The Equation Can Be Rewritten as Which of the Following Equations?

Learning Objectives

• 2.5.one Write the vector, parametric, and symmetric equations of a line through a given signal in a given direction, and a line through 2 given points.
• 2.5.2 Find the distance from a signal to a given line.
• 2.5.three Write the vector and scalar equations of a aeroplane through a given point with a given normal.
• two.5.iv Notice the distance from a point to a given plane.
• ii.5.5 Notice the angle between two planes.

By now, we are familiar with writing equations that describe a line in ii dimensions. To write an equation for a line, we must know two points on the line, or we must know the direction of the line and at least one point through which the line passes. In two dimensions, nosotros use the concept of slope to depict the orientation, or direction, of a line. In iii dimensions, nosotros describe the direction of a line using a vector parallel to the line. In this section, we examine how to use equations to describe lines and planes in infinite.

Equations for a Line in Space

Allow's showtime explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, $\text{u}$ and $\text{v},$ are parallel, we claim there must exist a scalar, $\mathrm{yard},$ such that $\text{u}=k\text{v}.$ If $\text{u}$ and $\text{5}$ take the aforementioned management, merely choose $\mathrm{thousand}=\frac{\Vert \text{u}\Vert }{\Vert \text{v}\Vert }.$ If $\text{u}$ and $\text{five}$ have opposite directions, cull $k=-\frac{\Vert \text{u}\Vert }{\Vert \text{five}\Vert }.$ Note that the converse holds as well. If $\text{u}=k\text{v}$ for some scalar $\text{k},$ then either $\text{u}$ and $\text{5}$ have the same management $\left(\mathrm{thousand}>0\right)$ or reverse directions $\left(\mathrm{thousand}<0\right),$ and then $\text{u}$ and $\text{v}$ are parallel. Therefore, 2 nonzero vectors $\text{u}$ and $\text{v}$ are parallel if and simply if $\text{u}=m\text{v}$ for some scalar $\text{k}.$ By convention, the cypher vector $0$ is considered to be parallel to all vectors.

As in 2 dimensions, we can describe a line in space using a signal on the line and the direction of the line, or a parallel vector, which we phone call the direction vector (Figure two.63). Let $\mathrm{Fifty}$ be a line in space passing through signal $P\left(x_0,y_0,z_0\right).$ Let $\text{v}=\langle a,b,c\rangle$ be a vector parallel to $\mathrm{50}.$ Then, for any point on line $Q\left(x,y,z\right),$ we know that $\overrightarrow{PQ}$ is parallel to $\text{5}.$ Thus, as we just discussed, there is a scalar, $t,$ such that $\overrightarrow{PQ}=t\text{5},$ which gives

$$\begin{array}{ccc}\hfill \overrightarrow{PQ}& =\hfill & t\text{v}\hfill \\ \hfill \langle \mathrm{10}-x_0,y-y_0,z-z_0\rangle & =\hfill & t\langle a,b,c\rangle \hfill \\ \hfill \langle x-{\mathrm{ten}}_0,y-y_0,z-z_0\rangle & =\hfill & \langle ta,tb,tc\rangle .\hfill \end{array}$$

(2.xi)

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Figure 2.63 Vector v is the direction vector for $\overrightarrow{PQ}.$

Using vector operations, nosotros can rewrite Equation two.eleven equally

$$\begin{array}{ccc}\hfill \langle x-x_0,y-y_0,z-z_0\rangle & =\hfill & \langle ta,tb,tc\rangle \hfill \\ \hfill \langle \mathrm{ten},y,z\rangle -\langle x_0,y_0,z_0\rangle & =\hfill & t\langle a,b,c\rangle \hfill \\ \hfill \langle x,y,z\rangle & =\hfill & \langle {\mathrm{ten}}_0,y_0,z_0\rangle +t\langle a,b,c\rangle .\hfill \end{array}$$

Setting $\text{r}=\langle \mathrm{ten},y,z\rangle$ and ${\text{r}}_0=\langle x_0,y_0,z_0\rangle ,$ we now have the vector equation of a line:

Equating components, Equation 2.11 shows that the post-obit equations are simultaneously truthful: $x-{\mathrm{10}}_0=ta,$ $y-y_0=tb,$ and $z-z_0=tc.$ If we solve each of these equations for the component variables $x,y,\text{and}z,$ nosotros become a gear up of equations in which each variable is defined in terms of the parameter t and that, together, draw the line. This set up of iii equations forms a set of parametric equations of a line:

$$\mathrm{10}=x_0+tay=y_0+tbz=z_0+tc.$$

If we solve each of the equations for $t$ assuming $a,b,\text{and}c$ are nonzero, we get a different clarification of the same line:

$$\frac{x-x_0}{a}=t\frac{y-y_0}{b}=t\frac{z-z_0}{c}=t.$$

Because each expression equals t, they all have the aforementioned value. Nosotros can set them equal to each other to create symmetric equations of a line:

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$

We summarize the results in the following theorem.

Theorem 2.11

Parametric and Symmetric Equations of a Line

A line $L$ parallel to vector $\text{v}=\langle a,b,c\rangle$ and passing through point $P\left({\mathrm{ten}}_0,y_0,z_0\right)$ tin can be described past the following parametric equations:

$$x={\mathrm{ten}}_0+ta,y=y_0+tb,\text{and}z=z_0+tc.$$

(2.thirteen)

If the constants $a,b,\text{and}c$ are all nonzero, then $L$ can be described by the symmetric equation of the line:

$$\frac{x-{\mathrm{10}}_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$

(2.14)

The parametric equations of a line are not unique. Using a different parallel vector or a different indicate on the line leads to a different, equivalent representation. Each set up of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is non unique either.

Example 2.45

Equations of a Line in Space

Discover parametric and symmetric equations of the line passing through points $\left(1,4,\mathrm{-two}\right)$ and $(\mathrm{-3},\mathrm{v},0).$

Checkpoint 2.43

Find parametric and symmetric equations of the line passing through points $\left(1,\mathrm{-3},\mathrm{two}\right)$ and $\left(5,\mathrm{-ii},\mathrm{eight}\right).$

Sometimes we don't desire the equation of a whole line, simply a line segment. In this case, we limit the values of our parameter $t.$ For case, allow $P\left(x_0,y_0,z_0\right)$ and $Q\left(x_1,y_{\mathrm{ane}},z_1\right)$ exist points on a line, and let $\text{p}=\langle {\mathrm{10}}_0,y_0,z_0\rangle$ and $\text{q}=\langle x_1,y_{\mathrm{ane}},z_1\rangle$ be the associated position vectors. In addition, let $\text{r}=\langle x,y,z\rangle .$ Nosotros desire to find a vector equation for the line segment between $P$ and $Q.$ Using $P$ equally our known betoken on the line, and $\overrightarrow{PQ}=\langle {\mathrm{10}}_1-{\mathrm{ten}}_0,y_1-y_0,z_1-z_0\rangle$ every bit the direction vector equation, Equation 2.12 gives

$$\text{r}=\text{p}+t\left(\overrightarrow{PQ}\right).$$

Using properties of vectors, so

$$\begin{array}{cc}\hfill \text{r}& =\text{p}+t\left(\overrightarrow{PQ}\right)\hfill \\ & =\langle x_0,y_0,z_0\rangle +t\langle x_1-{\mathrm{ten}}_0,y_1-y_0,z_1-z_0\rangle \hfill \\ & =\langle x_0,y_0,z_0\rangle +t\left(\langle x_1,y_1,z_1\rangle -\langle {\mathrm{10}}_0,y_0,z_0\rangle \right)\hfill \\ & =\langle x_0,y_0,z_0\rangle +t\langle x_1,y_1,z_1\rangle -t\langle {\mathrm{10}}_0,y_0,z_0\rangle \hfill \\ & =\left(\mathrm{one}-t\right)\langle x_0,y_0,z_0\rangle +t\langle x_{\mathrm{one}},y_1,z_1\rangle \hfill \\ & =\left(\mathrm{ane}-t\right)\text{p}+t\text{q}.\hfill \end{array}$$

Thus, the vector equation of the line passing through $P$ and $Q$ is

$$\text{r}=\left(1-t\right)\text{p}+t\text{q}.$$

Remember that we didn't want the equation of the whole line, but the line segment between $P$ and $Q.$ Find that when $t=0,$ nosotros have $r=p,$ and when $t=\mathrm{ane},$ we have $r=q.$ Therefore, the vector equation of the line segment between $P$ and $Q$ is

$$\text{r}=\left(1-t\right)\text{p}+t\text{q},0\le t\le \mathrm{i}.$$

(2.15)

Going back to Equation ii.12, we can also find parametric equations for this line segment. We take

$$\begin{array}{ccc}\hfill \text{r}& =\hfill & \text{p}+t\left(\overrightarrow{PQ}\right)\hfill \\ \hfill \langle \mathrm{ten},y,z\rangle & =\hfill & \langle x_0,y_0,z_0\rangle +t\langle x_1-x_0,y_{\mathrm{ane}}-y_0,z_1-z_0\rangle \hfill \\ & =\hfill & \langle x_0+t\left(x_1-x_0\right),y_0+t\left(y_1-y_0\right),z_0+t\left(z_{\mathrm{ane}}-z_0\right)\rangle .\hfill \end{array}$$

Then, the parametric equations are

$$\mathrm{10}={\mathrm{10}}_0+t\left({\mathrm{ten}}_1-x_0\right),y=y_0+t\left(y_{\mathrm{ane}}-y_0\right),z=z_0+t\left(z_1-z_0\right),0\le t\le \mathrm{i}.$$

(2.16)

Example two.46

Parametric Equations of a Line Segment

Observe parametric equations of the line segment between the points $P(\mathrm{ii},1,4)$ and $Q\left(3,\mathrm{-one},3\right).$

Checkpoint 2.44

Observe parametric equations of the line segment between points $P(\mathrm{-1},3,\mathrm{half\; dozen})$ and $Q\left(\mathrm{-eight},2,4\right).$

Distance betwixt a Point and a Line

We already know how to calculate the distance betwixt ii points in space. We at present aggrandize this definition to describe the distance between a bespeak and a line in space. Several real-world contexts exist when information technology is important to exist able to summate these distances. When building a home, for example, builders must consider "setback" requirements, when structures or fixtures take to be a certain distance from the property line. Air travel offers another case. Airlines are concerned most the distances between populated areas and proposed flight paths.

Let $L$ be a line in the airplane and let $M$ be any point non on the line. Then, nosotros define distance $d$ from $M$ to $\mathrm{Fifty}$ as the length of line segment $\stackrel{—}{MP},$ where $P$ is a point on $\mathrm{Fifty}$ such that $\stackrel{—}{MP}$ is perpendicular to $L$ (Effigy ii.64).

Figure 2.64 The altitude from betoken $\mathrm{Yard}$ to line $L$ is the length of $\stackrel{—}{MP}.$

When nosotros're looking for the distance between a line and a betoken in space, Effigy 2.64 still applies. We still ascertain the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear style to know which indicate on the line creates such a perpendicular line segment, then we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let $P$ be an arbitrary point on line $L$ and permit $\text{v}$ be a management vector for $\mathrm{50}$ (Figure 2.65).

Figure two.65 Vectors $\overrightarrow{P\mathrm{One\; thousand}}$ and v form 2 sides of a parallelogram with base of operations $\Vert \text{v}\Vert$ and height $d,$ which is the distance between a line and a betoken in infinite.

By Expanse of a Parallelogram, vectors $\overrightarrow{PM}$ and $\text{v}$ course ii sides of a parallelogram with expanse $\Vert \overrightarrow{P\mathrm{Thou}}\times \text{v}\Vert .$ Using a formula from geometry, the area of this parallelogram tin also be calculated as the product of its base and height:

$$\Vert \overrightarrow{P\mathrm{1000}}\times \text{five}\Vert =\Vert \text{v}\Vert d.$$

We can utilise this formula to observe a general formula for the altitude between a line in space and any point not on the line.

Theorem ii.12

Distance from a Point to a Line

Let $\mathrm{Fifty}$ exist a line in space passing through point $P$ with direction vector $\text{v}.$ If $M$ is any point not on $L,$ then the altitude from $G$ to $L$ is

$$d=\frac{\Vert \overrightarrow{PM}\times \text{5}\Vert }{\Vert \text{5}\Vert }.$$

Example two.47

Computing the Altitude from a Point to a Line

Observe the distance between t point $M=\left(1,\mathrm{one},\mathrm{three}\right)$ and line $\frac{\mathrm{ten}-3}{\mathrm{iv}}=\frac{y+1}{\mathrm{ii}}=z-\mathrm{three}.$

Checkpoint 2.45

Observe the distance between point $\left(0,3,6\right)$ and the line with parametric equations $x=\mathrm{i}-t,y=1+\mathrm{ii}t,z=\mathrm{v}+3t.$

Relationships between Lines

Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a unmarried indicate. In three dimensions, a quaternary instance is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to exist skew lines (Effigy 2.67).

Figure ii.67 In three dimensions, it is possible that two lines practise not cantankerous, even when they have different directions.

To allocate lines every bit parallel but non equal, equal, intersecting, or skew, nosotros demand to know 2 things: whether the direction vectors are parallel and whether the lines share a point (Effigy ii.68).

Figure ii.68 Determine the relationship between ii lines based on whether their management vectors are parallel and whether they share a betoken.

Example 2.48

Classifying Lines in Space

For each pair of lines, decide whether the lines are equal, parallel but not equal, skew, or intersecting.

1. $L_{\mathrm{ane}}:\mathrm{ten}=2s-\mathrm{i},y=\mathrm{south}-\mathrm{one},z=s-\mathrm{four}$
   $L_2:\mathrm{ten}=t-\mathrm{three},y=3t+8,z=5-2t$
2. $L_1\text{:}$ $\mathrm{ten}=\text{−}y=z$
   $L_2:\frac{\mathrm{ten}-\mathrm{iii}}{\mathrm{two}}=y=z-\mathrm{ii}$
3. $L_1:\mathrm{ten}=6\mathrm{south}-\mathrm{one},y=\mathrm{-ii}s,z=3s+\mathrm{ane}$
   ${\mathrm{Fifty}}_2:\frac{x-4}{\mathrm{vi}}=\frac{y+3}{\mathrm{-2}}=\frac{z-1}{3}$

Checkpoint 2.46

Describe the human relationship between the lines with the post-obit parametric equations:

$$\mathrm{ten}=1-\mathrm{four}t,y=\mathrm{three}+t,z=8-6t$$

$$\mathrm{ten}=2+\mathrm{iii}s,y=2s,z=\mathrm{-one}-\mathrm{iii}s.$$

Equations for a Plane

We know that a line is determined past two points. In other words, for any two distinct points, there is exactly i line that passes through those points, whether in ii dimensions or three. Similarly, given whatever three points that do not all lie on the same line, in that location is a unique plane that passes through these points. Just as a line is adamant by two points, a airplane is determined by three.

This may be the simplest fashion to characterize a plane, but we can utilize other descriptions equally well. For example, given two distinct, intersecting lines, at that place is exactly 1 plane containing both lines. A plane is also determined by a line and whatever indicate that does non lie on the line. These characterizations ascend naturally from the thought that a plane is determined by three points. Perchance the nigh surprising characterization of a plane is really the most useful.

Imagine a pair of orthogonal vectors that share an initial indicate. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, nosotros describe that concept mathematically. Let $\text{due north}=\langle a,b,c\rangle$ exist a vector and $P=\left(x_0,y_0,z_0\right)$ exist a point. And so the set up of all points $Q=\left(x,y,z\right)$ such that $\overrightarrow{PQ}$ is orthogonal to $\text{due north}$ forms a airplane (Figure two.69). Nosotros say that $\text{due north}$ is a normal vector, or perpendicular to the plane. Call up, the dot product of orthogonal vectors is goose egg. This fact generates the vector equation of a plane: $\text{n}·\overrightarrow{PQ}=0.$ Rewriting this equation provides boosted ways to draw the plane:

$$\begin{array}{}\\ \hfill \text{north}·\overrightarrow{PQ}& =\hfill & 0\hfill \\ \hfill \langle a,b,c\rangle ·\langle \mathrm{ten}-x_0,y-y_0,z-z_0\rangle & =\hfill & 0\hfill \\ \hfill a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)& =\hfill & 0.\hfill \end{array}$$

Figure two.69 Given a point $P$ and vector $\text{north},$ the gear up of all points $Q$ with $\overrightarrow{PQ}$ orthogonal to $\text{northward}$ forms a plane.

Definition

Given a point $P$ and vector $\text{n},$ the prepare of all points $Q$ satisfying the equation $\text{northward}·\overrightarrow{PQ}=0$ forms a plane. The equation

is known as the vector equation of a plane.

The scalar equation of a plane containing point $P=\left({\mathrm{10}}_0,y_0,z_0\right)$ with normal vector $\text{n}=\langle a,b,c\rangle$ is

$$a\left(x-{\mathrm{ten}}_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0.$$

(2.eighteen)

This equation can be expressed as $a\mathrm{10}+by+cz+d=0,$ where $d=\text{−}a{x}_0-b{y}_0-c{z}_0.$ This form of the equation is sometimes called the full general course of the equation of a aeroplane.

As described earlier in this section, any three points that do not all lie on the same line make up one's mind a airplane. Given iii such points, we can find an equation for the plane containing these points.

Case 2.49

Writing an Equation of a Plane Given 3 Points in the Airplane

Write an equation for the plane containing points $P=\left(1,1,\mathrm{-2}\right),$ $Q=\left(0,2,1\right),$ and $R=\left(\mathrm{-i},\mathrm{-1},0\right)$ in both standard and general forms.

The scalar equations of a plane vary depending on the normal vector and point chosen.

Example 2.fifty

Writing an Equation for a Plane Given a Point and a Line

Find an equation of the aeroplane that passes through indicate $\left(1,4,3\right)$ and contains the line given by $x=\frac{y-\mathrm{i}}{2}=z+\mathrm{one}.$

Checkpoint 2.47

Find an equation of the airplane containing the lines $L_1$ and $L_{\mathrm{two}}\text{:}$

$$\begin{array}{c}{L}_1:x=\text{−}y=z\hfill \\ L_{\mathrm{two}}:\frac{x-\mathrm{iii}}{\mathrm{two}}=y=z-\mathrm{ii.}\hfill \end{array}$$

Now that nosotros tin can write an equation for a plane, we can utilise the equation to find the distance $d$ betwixt a betoken $P$ and the plane. Information technology is defined as the shortest possible distance from $P$ to a point on the plane.

Effigy two.seventy We want to find the shortest distance from indicate P to the plane. Allow bespeak $R$ exist the point in the airplane such that, for whatever other point in the plane $Q,$ $\Vert \overrightarrow{RP}\Vert <\Vert \overrightarrow{QP}\Vert .$

Merely as nosotros find the two-dimensional distance between a indicate and a line by computing the length of a line segment perpendicular to the line, we discover the three-dimensional distance between a bespeak and a plane by calculating the length of a line segment perpendicular to the airplane. Let $R$ bet the indicate in the aeroplane such that $\overrightarrow{RP}$ is orthogonal to the plane, and let $Q$ be an arbitrary point in the plane. And then the projection of vector $\overrightarrow{QP}$ onto the normal vector describes vector $\overrightarrow{RP},$ as shown in Effigy 2.70.

Theorem two.xiii

The Distance between a Plane and a Point

Suppose a plane with normal vector $\text{north}$ passes through point $Q.$ The distance $d$ from the airplane to a point $P$ not in the plane is given by

$$d=\Vert {\text{proj}}_{\text{north}}\overrightarrow{QP}\Vert =\left|{\text{comp}}_{\text{n}}\overrightarrow{QP}\right|=\frac{\left|\overrightarrow{QP}·\text{due north}\right|}{\Vert \text{n}\Vert }.$$

(2.nineteen)

Instance 2.51

Distance between a Signal and a Plane

Detect the distance betwixt point $P=\left(3,1,\mathrm{two}\right)$ and the plane given past $x-\mathrm{two}y+z=5$ (see the post-obit figure).

Checkpoint 2.48

Find the distance between indicate $P=\left(5,\mathrm{-1},0\right)$ and the plane given by $\mathrm{four}x+2y-z=3.$

Parallel and Intersecting Planes

We have discussed the diverse possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the 2 distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line (Effigy two.71).

Figure ii.71 The intersection of ii nonparallel planes is ever a line.

We can use the equations of the two planes to find parametric equations for the line of intersection.

Instance 2.52

Finding the Line of Intersection for Ii Planes

Find parametric and symmetric equations for the line formed by the intersection of the planes given by $x+y+z=0$ and $\mathrm{ii}x-y+z=0$ (see the following effigy).

Checkpoint two.49

Find parametric equations for the line formed by the intersection of planes $x+y-z=3$ and $\mathrm{iii}\mathrm{10}-y+3z=5.$

In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed by the intersection of 2 planes. For case, builders constructing a house need to know the angle where dissimilar sections of the roof see to know whether the roof will wait skilful and drain properly. Nosotros can use normal vectors to calculate the angle between the two planes. We can do this because the angle betwixt the normal vectors is the same equally the bending between the planes. Effigy 2.72 shows why this is true.

Figure 2.72 The bending between two planes has the same measure out as the angle between the normal vectors for the planes.

We can find the measure of the bending θ between ii intersecting planes by outset finding the cosine of the angle, using the post-obit equation:

$$\text{cos}\theta =\frac{\left|{\text{n}}_{\mathrm{ane}}·{\text{n}}_2\right|}{\Vert {\text{n}}_1\Vert \Vert {\text{n}}_{\mathrm{ii}}\Vert }.$$

We can and then utilise the angle to determine whether 2 planes are parallel or orthogonal or if they intersect at some other angle.

Example two.53

Finding the Bending between Two Planes

Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but non orthogonal, find the measure of the angle between them. Give the respond in radians and circular to ii decimal places.

1. $x+\mathrm{two}y-z=8\text{and}2\mathrm{10}+4y-2z=10$
2. $2x-3y+2z=\mathrm{three}\text{and}6x+\mathrm{two}y-3z=1$
3. $x+y+z=4\text{and}x-3y+5z=\mathrm{i}$

Checkpoint two.50

Find the measure of the bending between planes $x+y-z=3$ and $3x-y+3z=5.$ Give the answer in radians and round to 2 decimal places.

When we find that ii planes are parallel, we may need to observe the distance betwixt them. To observe this distance, we just select a signal in 1 of the planes. The distance from this indicate to the other airplane is the distance between the planes.

Previously, nosotros introduced the formula for computing this distance in Equation 2.19:

$$d=\frac{\overrightarrow{QP}·\text{north}}{\Vert \text{northward}\Vert },$$

where $Q$ is a point on the aeroplane, $P$ is a indicate non on the plane, and $\text{northward}$ is the normal vector that passes through point $Q.$ Consider the altitude from point $\left({\mathrm{ten}}_0,y_0,z_0\right)$ to plane $ax+by+cz+k=0.$ Allow $\left(x_1,y_{\mathrm{ane}},z_1\right)$ exist any point in the aeroplane. Substituting into the formula yields

$$\begin{array}{cc}\hfill d& =\frac{\left|a\left(x_0-x_1\right)+b\left(y_0-y_1\right)+c\left(z_0-z_1\right)\right|}{\sqrt{a^{\mathrm{two}}+b^{\mathrm{two}}+c^2}}\hfill \\ & =\frac{\left|a{\mathrm{10}}_0+b{y}_0+c{z}_0+k\right|}{\sqrt{a^2+b^2+c^2}}.\hfill \end{array}$$

Nosotros state this result formally in the following theorem.

Theorem 2.14

Distance from a Signal to a Plane

Let $P\left({\mathrm{10}}_0,y_0,z_0\right)$ be a betoken. The distance from $P$ to airplane $ax+by+cz+k=0$ is given by

$$d=\frac{\left|a{x}_0+b{y}_0+c{z}_0+k\right|}{\sqrt{a^2+b^{\mathrm{two}}+c^2}}.$$

Example 2.54

Finding the Distance between Parallel Planes

Find the distance between the two parallel planes given by $2x+y-z=2$ and $2x+y-z=8.$

Checkpoint 2.51

Observe the distance betwixt parallel planes $\mathrm{v}\mathrm{ten}-2y+z=\mathrm{half-dozen}$ and $5\mathrm{10}-2y+z=\mathrm{-iii}.$

Student Project

Distance between Ii Skew Lines

Figure 2.73 Industrial pipe installations often characteristic pipes running in different directions. How can nosotros find the distance between 2 skew pipes?

Finding the distance from a betoken to a line or from a line to a plane seems like a pretty abstract procedure. Simply, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to mensurate. 1 way is to model the two pipes as lines, using the techniques in this chapter, so summate the altitude between them. The calculation involves forming vectors along the directions of the lines and using both the cantankerous production and the dot product.

The symmetric forms of two lines, ${\mathrm{Fifty}}_1$ and $L_2,$ are

$$\begin{array}{}\\ \\ L_1:\frac{x-{\mathrm{10}}_1}{a_{\mathrm{one}}}=\frac{y-y_1}{b_{\mathrm{ane}}}=\frac{z-z_1}{c_{\mathrm{one}}}\hfill \\ {\mathrm{50}}_2:\frac{x-x_2}{a_2}=\frac{y-y_{\mathrm{ii}}}{b_2}=\frac{z-z_{\mathrm{two}}}{c_{\mathrm{ii}}}.\hfill \end{array}$$

You are to develop a formula for the altitude $d$ between these 2 lines, in terms of the values $a_1,b_1,c_{\mathrm{one}};a_2,b_{\mathrm{ii}},c_{\mathrm{ii}};{\mathrm{ten}}_{\mathrm{ane}},y_{\mathrm{ane}},z_1;\text{and}{\mathrm{10}}_2,y_2,z_2.$ The distance between two lines is commonly taken to mean the minimum distance, then this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.

1. Starting time, write down two vectors, ${\text{v}}_1$ and ${\text{five}}_{\mathrm{two}},$ that prevarication along $L_{\mathrm{one}}$ and ${\mathrm{50}}_{\mathrm{two}},$ respectively.
2. Discover the cross product of these two vectors and call it $\text{N}.$ This vector is perpendicular to ${\text{v}}_1\text{and}{\text{v}}_{\mathrm{two}},$ and hence is perpendicular to both lines.
3. From vector $\text{N},$ form a unit vector $\text{due north}$ in the aforementioned management.
4. Use symmetric equations to find a convenient vector ${\text{v}}_{12}$ that lies betwixt any two points, one on each line. Again, this tin can exist done straight from the symmetric equations.
5. The dot product of two vectors is the magnitude of the projection of 1 vector onto the other—that is, $\text{A}·\text{B}=\Vert \text{A}\Vert \Vert \text{B}\Vert \text{cos}\theta ,$ where $\theta$ is the angle between the vectors. Using the dot production, find the projection of vector ${\text{5}}_{12}$ found in stride $4$ onto unit vector $\text{n}$ constitute in step iii. This project is perpendicular to both lines, and hence its length must be the perpendicular distance $d$ between them. Note that the value of $d$ may be negative, depending on your choice of vector ${\text{v}}_{12}$ or the gild of the cross production, so utilize absolute value signs around the numerator.
6. Check that your formula gives the right altitude of $\left|\mathrm{-25}\right|\text{/}\sqrt{198}\approx 1.78$ between the following ii lines:
   

   $$\begin{array}{}\\ \\ L_{\mathrm{ane}}:\frac{x-\mathrm{five}}{2}=\frac{y-3}{4}=\frac{z-1}{3}\hfill \\ L_2:\frac{x-\mathrm{six}}{3}=\frac{y-1}{\mathrm{v}}=\frac{z}{7}.\hfill \end{array}$$

7. Is your general expression valid when the lines are parallel? If not, why not? (Hint: What do you know near the value of the cross product of two parallel vectors? Where would that result show up in your expression for $d?)$
8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the management of $\text{north}$ and ${\text{v}}_{12}.$ What is the consequence of their dot product?
9. Consider the following application. Engineers at a refinery have determined they need to install support struts betwixt many of the gas pipes to reduce damaging vibrations. To minimize cost, they program to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.
   The rectangular frame structure has the dimensions $\mathrm{iv.0}\times 15.0\times 10.0\text{1000}$ (height, width, and depth). One sector has a pipe inbound the lower corner of the standard frame unit of measurement and exiting at the diametrically opposed corner (the 1 farthest away at the top); call this ${\mathrm{50}}_1.$ A second piping enters and exits at the two different opposite lower corners; call this ${\mathrm{50}}_2$ (Figure two.74).
   

   

   Figure 2.74 2 pipes cantankerous through a standard frame unit.

   
   Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector $\text{northward},$ define a vector that spans two points on each line, and finally determine the minimum altitude between the lines. (Take the origin to be at the lower corner of the first pipage.) Similarly, yous may also develop the symmetric equations for each line and substitute directly into your formula.

Section 2.5 Exercises

In the post-obit exercises, points $P$ and $Q$ are given. Allow $L$ be the line passing through points $P$ and $Q.$

1. Find the vector equation of line $L.$
2. Observe parametric equations of line $L.$
3. Find symmetric equations of line $\mathrm{50}.$
4. Observe parametric equations of the line segment determined by $P$ and $Q.$

243 .

$P\left(\mathrm{-iii},5,\mathrm{ix}\right),$ $Q\left(\mathrm{four},\mathrm{-seven},\mathrm{two}\right)$

244 .

$P\left(\mathrm{iv},0,\mathrm{five}\right),Q\left(2,\mathrm{iii},1\right)$

245 .

$P\left(\mathrm{-ane},0,5\right),$ $Q\left(4,0,\mathrm{iii}\right)$

246 .

$P\left(7,\mathrm{-2},6\right),$ $Q\left(\mathrm{-3},0,\mathrm{vi}\right)$

For the post-obit exercises, indicate $P$ and vector $\text{v}$ are given. Let $\mathrm{Fifty}$ be the line passing through point $P$ with direction $\text{v}.$

1. Find parametric equations of line $L.$
2. Find symmetric equations of line $L.$
3. Find the intersection of the line with the xy-plane.

247 .

$P\left(1,\mathrm{-2},\mathrm{iii}\right),$ $\text{5}=\langle 1,\mathrm{two},3\rangle$

248 .

$P(3,1,\mathrm{five}),$ $\text{five}=\langle \mathrm{one},1,1\rangle$

249 .

$P(\mathrm{three},1,5),$ $\text{v}=\overrightarrow{QR},$ where $Q(2,\mathrm{ii},3)$ and $R(3,\mathrm{ii},\mathrm{iii})$

250 .

$P(2,\mathrm{three},0),$ $\text{five}=\overrightarrow{QR},$ where $Q(0,4,\mathrm{five})$ and $R(0,4,6)$

For the following exercises, line $L$ is given.

1. Find indicate $P$ that belongs to the line and direction vector $\text{five}$ of the line. Express $\text{v}$ in component form.
2. Find the distance from the origin to line $L.$

251 .

$x=\mathrm{one}+t,y=\mathrm{iii}+t,z=\mathrm{five}+4t,$ $t\in ℝ$

252 .

$\text{−}x=y+1,z=\mathrm{two}$

253 .

Notice the distance between point $A\left(\mathrm{-iii},1,1\right)$ and the line of symmetric equations

$x=\text{−}y=\text{−}z.$

254 .

Observe the distance between point $A\left(4,2,5\right)$ and the line of parametric equations

$x=\mathrm{-1}-t,y=\text{−}t,z=2,$ $t\in ℝ.$

For the following exercises, lines $L_{\mathrm{i}}$ and $L_2$ are given.

1. Verify whether lines $L_1$ and $L_2$ are parallel.
2. If the lines ${\mathrm{50}}_1$ and $L_2$ are parallel, then discover the altitude between them.

255 .

$L_1:x=\mathrm{ane}+t,y=t,z=2+t,$ $t\in ℝ,$ $L_2:x-3=y-1=z-3$

256 .

${\mathrm{Fifty}}_1:x=2,y=1,z=t,$ $L_2:x=\mathrm{i},y=1,z=\mathrm{two}-3t,$ $t\in ℝ$

257 .

Show that the line passing through points $P\left(3,\mathrm{i},0\right)$ and $Q\left(1,4,\mathrm{-three}\right)$ is perpendicular to the line with equation $\mathrm{10}=3t,y=\mathrm{three}+\mathrm{eight}t,z=\mathrm{-7}+6t,$ $t\in ℝ.$

258 .

Are the lines of equations $x=\mathrm{-two}+\mathrm{ii}t,y=\mathrm{-6},z=2+6t$ and $x=\mathrm{-1}+t,y=\mathrm{i}+t,z=t,$ $t\in ℝ,$ perpendicular to each other?

259 .

Observe the point of intersection of the lines of equations $x=\mathrm{-2}y=\mathrm{three}z$ and $\mathrm{10}=\mathrm{-5}-t,y=\mathrm{-one}+t,z=t-\mathrm{eleven},$ $t\in ℝ.$

260 .

Find the intersection point of the x-axis with the line of parametric equations

$x=\mathrm{ten}+t,y=2-2t,z=\mathrm{-three}+3t,$ $t\in ℝ.$

For the following exercises, lines $L_1$ and $L_2$ are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting.

261 .

$L_1:x=y-\mathrm{i}=\text{−}z$ and ${\mathrm{Fifty}}_{\mathrm{ii}}:\mathrm{ten}-2=\text{−}y=\frac{z}{2}$

262 .

${\mathrm{50}}_{\mathrm{one}}:x=\mathrm{two}t,y=0,z=\mathrm{three},$ $t\in ℝ$ and ${\mathrm{Fifty}}_2:x=0,y=8+\mathrm{due\; south},z=\mathrm{vii}+s,$ $s\in ℝ$

263 .

${\mathrm{50}}_1:x=\mathrm{-1}+\mathrm{ii}t,y=\mathrm{one}+3t,z=7t,$ $t\in ℝ$ and $L_2:\mathrm{10}-1=\frac{2}{3}\left(y-\mathrm{iv}\right)=\frac{\mathrm{ii}}{\mathrm{vii}}z-\mathrm{two}$

264 .

$L_1:3\mathrm{10}=y+1=\mathrm{ii}z$ and ${\mathrm{Fifty}}_{\mathrm{ii}}:x=\mathrm{vi}+2t,y=17+\mathrm{six}t,z=\mathrm{nine}+3t,$ $t\in ℝ$

265 .

Consider line $L$ of symmetric equations $x-\mathrm{two}=\text{−}y=\frac{z}{2}$ and point $A(1,1,1).$

1. Find parametric equations for a line parallel to $L$ that passes through bespeak $A.$
2. Detect symmetric equations of a line skew to $L$ and that passes through betoken $A.$
3. Find symmetric equations of a line that intersects $\mathrm{50}$ and passes through point $A.$

266 .

Consider line $L$ of parametric equations $x=t,y=\mathrm{two}t,z=3,$ $t\in ℝ.$

1. Observe parametric equations for a line parallel to $L$ that passes through the origin.
2. Discover parametric equations of a line skew to $\mathrm{50}$ that passes through the origin.
3. Discover symmetric equations of a line that intersects $L$ and passes through the origin.

For the following exercises, point $P$ and vector $\text{n}$ are given.

1. Find the scalar equation of the aeroplane that passes through $P$ and has normal vector $\text{due north}.$
2. Find the general course of the equation of the aeroplane that passes through $P$ and has normal vector $\text{n}.$

267 .

$P(0,0,0),$ $\text{n}=3\text{i}-2\text{j}+\mathrm{iv}\text{chiliad}$

268 .

$P\left(\mathrm{three},\mathrm{two},2\right),$ $\text{north}=2\text{i}+3\text{j}-\text{k}$

269 .

$P\left(1,2,3\right),$ $\text{due north}=\langle 1,\mathrm{ii},3\rangle$

270 .

$P(0,0,0),$ $\text{due north}=\langle \mathrm{-iii},2,\mathrm{-1}\rangle$

For the following exercises, the equation of a plane is given.

1. Find normal vector $\text{n}$ to the airplane. Limited $\text{n}$ using standard unit vectors.
2. Find the intersections of the plane with the axes of coordinates.
3. Sketch the plane.

271 .

[T] $4x+5y+10z-20=0$

272 .

$3x+\mathrm{four}y-12=0$

273 .

$3\mathrm{10}-\mathrm{ii}y+4z=0$

274 .

$x+z=0$

275 .

Given point $P\left(\mathrm{i},2,3\right)$ and vector $\text{n}=\text{i}+\text{j},$ find point $Q$ on the x-centrality such that $\overrightarrow{PQ}$ and $\text{due north}$ are orthogonal.

276 .

Testify there is no plane perpendicular to $\text{n}=\text{i}+\text{j}$ that passes through points $P\left(1,2,3\right)$ and $Q\left(2,3,\mathrm{four}\right).$

277 .

Observe parametric equations of the line passing through point $P(\mathrm{-two},1,3)$ that is perpendicular to the airplane of equation $2x-\mathrm{iii}y+z=7.$

278 .

Find symmetric equations of the line passing through indicate $P(\mathrm{ii},\mathrm{v},\mathrm{four})$ that is perpendicular to the aeroplane of equation $\mathrm{ii}x+\mathrm{three}y-\mathrm{v}z=0.$

279 .

Show that line $\frac{\mathrm{10}-\mathrm{ane}}{\mathrm{two}}=\frac{y+1}{3}=\frac{z-2}{4}$ is parallel to plane $x-2y+z=6.$

280 .

Find the real number $\alpha$ such that the line of parametric equations $x=t,y=2-t,z=\mathrm{three}+t,$ $t\in ℝ$ is parallel to the plane of equation $\alpha x+5y+z-10=0.$

For the following exercises, points $P,Q,\text{and}R$ are given.

1. Observe the full general equation of the plane passing through $P,Q,\text{and}R.$
2. Write the vector equation $\text{n}·\overrightarrow{P\mathrm{Due\; south}}=0$ of the plane at a., where $S\left(x,y,z\right)$ is an arbitrary point of the plane.
3. Find parametric equations of the line passing through the origin that is perpendicular to the airplane passing through $P,Q,\text{and}R.$

281 .

$P(1,1,\mathrm{i}),Q(2,4,3),$ and $R(\mathrm{-ane},\mathrm{-2},\mathrm{-ane})$

282 .

$P\left(\mathrm{-2},1,\mathrm{four}\right),Q\left(3,1,3\right),$ and $R\left(\mathrm{-two},1,0\right)$

283 .

Consider the planes of equations $x+y+z=1$ and $x+z=0.$

1. Testify that the planes intersect.
2. Find symmetric equations of the line passing through point $P(\mathrm{i},\mathrm{four},6)$ that is parallel to the line of intersection of the planes.

284 .

Consider the planes of equations $\text{−}y+z-\mathrm{ii}=0$ and $x-y=0.$

1. Evidence that the planes intersect.
2. Observe parametric equations of the line passing through betoken $P(\mathrm{-8},0,\mathrm{two})$ that is parallel to the line of intersection of the planes.

285 .

Find the scalar equation of the plane that passes through point $P(\mathrm{-ane},2,\mathrm{i})$ and is perpendicular to the line of intersection of planes $x+y-z-2=0$ and $2\mathrm{ten}-y+3z-1=0.$

286 .

Notice the general equation of the plane that passes through the origin and is perpendicular to the line of intersection of planes $\text{−}x+y+2=0$ and $z-3=0.$

287 .

Decide whether the line of parametric equations $x=\mathrm{i}+\mathrm{ii}t,y=\mathrm{-2}t,z=2+t,$ $t\in ℝ$ intersects the plane with equation $3x+4y+6z-\mathrm{vii}=0.$ If it does intersect, find the bespeak of intersection.

288 .

Determine whether the line of parametric equations $x=\mathrm{v},y=4-t,z=\mathrm{ii}t,$ $t\in ℝ$ intersects the aeroplane with equation $2\mathrm{ten}-y+z=5.$ If it does intersect, discover the point of intersection.

289 .

Find the distance from point $P\left(1,5,\mathrm{-4}\right)$ to the plane of equation $3\mathrm{ten}-y+2z-6=0.$

290 .

Find the distance from point $P\left(\mathrm{i},\mathrm{-ii},3\right)$ to the plane of equation $\left(x-3\right)+2\left(y+1\right)-4z=0.$

For the post-obit exercises, the equations of 2 planes are given.

1. Determine whether the planes are parallel, orthogonal, or neither.
2. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer.

291 .

[T] $x+y+z=0,$ $\mathrm{two}x-y+z-\mathrm{vii}=0$

292 .

$\mathrm{five}x-\mathrm{three}y+z=4,$ $x+\mathrm{iv}y+7z=1$

293 .

$x-5y-z=1,$ $5\mathrm{ten}-25y-5z=\mathrm{-3}$

294 .

[T] $x-3y+\mathrm{half-dozen}z=4,$ $5x+y-z=\mathrm{iv}$

295 .

Show that the lines of equations $\mathrm{10}=t,y=1+t,z=2+t,$ $t\in ℝ\text{,}$ and $\frac{\mathrm{10}}{\mathrm{ii}}=\frac{y-1}{3}=z-\mathrm{three}$ are skew, and discover the distance between them.

296 .

Evidence that the lines of equations $\mathrm{ten}=\mathrm{-1}+t,y=\mathrm{-ii}+t,z=\mathrm{three}t,$ $t\in ℝ,$ and $\mathrm{10}=\mathrm{v}+s,y=\mathrm{-8}+2\mathrm{due\; south},z=7s,$ $s\in ℝ$ are skew, and find the distance between them.

297 .

Consider point $C\left(\mathrm{-three},\mathrm{two},4\right)$ and the aeroplane of equation $2\mathrm{10}+4y-3z=\mathrm{eight}.$

1. Find the radius of the sphere with middle $C$ tangent to the given plane.
2. Find point P of tangency.

298 .

Consider the plane of equation $x-y-z-8=0.$

1. Find the equation of the sphere with heart $C$ at the origin that is tangent to the given plane.
2. Find parametric equations of the line passing through the origin and the signal of tangency.

299 .

Ii children are playing with a ball. The girl throws the ball to the boy. The ball travels in

the air, curves $3$ ft to the correct, and falls $\mathrm{five}$ ft away from the girl (see the following effigy). If the aeroplane that contains the trajectory of the brawl is perpendicular to the footing, detect its equation.

300 .

[T] John allocates $d$ dollars to consume monthly three goods of prices $a,b,\text{and}c.$ In this context, the budget equation is defined as $ax+by+cz=d,$ where $x\ge 0,y\ge 0,$ and $z\ge 0$ correspond the number of items bought from each of the goods. The budget ready is given by $\left\{\left(x,y,z\right)|ax+by+cz\le d,x\ge 0,y\ge 0,z\ge 0\right\},$ and the budget plane is the role of the plane of equation $ax+by+cz=d$ for which $x\ge 0,y\ge 0,$ and $z\ge 0.$ Consider $a=\text{\$}8,$ $b=\text{\$}\mathrm{five},$ $c=\text{\$}10,$ and $d=\text{\$}500.$

1. Use a CAS to graph the budget set and upkeep plane.
2. For $z=25,$ observe the new upkeep equation and graph the budget ready in the same system of coordinates.

301 .

[T] Consider $\text{r}(t)=\langle \text{sin}t,\text{cos}t,\mathrm{two}t\rangle$ the position vector of a particle at time $t\in [0,3],$ where the components of r are expressed in centimeters and fourth dimension is measured in seconds. Let $\overrightarrow{OP}$ be the position vector of the particle after $1$ sec.

1. Decide the velocity vector $\text{v}(1)$ of the particle later on $\mathrm{ane}$ sec.
2. Find the scalar equation of the aeroplane that is perpendicular to $\text{five}(1)$ and passes through signal $P.$ This plane is chosen the normal plane to the path of the particle at point $P.$
3. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point $P.$

302 .

[T] A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) $A(\mathrm{eight},0,0),$ $B(8,18,0),$ $C(0,18,8),$ and $D(0,0,8)$ (see the following figure).

1. Find the general grade of the equation of the plane that contains the solar panel by using points $A,B,\text{and}C,$ and evidence that its normal vector is equivalent to $\overrightarrow{AB}\times \overrightarrow{AD}.$
2. Discover parametric equations of line ${\mathrm{50}}_1$ that passes through the middle of the solar panel and has management vector $\text{s}=\frac{1}{\sqrt{\mathrm{iii}}}\text{i}+\frac{\mathrm{one}}{\sqrt{3}}\text{j}+\frac{1}{\sqrt{\mathrm{iii}}}\text{thou},$ which points toward the position of the Dominicus at a item time of twenty-four hour period.
3. Find symmetric equations of line ${\mathrm{50}}_{\mathrm{two}}$ that passes through the center of the solar console and is perpendicular to it.
4. Decide the angle of elevation of the Sun above the solar panel by using the angle betwixt lines $L_{\mathrm{one}}$ and $L_2.$

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Source: https://openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space

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