, 0) = (8, 0, 1) \\[5pt] \vec{P_oP_2} = (4, -4, -2) - (-2, -1, 0) = (6, -3, -2) \end{align}$$

Now we can find a normal to the plane by calculating the cross product of these two vectors (we'll do this by minors):

$\vec{P_oP_1} \times \vec{P_oP_2} =$

$$= \hat{i} \left| \begin{matrix} 0 & 1 \\ -3 & -2 \end{matrix} \right| - \hat{j} \left| \begin{matrix} 8 & 1 \\ 6 & -2 \end{matrix} \right| + \hat{k} \left| \begin{matrix} 8 & 0 \\ 6 & -3 \end{matrix} \right|$$

$$ \begin{align} &= \hat{i}(0 + 3) - \hat{j}(-16 - 6) + \hat{k}(-24 - 0) \\[5pt] \vec{N} &= 3 \hat{i} + 22 \hat{j} - 24 \hat{k} \end{align}$$

Now for the last step, let's translate this plane so that includes one of our three points. If we choose P_{0}, we want the dot product of $\vec{N}$ with $(x + 2, y + 1, z - 0):$

$$ \begin{align} (3, 22, -24) \cdot (x + 2, y + 1, z) &= 0 \\[5pt] 3x + 6 + 22y + 22 - 24z &= 0 \\[5pt] 3x + 22y - 24z &= -28 \end{align}$$

We can confirm that all three points satisfy this equation of our plane:

$$ \begin{align} (-2, -1, 0): \; -6 - 22 = -28 \; \; \checkmark \\[5pt] (6, -1, 1): \; 18 - 22 - 24 = -28 \; \; \checkmark \\[5pt] (4, -4, -2): \; 12 - 88 + 48 = -28 \; \; \checkmark \end{align}$$

Here is a Mathematica plot of our plane so you can see what it actually looks like on a 3D axis system.

Find an equation of the plane defined by points $(1, 2, 3), \; (-1, -1, 4)$ and $(2, -1, 3).$

First we need two vectors. We'll use the first point, (1, 2, 3) as our reference point:

$$ \begin{align} \vec{P_oP_1} = (1, 2, 3) - (-1, -1, 4) = (2, 3, -1) \\[5pt] \vec{P_oP_2} = (1, 2, 3) - (2, -1, 3) = (-1, 3, 0) \end{align}$$

Now find the cross product of these two vectors (we'll do this by minors):

$\vec{v}_1 \times \vec{v}_2 =$

$$= \hat{i} \left| \begin{matrix} 3 & -1 \\ 3 & 0 \end{matrix} \right| - \hat{j} \left| \begin{matrix} 2 & -1 \\ -1 & 0 \end{matrix} \right| + \hat{k} \left| \begin{matrix} 2 & 3 \\ -1 & 3 \end{matrix} \right|$$

$$ \begin{align} &= \hat{i}(0 + 3) - \hat{j}(-16 - 6) + \hat{k}(-24 - 0) \\[5pt] \vec{N} &= 3 \hat{i} + \hat{j} + 9 \hat{k} \end{align}$$

Now for the last step, let's translate this plane so that includes one of our three points. If we choose P_{0}, we want the dot product of $\vec{N}$ with $(x - 1, y - 2, z - 3):$

$$ \begin{align} (3, 1, 9) \cdot (x - 1, y - 2, z - 3) &= 0 \\[5pt] 3x - 3 + y - 2 + 9z - 27 &= 0 \\[5pt] 3x + y + 9z &= 32 \end{align}$$

We can confirm that all three points satisfy this equation of our plane:

$$ \begin{align} (1, 2, 3): \; 3 + 2 + 27 &= 32 \; \; \checkmark \\[5pt] (-1, -1, 4): \; -3 - 1 + 36 &= 32 \; \; \checkmark \\[5pt] (2, -1, 3): \; 6 - 1 + 27 &= 32 \; \; \checkmark \end{align}$$

Here is a Mathematica plot of our plane so you can see what it actually looks like on a 3D axis system.

Now let's solve for the intersection of three planes. This will amount to solving a system of three equations, each with three unknowns — three planes. We'll do it using 3×3 matrices. There are several possibilities for the intersecion of three planes. Here's an illustration.

A few of the possibilities are

- Three planes intersect at one point.
- Three planes intersect in a single line.
- Three parallel planes do not intersect (are parallel), therefore there will be no lolutions to the problem.
- Three planes intersect in two lines (two linear intersections of two planes).
- Three planes intersect in three lines. In these last two cases, a 3×3 system has no solutions, although there are an infinite number of solutions to the intersection of some pairs of two planes.

There are several methods for solving systems of three equations and three unknowns (3×3 systems). They include:

- Gaussian or Gauss-Jordan elimination, with or without matrices. This is the most common way that newer students of linear systems solve them, but it is the least useful as systems get larger (more variables, more equations).
- Cramer's rule, discussed here. Cramer's rule is useful and quick for 2×2 and 3×3 systems.
- Using the inverse matrix to solve the linear system $A \vec{x} = y: \; \; A^{-1} A \vec{x} = A^{-1} \vec{y},$ where
**A**is the inverse of^{-1}**A**. This is the more general method and is scalable in computer programs to handle larger systems.

Find the point of intersection of these planes, if it exists:

$$ \begin{align} x + 2y - z = 2 \\[5pt] 2x + y + z = 3 \\[5pt] 2x - y - 3z = -2 \end{align}$$

**Solution**

$$ \left( \begin{matrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 2 & -1 & -3 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 2 \\ 3 \\ -2 \end{matrix} \right)$$

Let's use Cramer's rule for this one. Note that I'm using straight ("absolute value") bars to denote determinants. First we'll need the determinant of the coefficient matrix:

$$\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 2 & -1 & -3 \end{matrix} \right| = 18$$

Now calculate x, y and z using Cramer's rule:

$$x = \frac{\left| \begin{matrix} 2 & 2 & -1 \\ 3 & 1 & 1 \\ -2 & -1 & -3 \end{matrix} \right|}{18} = \frac{11}{18}$$

$$y = \frac{\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 2 & -2 & -3 \end{matrix} \right|}{18} = \frac{19}{18}$$

$$z = \frac{\left| \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 3 \\ 2 & -1 & -2 \end{matrix} \right|}{18} = \frac{13}{18}$$

So these planes intersect at in a single point,

$$(x, y, z) = \left( \frac{11}{18}, \frac{19}{18}, \frac{13}{18} \right)$$

That is, this single point satisfies all three equations. We could also have solved this system by elimination of variables between equations or by inversion of the coefficient matrix. It's your choice.

Find the point of intersection of these planes, if it exists:

$$ \begin{align} x + y + 2z &= 2 \\[5pt] 3x - y + 14z &= 6 \\[5pt] x + 2y \phantom{0000} &= -2 \end{align}$$

**Solution**

$$ \left| \begin{matrix} 1 & 1 & 2 \\ 3 & -1 & 14 \\ 1 & 2 & 0 \end{matrix} \right| = 0 $$

So this system does not have a unique solution. That is, these three planes don't intersect at a single point. Let's continue to explore it, though, and see what's happening.

First, recall that the normal vectors for these planes are just the coefficients of x, y and z:

$$ \begin{align} \vec{N}_1 &= (1, 1, 2) \\[4pt] \vec{N}_2 &= (3, -1, 14) \\[4pt] \vec{N}_3 &= (1, 2, 0) \end{align}$$

Now if these normal vectors all lie in the same plane, then the three planes intersect in a line normal to that plane, thus all combinations of cross products of normal vectors will yield a vector along that line. They are

$$ \begin{align} (1, 1, 2) \times (3, -1, 4) &= 16 \hat{i} - 8 \hat{j} - 4 \hat{k} \\[4pt] (1, 1, 2) \times (1, 2, 0) &= -4 \hat{i} + 2 \hat{j} + \hat{k} \\[4pt] (3, -1, 14) \times (1, 2, 0) &= -4 \hat{i} + 2 \hat{j} + \hat{k} \end{align}$$

Now notice that these vectors are all the same to within a multiplicative constant (which just changes the vector length), so they're all the same, thus these vectors lie on the same plane, and the line perpendicular to them is the line through which our planes intersect.

$$y = \frac{\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 2 & -2 & -3 \end{matrix} \right|}{18} = \frac{19}{18}$$

$$z = \frac{\left| \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 3 \\ 2 & -1 & -2 \end{matrix} \right|}{18} = \frac{13}{18}$$

So these planes intersect at in a single point,

$$(x, y, z) = \left( \frac{11}{18}, \frac{19}{18}, \frac{13}{18} \right)$$

That is, this single point satisfies all three equations. We could also have solved this system by elimination of variables between equations or by inversion of the coefficient matrix. It's your choice.

We can find the angle between two planes by calculating the angle between their normal vectors.

The basic geometry of how this works is shown below. Our normal vectors and the plane surfaces form a quadrilateral, with the angle formed by the normal vectors $\vec{N}_1$ and $\vec{N}_2$ labeled θ. Two of the quadrilateral angles are right angles, so the final angle is 180˚ minus θ Because 180˚-θ is supplementary to θ, we have our angle.

Note, however, that there are two angles between intersecting planes. It isn't always clear without some more careful work which of the two we'll end up calculating with this method, but the good news is that the angles are supplementary.

Here's an example: Find the angle between planes

$$ \begin{align} x - 2y + 3z &= 4 \\[4pt] 2x - y - 3z &= 8 \end{align}$$

$$ \begin{align} \vec{N}_1 &= (1, -2, 3) \\[4pt] \vec{N}_2 &= (2, -1, -3) \end{align}$$

Now the angle between these vectors is found using the definition of the dot product:

$$\theta = cos^{-1} \left( \frac{\vec{N}_1 \cdot \vec{N}_2}{|\vec{N}_1||\vec{N}_2|} \right)$$

which gives $\theta = 1.2 \; \text{rad} = 69˚.$

Here is a graph of our two planes with the angle between them drawn in.

1. |
Determine the direction cosines and the angles these $\mathbb{R}^3$ vectors form with the x, y and z axes. - $\vec{a} = (-1, \, 0, \, 1)$
- $\vec{a} = (1, \, 2, \, -5)$
- $\vec{a} = (1, \, 2, \, 2)$
## Solution(a) Vector length: $$|\vec{a}| = \sqrt{1 + 0 + 1} = \sqrt{2}$$ Cosines and angles: $$ \begin{align} cos(\alpha) &= \frac{a_1}{|\vec{a}|} = \frac{-1}{\sqrt{2}} \\[5pt] \alpha &= 45˚ \\[5pt] cos(\beta) &= \frac{a_2}{|\vec{a}|} = \frac{0}{\sqrt{2}} \\[5pt] \beta &= 90˚ \\[5pt] cos(\gamma) &= \frac{a_3}{|\vec{a}|} = \frac{1}{\sqrt{2}} \\[5pt] \gamma &= 45˚ \\[5pt] \end{align}$$ Recall that cos(x) is an even function, so f(-x) = f(x) (b) Vector length: $$|\vec{a}| = \sqrt{1 + 4 + 25} = \sqrt{30}$$ Cosines and angles: $$ \begin{align} cos(\alpha) &= \frac{1}{\sqrt{30}} \\[5pt] \alpha &= 79.5˚ \\[5pt] cos(\beta) &= \frac{2}{\sqrt{30}} \\[5pt] \beta &= 43˚ \\[5pt] cos(\gamma) &= \frac{-5}{\sqrt{30}} \\[5pt] \gamma &= 156˚ \\[5pt] \end{align}$$ (c) Vector length: $$|\vec{a}| = \sqrt{1 + 4 + 4} = 3$$ Cosines and angles: $$ \begin{align} cos(\alpha) &= \frac{1}{3} \\[5pt] \alpha &= 70.5˚ \\[5pt] cos(\beta) &= \frac{2}{3} \\[5pt] \beta &= 48˚ \\[5pt] cos(\gamma) &= \frac{2}{3} \\[5pt] \gamma &= 48˚ \\[5pt] \end{align}$$ |

2. |
Calculate the scalar and vector projections of $\vec{b}$ onto $\vec{a}.$ - $\vec{a} = (-1, \, 2)$ $\vec{b} = (2, \, 3)$
- $\vec{a} = (-1, \, 2, \, 3)$ $\vec{b} = (1, \, 2, \, 4)$
- $\vec{a} = (2, \, 5, \, -2)$ $\vec{b} = (1, \, 1, \, 2)$
## Solution(a) Length of vector a: $$|\vec{a}| = \sqrt{1 + 4} = \sqrt{5}$$ Scalar projection: $$ \begin{align} s &= \frac{\vec{a}}{|\vec{a}|} \cdot b \\[5pt] &= \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)\cdot (2, 3) \\[5pt] &= \frac{-2}{\sqrt{5}} + \frac{6}{\sqrt{5}} = \frac{4}{\sqrt{5}} \end{align}$$ Now the vector projection: $$ \begin{align} \vec{s} &= s \frac{\vec{a}}{|\vec{a}|} \\[5pt] &= \frac{4}{\sqrt{5}} \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \\[5pt] &= \left( \frac{-4}{5}, \frac{8}{5} \right) \end{align}$$ (b) Length of vector a: $$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$$ Scalar projection: $$ \begin{align} s &= \frac{\vec{a}}{|\vec{a}|} \cdot b \\[5pt] &= \left( \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right)\cdot (1, 2, 4) \\[5pt] &= \frac{-1}{\sqrt{14}} + \frac{2}{\sqrt{14}} + \frac{12}{\sqrt{14}} \\[5pt] &= \frac{13}{\sqrt{14}} \end{align}$$ Now the vector projection: $$ \begin{align} \vec{s} &= s \frac{\vec{a}}{|\vec{a}|} \\[5pt] &= \frac{13}{\sqrt{14}} \left( \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right) \\[5pt] &= \left( \frac{-13}{14}, \frac{26}{14}, \frac{39}{14} \right) \\[5pt] &= \left( \frac{-13}{14}, \frac{13}{7}, \frac{39}{14} \right) \end{align}$$ (c) Length of vector a: $$|\vec{a}| = \sqrt{4 + 25 + 4} = \sqrt{33}$$ Scalar projection: $$ \begin{align} s &= \frac{\vec{a}}{|\vec{a}|} \cdot b \\[5pt] &= \left( \frac{2}{\sqrt{33}}, \frac{5}{\sqrt{33}}, \frac{-2}{\sqrt{33}} \right)\cdot (1, 1, 2) \\[5pt] &= \frac{2}{\sqrt{33}} + \frac{5}{\sqrt{33}} - \frac{2}{\sqrt{33}} \\[5pt] &= \frac{5}{\sqrt{33}} \end{align}$$ Now the vector projection: $$ \begin{align} \vec{s} &= s \frac{\vec{a}}{|\vec{a}|} \\[5pt] &= \frac{5}{\sqrt{33}} \left( \frac{2}{\sqrt{33}}, \frac{5}{\sqrt{33}}, \frac{-2}{\sqrt{33}} \right) \\[5pt] &= \left( \frac{10}{33}, \frac{25}{33}, \frac{-10}{33} \right) \end{align}$$ |

3. |
Calculate the scalar projection of vector $\vec{b} = (1, 0, 1)$ on vector $\vec{a} = (1, 1, -1),$ and explain the result. ## SolutionFirst calculate the length of $\vec{a}$: $$|\vec{a}| = \sqrt{1 + 1 + 1} = \sqrt{3}$$ Now the scalar projection is $$ \begin{align} \frac{\vec{a}}{|\vec{a}|} \cdot \vec{b} &= \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \right) \cdot (1, 0, 1) \\[5pt] &= \frac{1}{\sqrt{3}} + 0 - \frac{1}{\sqrt{3}} = 0 \end{align}$$ The projection has a length of zero because vectors a and b are orthogonal. You can confirm this by calculating the dot product, which is zero. The "shadow" of vector b on vector a is a point, like the shadow of a pole in the noon-sun. |

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