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Equations of Lines and Planes in Space

Write the vector, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.

Equations for a Line in Space

Let’s first 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 be a scalar, \(k,\) such that \(\text{u}=k\text{v}.\) If \(\text{u}\) and \(\text{v}\) have the same direction, simply choose \(k=\frac{‖\text{u}‖}{‖\text{v}‖}.\) If \(\text{u}\) and \(\text{v}\) have opposite directions, choose \(k=-\frac{‖\text{u}‖}{‖\text{v}‖}.\) Note that the converse holds as well. If \(\text{u}=k\text{v}\) for some scalar \(\text{k},\) then either \(\text{u}\) and \(\text{v}\) have the same direction \((k>0)\) or opposite directions \((k<0),\) so \(\text{u}\) and \(\text{v}\) are parallel. Therefore, two nonzero vectors \(\text{u}\) and \(\text{v}\) are parallel if and only if \(\text{u}=k\text{v}\) for some scalar \(\text{k}.\) By convention, the zero vector \(0\) is considered to be parallel to all vectors.

As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (). Let \(L\) be a line in space passing through point \(P({x}_{0},{y}_{0},{z}_{0}).\) Let \(\text{v}=〈a,b,c〉\) be a vector parallel to \(L.\) Then, for any point \(Q(x,y,z)\) on line \(L\), we know that \(\overset{\to}{PQ}\) is parallel to \(\text{v}.\) Thus, as we just discussed, there is a scalar, \(t,\) such that \(\overset{\to}{PQ}=t\text{v},\) which gives

\[\begin{array}{lll}\overset{\to}{PQ} & = & t\text{v} \\ 〈x-{x}_{0},y-{y}_{0},z-{z}_{0}〉 & = & t〈a,b,c〉 \\ 〈x-{x}_{0},y-{y}_{0},z-{z}_{0}〉 & = & 〈ta,tb,tc〉.\end{array}\]

Using vector operations, we can rewrite as

\[\begin{array}{lll}〈x-{x}_{0},y-{y}_{0},z-{z}_{0}〉 & = & 〈ta,tb,tc〉 \\ 〈x,y,z〉-〈{x}_{0},{y}_{0},{z}_{0}〉 & = & t〈a,b,c〉 \\ 〈x,y,z〉 & = & 〈{x}_{0},{y}_{0},{z}_{0}〉+t〈a,b,c〉.\end{array}\]

Setting \(\text{r}=〈x,y,z〉\) and \({\text{r}}_{0}=〈{x}_{0},{y}_{0},{z}_{0}〉,\) we now have the vector equation of a line:

\[\text{r}={\text{r}}_{0}+t\text{v}.\]

Equating components, shows that the following equations are simultaneously true: \(x-{x}_{0}=ta,\) \(y-{y}_{0}=tb,\) and \(z-{z}_{0}=tc.\) If we solve each of these equations for the component variables \(x,y,\ \text{and}\ z,\) we get a set of equations in which each variable is defined in terms of the parameter t and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

\[x={x}_{0}+ta\ y={y}_{0}+tb\ z={z}_{0}+tc.\]

If we solve each of the equations for \(t\) assuming \(a,b,\ \text{and}\ c\) are nonzero, we get a different description 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 same value. We can set them equal to each other to create symmetric equations of a line:

\[\frac{x-{x}_{0}}{a}=\frac{y-{y}_{0}}{b}=\frac{z-{z}_{0}}{c}.\]\[\text{r}=\text{p}+t(\overset{\to}{PQ}).\]\[\text{r}=(1-t)\text{p}+t\text{q}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Distance between a Point and a Line

We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider “setback” requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.

Let \(L\) be a line in the plane and let \(M\) be any point not on the line. Then, we define distance \(d\) from \(M\) to \(L\) as the length of line segment \(\overset{—}{MP},\) where \(P\) is a point on \(L\) such that \(\overset{—}{MP}\) is perpendicular to \(L\) ().

When we’re looking for the distance between a line and a point in space, still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so 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 let \(\text{v}\) be a direction vector for \(L\) ().

By , vectors \(\overset{\to}{PM}\) and \(\text{v}\) form two sides of a parallelogram with area \(‖\overset{\to}{PM}\ \times \ \text{v}‖.\) Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

\[‖\overset{\to}{PM}\ \times \ \text{v}‖=‖\text{v}‖d.\]

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Condensed — the full section is in OpenStax Calculus Volume 3.

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 single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines ().

To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point ().

Example

Try it.

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

  1. \({L}_{1}:x=2s-1,y=s-1,z=s-4\)
    \({L}_{2}:x=t-3,y=3t+8,z=5-2t\)
  2. \({L}_{1}\text{:}\) \(x=\text{-}y=z\)
    \({L}_{2}:\frac{x-3}{2}=y=z-2\)
  3. \({L}_{1}:x=6s-1,y=-2s,z=3s+1;s\ne 0\)
    \({L}_{2}:\frac{x-4}{6}=\frac{y+3}{-2}=\frac{z-1}{3}\)
Solution
  1. Line \({L}_{1}\) has direction vector \({\text{v}}_{1}=〈2,1,1〉;\) line \({L}_{2}\) has direction vector \({\text{v}}_{2}=〈1,3,-2〉.\) Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, \((x,y,z),\) that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
    \[2s-1=t-3;\ s-1=3t+8;\ s-4=5-2t.\]
    By the first equation, \(t=2s+2.\) Substituting into the second equation yields
    \[\begin{array}{lll}s-1 & = & 3(2s+2)+8 \\ s-1 & = & 6s+6+8 \\ 5s & = & -15 \\ s & = & -3.\end{array}\]
    Substitution into the third equation, however, yields a contradiction:
    \[\begin{array}{lll}s-4 & = & 5-2(2s+2) \\ s-4 & = & 5-4s-4 \\ 5s & = & 5 \\ s & = & 1.\end{array}\]
    There is no single point that satisfies the parametric equations for \({L}_{1}\ \text{and}\ {L}_{2}\) simultaneously. These lines do not intersect, so they are skew (see the following figure).
  2. Line L1 has direction vector \({\text{v}}_{1}=〈1,-1,1〉\) and passes through the origin, \((0,0,0).\) Line \({L}_{2}\) has a different direction vector, \({\text{v}}_{2}=〈2,1,1〉,\) so these lines are not parallel or equal. Let \(r\) represent the parameter for line \({L}_{1}\) and let \(s\) represent the parameter for \({L}_{2}\text{:}\)
    \[\begin{array}{llllllllll}\begin{array}{ll}x & =r \\ y & =\text{-}r \\ z & =r\end{array} & & & \begin{array}{ll}x & =2s+3 \\ y & =s \\ z & =s+2.\end{array}\end{array}\]
    Solve the system of equations to find \(r=1\) and \(s=-1.\) If we need to find the point of intersection, we can substitute these parameters into the original equations to get \((1,-1,1)\) (see the following figure).
  3. Lines \({L}_{1}\) and \({L}_{2}\) have equivalent direction vectors: \(\text{v}=〈6,-2,3〉.\) These two lines are parallel (see the following figure).

Condensed — the full section is in OpenStax Calculus Volume 3.

Equations for a Plane

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

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let \(\text{n}=〈a,b,c〉\) be a vector and \(P=({x}_{0},{y}_{0},{z}_{0})\) be a point. Then the set of all points \(Q=(x,y,z)\) such that \(\overset{\to}{PQ}\) is orthogonal to \(\text{n}\) forms a plane (). We say that \(\text{n}\) is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane: \(\text{n}\cdot \overset{\to}{PQ}=0.\) Rewriting this equation provides additional ways to describe the plane:

\[\begin{array}{lll} \\ \text{n}\cdot \overset{\to}{PQ} & = & 0 \\ 〈a,b,c〉\cdot 〈x-{x}_{0},y-{y}_{0},z-{z}_{0}〉 & = & 0 \\ a(x-{x}_{0})+b(y-{y}_{0})+c(z-{z}_{0}) & = & 0.\end{array}\]

As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.

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

Condensed — the full section is in OpenStax Calculus Volume 3.

Parallel and Intersecting Planes

We have discussed the various 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 two 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 ().

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

Example

Try it.

Find parametric and symmetric equations for the line formed by the intersection of the planes given by \(x+y+z=0\) and \(2x-y+z=0\) (see the following figure).

Solution

Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, \(y\text{:}\)

\[\begin{array}{lllllllllllll}\underset{\text{____________________}}{\begin{array}{lllllll}x & + & y & + & z & = & 0 \\ 2x & - & y & + & z & = & 0\end{array}} \\ \\ 3x\ +2z\ =\ 0.\end{array}\]

This gives us \(x=-\frac{2}{3}z.\) We substitute this value into the first equation to express \(y\) in terms of \(z\text{:}\)

\[\begin{array}{lll}x+y+z & = & 0 \\ -\frac{2}{3}z+y+z & = & 0 \\ y+\frac{1}{3}z & = & 0 \\ y & = & -\frac{1}{3}z.\end{array}\]

We now have the first two variables, \(x\) and \(y,\) in terms of the third variable, \(z.\) Now we define \(z\) in terms of \(t.\) To eliminate the need for fractions, we choose to define the parameter \(t\) as \(t=-\frac{1}{3}z.\) Then, \(z=-3t.\) Substituting the parametric representation of \(z\) back into the other two equations, we see that the parametric equations for the line of intersection are \(x=2t,y=t,z=-3t.\) The symmetric equations for the line are \(\frac{x}{2}=y=\frac{z}{-3}.\)

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 two planes. For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes. shows why this is true.

\[\text{cos}\ \theta =\frac{|{\text{n}}_{1}\cdot {\text{n}}_{2}|}{‖{\text{n}}_{1}‖‖{\text{n}}_{2}‖}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • In three dimensions, the direction of a line is described by a direction vector. The vector equation of a line with direction vector \(\text{v}=〈a,b,c〉\) passing through point \(P=({x}_{0},{y}_{0},{z}_{0})\) is \(\text{r}={\text{r}}_{0}+t\text{v},\) where \({\text{r}}_{0}=〈{x}_{0},{y}_{0},{z}_{0}〉\) is the position vector of point \(P.\) This equation can be rewritten to form the parametric equations of the line: \(x={x}_{0}+ta,\) \(y={y}_{0}+tb,\) and \(z={z}_{0}+tc.\) The line can also be described with the symmetric equations \(\frac{x-{x}_{0}}{a}=\frac{y-{y}_{0}}{b}=\frac{z-{z}_{0}}{c}.\)
  • Let \(L\) be a line in space passing through point \(P\) with direction vector \(\text{v}.\) If \(Q\) is any point not on \(L,\) then the distance from \(Q\) to \(L\) is \(d=\frac{‖\overset{\to}{PQ}\ \times \ \text{v}‖}{‖\text{v}‖}.\)
  • In three dimensions, two lines may be parallel but not equal, equal, intersecting, or skew.
  • Given a point \(P\) and vector \(\text{n},\) the set of all points \(Q\) satisfying equation \(\text{n}\cdot \overset{\to}{PQ}=0\) forms a plane. Equation \(\text{n}\cdot \overset{\to}{PQ}=0\) is known as the vector equation of a plane.
  • The scalar equation of a plane containing point \(P=({x}_{0},{y}_{0},{z}_{0})\) with normal vector \(\text{n}=〈a,b,c〉\) is \(a(x-{x}_{0})+b(y-{y}_{0})+c(z-{z}_{0})=0.\) This equation can be expressed as \(ax+by+cz+d=0,\) where \(d=\text{-}a{x}_{0}-b{y}_{0}-c{z}_{0}.\) This form of the equation is sometimes called the general form of the equation of a plane.
  • Suppose a plane with normal vector n passes through point \(Q.\) The distance \(D\) from the plane to point \(P\) not in the plane is given by
    \[D=‖{\text{proj}}_{\text{n}}\overset{\to}{QP}‖=|{\text{comp}}_{\text{n}}\overset{\to}{QP}|=\frac{|\overset{\to}{QP}\cdot \text{n}|}{‖\text{n}‖}.\]
  • The normal vectors of parallel planes are parallel. When two planes intersect, they form a line.
  • The measure of the angle \(\theta\) between two intersecting planes can be found using the equation: \(\text{cos}\ \theta =\frac{|{\text{n}}_{1}\cdot {\text{n}}_{2}|}{‖{\text{n}}_{1}‖‖{\text{n}}_{2}‖},\) where \({\text{n}}_{1}\) and \({\text{n}}_{2}\) are normal vectors to the planes.
  • The distance \(D\) from point \(({x}_{0},{y}_{0},{z}_{0})\) to plane \(ax+by+cz+d=0\) is given by
    \[D=\frac{|a({x}_{0}-{x}_{1})+b({y}_{0}-{y}_{1})+c({z}_{0}-{z}_{1})|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}=\frac{|a{x}_{0}+b{y}_{0}+c{z}_{0}+d|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}.\]

Key Equations

Vector Equation of a Line\(\text{r}={\text{r}}_{0}+t\text{v}\)
Parametric Equations of a Line\(\frac{x-{x}_{0}}{a}=\frac{y-{y}_{0}}{b}=\frac{z-{z}_{0}}{c}\)
Vector Equation of a Plane\(\text{n}\cdot \overset{\to}{PQ}=0\)
Scalar Equation of a Plane\(a(x-{x}_{0})+b(y-{y}_{0})+c(z-{z}_{0})=0\)
Distance between a Plane and a Point\(d=‖{\text{proj}}_{\text{n}}\overset{\to}{QP}‖=|{\text{comp}}_{\text{n}}\overset{\to}{QP}|=\frac{|\overset{\to}{QP}\cdot \text{n}|}{‖\text{n}‖}\)

Equations of Lines and Planes in Space

In the following exercises, points \(P\) and \(Q\) are given. Let \(L\) be the line passing through points \(P\) and \(Q.\)

  1. Find the vector equation of line \(L.\)
  2. Find parametric equations of line \(L.\)
  3. Find symmetric equations of line \(L.\)
  4. Find parametric equations of the line segment determined by \(P\) and \(Q.\)

For the following exercises, point \(P\) and vector \(\text{v}\) are given. Let \(L\) 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.

For the following exercises, line \(L\) is given.

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

For the following exercises, lines \({L}_{1}\) and \({L}_{2}\) are given.

  1. Verify whether lines \({L}_{1}\) and \({L}_{2}\) are parallel.
  2. If the lines \({L}_{1}\) and \({L}_{2}\) are parallel, then find the distance between them.


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.

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

  1. Find the scalar equation of the plane that passes through \(P\) and has normal vector \(\text{n}.\)
  2. Find the general form of the equation of the plane that passes through \(P\) and has normal vector \(\text{n}.\)
  1. Find normal vector \(\text{n}\) to the plane. Express \(\text{n}\) using standard unit vectors.
  2. Find the intersections of the plane with the coordinate axes.
  3. Sketch the plane.
  1. Find the general equation of the plane passing through \(P,Q,\ \text{and}\ R.\)
  2. Write the vector equation \(\text{n}\cdot \overset{\to}{PS}=0\) of the plane at a., where \(S(x,y,z)\) is an arbitrary point of the plane.
  3. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through \(P,Q,\ \text{and}\ R.\)
  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.

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Find parametric and symmetric equations of the line passing through points \((1,4,-2)\) and \((-3,5,0).\)

    كشفت الإجابة

    First, identify a vector parallel to the line:

    \[\text{v}=〈-3-1,5-4,0-(-2)〉=〈-4,1,2〉.\]

    Use either of the given points on the line to complete the parametric equations:

    \[x=1-4t,y=4+t,\ \text{and}\ z=-2+2t.\]

    Solve each equation for \(t\) to create the symmetric equation of the line:

    \[\frac{x-1}{-4}=y-4=\frac{z+2}{2}.\]

  2. Find parametric and symmetric equations of the line passing through points \((1,-3,2)\) and \((5,-2,8).\)

    كشفت الإجابة

    Possible set of parametric equations: \(x=1+4t,y=-3+t,z=2+6t;\)

    related set of symmetric equations: \(\frac{x-1}{4}=y+3=\frac{z-2}{6}\)

  3. Find parametric equations of the line segment between the points \(P(2,1,4)\) and \(Q(3,-1,3).\)

    كشفت الإجابة

    By , we have

    \[x={x}_{0}+t({x}_{1}-{x}_{0}),y={y}_{0}+t({y}_{1}-{y}_{0}),z={z}_{0}+t({z}_{1}-{z}_{0}),0\le t\le 1.\]

    Working with each component separately, we get

    \[\begin{array}{ll}x & ={x}_{0}+t({x}_{1}-{x}_{0}) \\ & =2+t(3-2) \\ & =2+t,\end{array}\]\[\begin{array}{ll}y & ={y}_{0}+t({y}_{1}-{y}_{0}) \\ & =1+t(-1-1) \\ & =1-2t,\end{array}\]

    and

    \[\begin{array}{ll}z & ={z}_{0}+t({z}_{1}-{z}_{0}) \\ & =4+t(3-4) \\ & =4-t.\end{array}\]

    Therefore, the parametric equations for the line segment are

    \[x=2+t,y=1-2t,z=4-t,0\le t\le 1.\]

  4. Find parametric equations of the line segment between points \(P(-1,3,6)\) and \(Q(-8,2,4).\)

    كشفت الإجابة

    \(x=-1-7t,y=3-t,z=6-2t,0\le t\le 1\)

  5. Find the distance between the point \(M=(1,1,3)\) and line \(\frac{x-3}{4}=\frac{y+1}{2}=z-3.\)

    كشفت الإجابة

    From the symmetric equations of the line, we know that vector \(\text{v}=〈4,2,1〉\) is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point \(P(3,-1,3)\) lies on the line. Then,

    \[\overset{\to}{PM}=〈1-3,1-(-1),3-3〉=〈-2,2,0〉.\]

    To calculate the distance, we need to find \(\overset{\to}{PM}\ \times \ \text{v}\text{:}\)

    \[\begin{array}{llllllll}\overset{\to}{PM}\ \times \ \text{v} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -2 & 2 & 0 \\ 4 & 2 & 1\end{array}| \\ & =(2-0)\text{i}-(-2-0)\text{j}+(-4-8)\text{k} \\ & =2\text{i}+2\text{j}-12\text{k}.\end{array}\]

    Therefore, the distance between the point and the line is ()

    \[\begin{array}{ll}d & =\frac{‖\overset{\to}{PM}\ \times \ \text{v}‖}{‖\text{v}‖} \\ & =\frac{\sqrt{{2}^{2}+{2}^{2}+{12}^{2}}}{\sqrt{{4}^{2}+{2}^{2}+{1}^{2}}} \\ & =\frac{2\sqrt{38}}{\sqrt{21}}.\end{array}\]

  6. Find the distance between point \((0,3,6)\) and the line with parametric equations \(x=1-t,y=1+2t,z=5+3t.\)

    كشفت الإجابة

    \(\sqrt{\frac{10}{7}}\)

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

    1. \({L}_{1}:x=2s-1,y=s-1,z=s-4\)
      \({L}_{2}:x=t-3,y=3t+8,z=5-2t\)
    2. \({L}_{1}\text{:}\) \(x=\text{-}y=z\)
      \({L}_{2}:\frac{x-3}{2}=y=z-2\)
    3. \({L}_{1}:x=6s-1,y=-2s,z=3s+1;s\ne 0\)
      \({L}_{2}:\frac{x-4}{6}=\frac{y+3}{-2}=\frac{z-1}{3}\)
    كشفت الإجابة
    1. Line \({L}_{1}\) has direction vector \({\text{v}}_{1}=〈2,1,1〉;\) line \({L}_{2}\) has direction vector \({\text{v}}_{2}=〈1,3,-2〉.\) Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, \((x,y,z),\) that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
      \[2s-1=t-3;\ s-1=3t+8;\ s-4=5-2t.\]
      By the first equation, \(t=2s+2.\) Substituting into the second equation yields
      \[\begin{array}{lll}s-1 & = & 3(2s+2)+8 \\ s-1 & = & 6s+6+8 \\ 5s & = & -15 \\ s & = & -3.\end{array}\]
      Substitution into the third equation, however, yields a contradiction:
      \[\begin{array}{lll}s-4 & = & 5-2(2s+2) \\ s-4 & = & 5-4s-4 \\ 5s & = & 5 \\ s & = & 1.\end{array}\]
      There is no single point that satisfies the parametric equations for \({L}_{1}\ \text{and}\ {L}_{2}\) simultaneously. These lines do not intersect, so they are skew (see the following figure).
    2. Line L1 has direction vector \({\text{v}}_{1}=〈1,-1,1〉\) and passes through the origin, \((0,0,0).\) Line \({L}_{2}\) has a different direction vector, \({\text{v}}_{2}=〈2,1,1〉,\) so these lines are not parallel or equal. Let \(r\) represent the parameter for line \({L}_{1}\) and let \(s\) represent the parameter for \({L}_{2}\text{:}\)
      \[\begin{array}{llllllllll}\begin{array}{ll}x & =r \\ y & =\text{-}r \\ z & =r\end{array} & & & \begin{array}{ll}x & =2s+3 \\ y & =s \\ z & =s+2.\end{array}\end{array}\]
      Solve the system of equations to find \(r=1\) and \(s=-1.\) If we need to find the point of intersection, we can substitute these parameters into the original equations to get \((1,-1,1)\) (see the following figure).
    3. Lines \({L}_{1}\) and \({L}_{2}\) have equivalent direction vectors: \(\text{v}=〈6,-2,3〉.\) These two lines are parallel (see the following figure).

  8. Describe the relationship between the lines with the following parametric equations:

    \[x=1-4t,y=3+t,z=8-6t\]\[x=2+3s,y=2s,z=-1-3s.\]
    كشفت الإجابة

    These lines are skew because their direction vectors are not parallel and there is no point \((x,y,z)\) that lies on both lines.

  9. Write an equation for the plane containing points \(P=(1,1,-2),\) \(Q=(0,2,1),\) and \(R=(-1,-1,0)\) in both standard and general forms.

    كشفت الإجابة

    To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane:

    \[\begin{array}{lll}\overset{\to}{PQ} & = & 〈0-1,2-1,1-(-2)〉=〈-1,1,3〉 \\ \overset{\to}{QR} & = & 〈-1-0,-1-2,0-1〉=〈-1,-3,-1〉.\end{array}\]

    The cross product \(\overset{\to}{PQ}\ \times \ \overset{\to}{QR}\) is orthogonal to both \(\overset{\to}{PQ}\) and \(\overset{\to}{QR},\) so it is normal to the plane that contains these two vectors:

    \[\begin{array}{llllllll}\text{n} & =\overset{\to}{PQ}\ \times \ \overset{\to}{QR} \\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -1 & 1 & 3 \\ -1 & -3 & -1\end{array}| \\ & =(-1+9)\text{i}-(1+3)\text{j}+(3+1)\text{k} \\ & =8\text{i}-4\text{j}+4\text{k}.\end{array}\]

    Thus, \(\text{n}=〈8,-4,4〉,\) and we can choose any of the three given points to write an equation of the plane:

    \[\begin{array}{lll}8(x-1)-4(y-1)+4(z+2) & = & 0 \\ 8x-4y+4z+4 & = & 0.\end{array}\]

  10. Find an equation of the plane that passes through point \((1,4,3)\) and contains the line given by \(x=\frac{y-1}{2}=z+1.\)

    كشفت الإجابة

    Symmetric equations describe the line that passes through point \((0,1,\text{-}1)\) parallel to vector \({\text{v}}_{1}=〈1,2,1〉\) (see the following figure). Use this point and the given point, \((1,4,3),\) to identify a second vector parallel to the plane:

    \[{\text{v}}_{2}=〈1-0,4-1,3-(-1)〉=〈1,3,4〉.\]

    Use the cross product of these vectors to identify a normal vector for the plane:

    \[\begin{array}{llllllll}\text{n} & ={\text{v}}_{1}\ \times \ {\text{v}}_{2} \\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ 1 & 2 & 1 \\ 1 & 3 & 4\end{array}| \\ & =(8-3)\text{i}-(4-1)\text{j}+(3-2)\text{k} \\ & =5\text{i}-3\text{j}+\text{k}.\end{array}\]

    The scalar equations for the plane are \(5x-3(y-1)+(z+1)=0\) and \(5x-3y+z+4=0.\)

  11. Find an equation of the plane containing the lines \({L}_{1}\) and \({L}_{2}\text{:}\)

    \[\begin{array}{l}{L}_{1}:x=\text{-}y=z \\ {L}_{2}:\frac{x-3}{2}=y=z-2.\end{array}\]
    كشفت الإجابة

    \(-2(x-1)+(y+1)+3(z-1)=0\) or \(-2x+y+3z=0\)

  12. Find the distance between point \(P=(3,1,2)\) and the plane given by \(x-2y+z=5\) (see the following figure).

    كشفت الإجابة

    The coefficients of the plane’s equation provide a normal vector for the plane: \(\text{n}=〈1,-2,1〉.\) To find vector \(\overset{\to}{QP},\) we need a point in the plane. Any point will work, so set \(y=z=0\) to see that point \(Q=(5,0,0)\) lies in the plane. Find the component form of the vector from \(Q\ \text{to}\ P\text{:}\)

    \[\overset{\to}{QP}=〈3-5,1-0,2-0〉=〈-2,1,2〉.\]

    Apply the distance formula from :

    \[\begin{array}{ll}d & =\frac{|\overset{\to}{QP}\cdot \text{n}|}{‖\text{n}‖} \\ & =\frac{|〈-2,1,2〉\cdot 〈1,-2,1〉|}{\sqrt{{1}^{2}+{(-2)}^{2}+{1}^{2}}} \\ & =\frac{|-2-2+2|}{\sqrt{6}} \\ & =\frac{2}{\sqrt{6}}.\end{array}\]

  13. Find the distance between point \(P=(5,-1,0)\) and the plane given by \(4x+2y-z=3.\)

    كشفت الإجابة

    \(\frac{15}{\sqrt{21}}\)

  14. Find parametric and symmetric equations for the line formed by the intersection of the planes given by \(x+y+z=0\) and \(2x-y+z=0\) (see the following figure).

    كشفت الإجابة

    Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, \(y\text{:}\)

    \[\begin{array}{lllllllllllll}\underset{\text{____________________}}{\begin{array}{lllllll}x & + & y & + & z & = & 0 \\ 2x & - & y & + & z & = & 0\end{array}} \\ \\ 3x\ +2z\ =\ 0.\end{array}\]

    This gives us \(x=-\frac{2}{3}z.\) We substitute this value into the first equation to express \(y\) in terms of \(z\text{:}\)

    \[\begin{array}{lll}x+y+z & = & 0 \\ -\frac{2}{3}z+y+z & = & 0 \\ y+\frac{1}{3}z & = & 0 \\ y & = & -\frac{1}{3}z.\end{array}\]

    We now have the first two variables, \(x\) and \(y,\) in terms of the third variable, \(z.\) Now we define \(z\) in terms of \(t.\) To eliminate the need for fractions, we choose to define the parameter \(t\) as \(t=-\frac{1}{3}z.\) Then, \(z=-3t.\) Substituting the parametric representation of \(z\) back into the other two equations, we see that the parametric equations for the line of intersection are \(x=2t,y=t,z=-3t.\) The symmetric equations for the line are \(\frac{x}{2}=y=\frac{z}{-3}.\)

  15. Find parametric equations for the line formed by the intersection of planes \(x+y-z=3\) and \(3x-y+3z=5.\)

    كشفت الإجابة

    \(x=t,y=7-3t,z=4-2t\)

  16. Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.

    1. \(x+2y-z=8\ \text{and}\ 2x+4y-2z=10\)
    2. \(2x-3y+2z=3\ \text{and}\ 6x+2y-3z=1\)
    3. \(x+y+z=4\ \text{and}\ x-3y+5z=1\)
    كشفت الإجابة
    1. The normal vectors for these planes are \({\text{n}}_{1}=〈1,2,-1〉\) and \({\text{n}}_{2}=〈2,4,-2〉.\) These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.
    2. The normal vectors for these planes are \({\text{n}}_{1}=〈2,-3,2〉\) and \({\text{n}}_{2}=〈6,2,-3〉.\) Taking the dot product of these vectors, we have
      \[{\text{n}}_{1}\cdot {\text{n}}_{2}=〈2,-3,2〉\cdot 〈6,2,-3〉=2(6)-3(2)+2(-3)=0.\]
      The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.
    3. The normal vectors for these planes are \({\text{n}}_{1}=〈1,1,1〉\) and \({\text{n}}_{2}=〈1,-3,5〉\text{:}\)
      \[\begin{array}{ll}\text{cos}\ \theta & =\frac{|{\text{n}}_{1}\cdot {\text{n}}_{2}|}{‖{\text{n}}_{1}‖‖{\text{n}}_{2}‖} \\ & =\frac{|〈1,1,1〉\cdot 〈1,-3,5〉|}{\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\ \sqrt{{1}^{2}+{(-3)}^{2}+{5}^{2}}} \\ & =\frac{3}{\sqrt{105}}.\end{array}\]
      The angle between the two planes is \(1.27\) rad, or approximately \(73\text{^{\circ}}.\)

  17. Find the measure of the angle between planes \(x+y-z=3\) and \(3x-y+3z=5.\) Give the answer in radians and round to two decimal places.

    كشفت الإجابة

    \(1.44\) rad

  18. Find the distance between the two parallel planes given by \(2x+y-z=2\) and \(2x+y-z=8.\)

    كشفت الإجابة

    Point \((1,0,0)\) lies in the first plane. The desired distance, then, is

    \[\begin{array}{ll}d & =\frac{|a{x}_{0}+b{y}_{0}+c{z}_{0}+k|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}} \\ & =\frac{|2(1)+1(0)+(-1)(0)+(-8)|}{\sqrt{{2}^{2}+{1}^{2}+{(-1)}^{2}}} \\ & =\frac{6}{\sqrt{6}}=\sqrt{6}.\end{array}\]

  19. Find the distance between parallel planes \(5x-2y+z=6\) and \(5x-2y+z=-3.\)

    كشفت الإجابة

    \(\frac{9}{\sqrt{30}}\)

  20. \(P(-3,5,9),\) \(Q(4,-7,2)\)

    كشفت الإجابة

    a. \(\text{r}=〈-3,5,9〉+t〈7,-12,-7〉,\) \(t\in ℝ\text{;}\) b. \(x=-3+7t,y=5-12t,z=9-7t,\) \(t\in ℝ\text{;}\) c. \(\frac{x+3}{7}=\frac{y-5}{-12}=\frac{z-9}{-7};\) d. \(x=-3+7t,y=5-12t,z=9-7t,\) \(t\in [0,1]\)

  21. \(P(4,0,5),Q(2,3,1)\)

  22. \(P(-1,0,5),\) \(Q(4,0,3)\)

    كشفت الإجابة

    a. \(\text{r}=〈-1,0,5〉+t〈5,0,-2〉,\) \(t\in ℝ;\) b. \(x=-1+5t,y=0,z=5-2t,\) \(t\in ℝ;\) c. \(\frac{x+1}{5}=\frac{z-5}{-2},y=0;\) d. \(x=-1+5t,y=0,z=5-2t,\) \(t\in [0,1]\)

  23. \(P(7,-2,6),\) \(Q(-3,0,6)\)

  24. \(P(1,-2,3),\) \(\text{v}=〈1,2,3〉\)

    كشفت الإجابة

    a. \(x=1+t,y=-2+2t,z=3+3t,\) \(t\in ℝ\text{;}\) b. \(\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{3};\) c. \((0,-4,0)\)

  25. \(P(3,1,5),\) \(\text{v}=〈1,1,1〉\)

  26. \(P(3,1,5),\) \(\text{v}=\overset{\to}{QR},\) where \(Q(2,2,3)\) and \(R(3,2,3)\)

    كشفت الإجابة

    a. \(x=3+t,y=1,z=5,\) \(t\in ℝ;\) b. \(y=1,z=5;\) c. The line does not intersect the xy-plane.

  27. \(P(2,3,0),\) \(\text{v}=\overset{\to}{QR},\) where \(Q(0,4,5)\) and \(R(0,4,6)\)

  28. \(x=1+t,y=3+t,z=5+4t,\) \(t\in ℝ\)

    كشفت الإجابة

    a. \(P(1,3,5),\) \(v=〈1,1,4〉;\) b. \(\sqrt{3}\)

  29. \(\text{-}x=y+1,z=2\)

  30. Find the distance between point \(A(-3,1,1)\) and the line of symmetric equations

    \(x=\text{-}y=\text{-}z.\)

    كشفت الإجابة

    \(\frac{2\sqrt{2}}{\sqrt{3}}\)

  31. Find the distance between point \(A(4,2,5)\) and the line of parametric equations

    \(x=-1-t,y=\text{-}t,z=2,\) \(t\in ℝ.\)

  32. \({L}_{1}:x=1+t,y=t,z=2+t,\) \(t\in ℝ,\) \({L}_{2}:x-3=y-1=z-3\)

    كشفت الإجابة

    a. Parallel; b. \(\frac{\sqrt{2}}{\sqrt{3}}\)

  33. \({L}_{1}:x=2,y=1,z=t,\) \({L}_{2}:x=1,y=1,z=2-3t,\) \(t\in ℝ\)

  34. Show that the line passing through points \(P(3,1,0)\) and \(Q(1,4,-3)\) is perpendicular to the line with equations \(x=3+3t,y=1+8t,z=6t,\) \(t\in ℝ.\)

  35. Are the lines of equations \(x=-2+2t,y=-6,z=2+6t\) and \(x=-1+t,y=1+t,z=t,\) \(t\in ℝ,\) perpendicular to each other?

  36. Find the point of intersection of the lines of equations \(x=-2y=3z\) and \(x=-5-t,y=-1+t,z=t-11,\) \(t\in ℝ.\)

    كشفت الإجابة

    \((-12,6,-4)\)

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

    \(x=10+t,y=2-2t,z=-3+3t,\) \(t\in ℝ.\)

  38. \({L}_{1}:x=y-1=\text{-}z\) and \({L}_{2}:x-2=\text{-}y=\frac{z}{2}\)

    كشفت الإجابة

    The lines are skew.

  39. \({L}_{1}:x=2t,y=0,z=3,\) \(t\in ℝ\) and \({L}_{2}:x=0,y=8+s,z=7+s,\) \(s\in ℝ\)

  40. \({L}_{1}:x=-1+2t,y=1+3t,z=7t,\) \(t\in ℝ\) and \({L}_{2}:x-1=\frac{2}{3}(y-4)=\frac{2}{7}z-2\)

    كشفت الإجابة

    The lines are equal.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\theta
theta
The usual name for an angle.
P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
\neq
not equal
The two sides are different.
\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Equations of Lines and Planes in Space

  1. Write the vector, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.
  2. Find the distance from a point to a given line.
  3. Write the vector and scalar equations of a plane through a given point with a given normal.
  4. Find the distance from a point to a given plane.
  5. Find the angle between two planes.
  6. Line
  7. Line
  8. Lines

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

جرّب نفسك

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

أكثر في Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems