Lesson Explainer: Parallel, Perpendicular, and Intersecting Lines in Space Mathematics

In this explainer, we will learn how to find the equation of a straight line that is parallel or perpendicular to another one in space and find the intersection point between two lines.

We know that a nonzero vector parallel to the line and a point on the line completely characterizes a line in three dimensions. Naturally, the components of the direction vector and the coordinates of the point explicitly appear in the equation of a line in various forms. Let us recall different forms for the equation of the line.

Definition: Equation of a Line in Three Dimensions

If a line passes the point (𝑥,𝑦,𝑧) and is parallel to the nonzero vector (𝑎,𝑏,𝑐), then its equation

  • in the vector form is 𝑟(𝑡)=(𝑥,𝑦,𝑧)+𝑡(𝑎,𝑏,𝑐), where (𝑥,𝑦,𝑧) is the position vector of the point that belongs to the line;
  • in the Cartesian form is 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐,𝑎0,𝑏0,𝑐0;giventhatand
  • in the parametric form is 𝑥=𝑥+𝑎𝑡,𝑦=𝑦+𝑏𝑡,𝑧=𝑧+𝑐𝑡.

Let us consider an example where we find the vector form of the equation of a line, given a point on the line and a parallel vector.

Example 1: Determining the Vector Form of the Equation of a Straight Line

Determine the vector form of the equation of the straight line passing through the point (1,5,4) and parallel to the vector (3,5,1).

Answer

We recall that the vector form of the equation of the line through the point 𝑃 parallel to the nonzero vector 𝑑 is given by 𝑟(𝑡)=𝑂𝑃+𝑡𝑑, where 𝑂𝑃 is the position vector of point 𝑃(1,5,4) and 𝑂 is the origin point (0,0,0). We are given that our line passes through the point 𝑃(1,5,4), so we get 𝑂𝑃=(1,5,4).

We are also given that our line is parallel to the vector (3,5,1). So, the vector form of the equation of this line is 𝑟=(1,5,4)+𝑡(3,5,1).

A nonzero vector that is parallel to a line is called a direction vector of the line. We know that the choice of direction vectors in the equation of a line is not unique and that any nonzero parallel vectors can replace a given direction vector.

Even though the equation of the line would differ when a different direction vector is used, the new equation would still represent the same line.

We define parallel lines below.

Definition: Parallel Lines in Three Dimensions

Two lines are parallel if their direction vectors are parallel.

In our next example, we will find the vector form of the equation of a line which is parallel to a given line.

Example 2: Finding the Equation of a Straight Line Passing through a Given Point and Parallel to a Given Straight Line in Vector Form

Find the vector form of the equation of the straight line passing through the point 𝐴(2,5,5) and parallel to the straight line passing through the two points 𝐵(3,2,6) and 𝐶(5,0,9).

Answer

We recall that the vector form of the equation of the line through the point 𝐴 parallel to the vector 𝑑 is given by 𝑟(𝑡)=𝑂𝐴+𝑡𝑑, where 𝑂𝐴 is the position vector of point 𝐴 and point 𝑂 is the origin.

We are given that our line passes through the point 𝐴(2,5,5), which gives us 𝑂𝐴=(2,5,5). So, it remains to find the direction vector 𝑑 to complete the equation of our line.

We recall that two lines are parallel if they share a direction vector. Since the two lines are parallel, they share the direction vector. So, we can use the direction vector of the given line as 𝑑 in the vector equation of our line.

The given line goes through 𝐵(3,2,6) and 𝐶(5,0,9), so it is parallel to the vector 𝐵𝐶=𝑂𝐶𝑂𝐵=(5,0,9)(3,2,6)=(5(3),0(2),9(6))=(8,2,3).

This gives us 𝑑=(8,2,3).

Hence, the equation of our line in the vector form is 𝑟=(2,5,5)+𝑡(8,2,3).

If two lines intersect in three dimensions, then they share a common point. One convenient way to check for a common point between two lines is to use the parametric form of the equations of the two lines. In the parametric form, each coordinate of a point is given in terms of the parameter, say 𝑡. If two lines pass through the same point, then there must be a parameter value 𝑡=𝑡 which produces the coordinates of this point in the first line and another value 𝑡=𝑡 producing these coordinates in the second line.

In our next example, we will use this idea to find an unknown constant in the equations of two intersecting lines.

Example 3: Finding Unknown Coefficients in the Equation of Two Straight Lines in Three Dimensions given That They Intersect

For what value of 𝑎 do the lines 𝑥5=𝑦21=𝑧2 and 𝑥1𝑎=𝑦+24=𝑧+14 intersect?

Answer

We recall that we can convert the equation of a line in the Cartesian form to the parametric form by setting each component of the equation equal to 𝑡 and solving for the variables 𝑥, 𝑦, and 𝑧. From the equation of the first line, we get 𝑥5=𝑡,𝑥=5𝑡;𝑦21=𝑡,𝑦=2𝑡;𝑧2=𝑡,𝑧=2𝑡.whichleadstowhichleadstowhichleadsto

So, the equation of the first line in the parametric form is

𝑥=5𝑡,𝑦=2𝑡,𝑧=2𝑡.(1)

From the equation of the second line, we have 𝑥1𝑎=𝑡,𝑥=1+𝑎𝑡;𝑦+24=𝑡,𝑦=2+4𝑡;𝑧+14=𝑡,𝑧=1+4𝑡.whichleadstowhichleadstowhichleadsto

Hence, the equation of the second line in the parametric form is

𝑥=1+𝑎𝑡,𝑦=2+4𝑡,𝑧=1+4𝑡.(2)

If the two lines share a common point (𝑥,𝑦,𝑧), then there must be parameter values 𝑡 and 𝑡 that correspond to these coordinates in both lines. In other words, we substitute 𝑡=𝑡 into the parametric equations (1) and 𝑡=𝑡 into the parametric equations (2) and then equate the corresponding variables from either set of equations. This gives us the following system of equations: 5𝑡=1+𝑎𝑡,2𝑡=2+4𝑡,2𝑡=1+4𝑡..

Since the first equation involves an unknown constant, we will use the second and the third equations to identify 𝑡 and 𝑡. Then, we can substitute these values into the first equation to find the unknown constant 𝑎.

We can subtract the third equation from the second equation to get 2𝑡=2+4𝑡2𝑡=1+4𝑡2+𝑡=1

Solving this equation for 𝑡, we get 𝑡=3. We can substitute this into 2𝑡=1+4𝑡2×(3)=1+4𝑡.toget

Solving this equation for 𝑡, we obtain 𝑡=74. This gives us the parameter values 𝑡=3 and 𝑡=74. Substituting these values into 5𝑡=1+𝑎𝑡5×(3)=1+𝑎74.leadsto

Solving this equation for 𝑎 gives us 𝑎=47×14=8.

Hence, the two lines will intersect when 𝑎=8.

In our previous example, we identified an unknown constant in the equations of two intersecting lines. Let us consider an example where we find the coordinates of the point of the intersection between two lines and use this to find the Cartesian form of the equation of a line.

Example 4: Finding the Equation of a Line in the Cartesian Form Using the Intersection Point of Two Other Lines

Find the Cartesian form of the equation of the straight line passing through the origin and the intersection point of the two straight lines 𝐿𝑟=(1,1,2)+𝑡(1,4,3) and 𝐿𝑥=3,𝑦54=𝑧31.

Answer

To find the Cartesian form for the equation of a line, we need its direction vector and a point on the line. Since we are given that our line passes through the origin and the point of intersection, we can define the direction vector of our line to be the vector from the origin ending at the point of intersection. So, we begin by identifying the point of intersection.

To find the point of intersection between the two lines 𝐿 and 𝐿, we will use the parametric form of the equations of the lines. Let us begin by deriving the parametric form of the equations of both lines.

From the given vector form of the equation of 𝐿, we have 𝑟=(1,1,2)+𝑡(1,4,3)=(1+𝑡,1+4𝑡,2+3𝑡).

Then, the equation of 𝐿 in the parametric form is 𝐿𝑥=1+𝑡,𝑦=1+4𝑡,𝑧=2+3𝑡.

Next, let us convert the equation of 𝐿 to the parametric form. From the given Cartesian form, we note that 𝑥=3. For the second equation, we can set both sides equal to 𝑡 and solve for the variables 𝑦 and 𝑧. 𝑦54=𝑡𝑦=54𝑡,𝑧31=𝑡𝑧=3𝑡.leadstoleadsto

Then, the equation of 𝐿 in the parametric form is 𝐿𝑥=3,𝑦=54𝑡,𝑧=3𝑡.

Say the lines 𝐿 and 𝐿 intersect at the point (𝑥,𝑦,𝑧). Then, this point must be given by some parameter value 𝑡=𝑡 for 𝐿 and some other parameter value 𝑡=𝑡 for 𝐿. So, we can substitute these values into the parametric form of the equation of either line and set each corresponding coordinate equal to each other. This gives us 1+𝑡=3,1+4𝑡=54𝑡,2+3𝑡=3𝑡.

We can rearrange the first equation to get 𝑡=2. We can substitute this value into the second equation and rearrange: 1+4×2=54𝑡,𝑡=1.whichleadsto

Hence, we have obtained 𝑡=2 and 𝑡=1. Let us see whether these parameter values produce the same set of coordinates in 𝐿 and 𝐿. Substituting 𝑡=2 into the parametric equations of 𝐿, we get 𝐿𝑥=1+2=3,𝑦=1+4×2=9,𝑧=2+3×2=4.

So, this produces the coordinates (3,9,4) in 𝐿.

Next, we substitute 𝑡=1 into the parametric equations of 𝐿: 𝐿𝑥=3,𝑦=54×(1)=9,𝑧=3(1)=4.

This leads to the same coordinates (3,9,4) for the line 𝐿. Thus, the lines 𝐿 and 𝐿 intersect at the point 𝑃(3,9,4). Since our line also passes through the origin, our line has the direction vector 𝑂𝑃=(3,9,4).

Using the coordinates (0,0,0) of the origin, we obtain the Cartesian form of the equation of our line: 𝑥03=𝑦09=𝑧04.

Hence, the Cartesian form of the equation of the line which passes through the origin and the intersection point of the lines 𝐿 and 𝐿 is 𝑥3=𝑦9=𝑧4.

If two lines intersect, we can check whether they are perpendicular by using the direction vectors of either lines. We define perpendicular lines below.

Definition: Perpendicular Lines in Three Dimensions

Two lines are perpendicular if they intersect at a point and their direction vectors are perpendicular.

We recall that two vectors are perpendicular if the angle between them is 90, which is the same as saying that their dot product is equal to zero. Thus, two intersecting lines are perpendicular if the dot product of their direction vectors is equal to zero.

In our next example, we will determine whether the given two lines are parallel or perpendicular.

Example 5: Determining Whether Two Straight Lines are Parallel or Perpendicular

Consider the two lines 𝑥=4+2𝑡, 𝑦=6+𝑡, 𝑧=22𝑡 and 𝑟=(6,7,0)+𝑡(5,4,7). Determine whether they are parallel or perpendicular.

Answer

We recall that two lines are parallel if their direction vectors are parallel, and also that they are perpendicular if they intersect and their direction vectors are perpendicular. Let us begin with checking whether their direction vectors are parallel or perpendicular. If they are perpendicular, we will also need to check whether they intersect at a point.

From the parametric form of the equation of the first line, we obtain the direction vector (2,1,2). From the vector form of the equation of the second line, we obtain the direction vector (5,4,7).

Recall that two vectors 𝑢 and 𝑣 are parallel if there a scalar 𝑐 satisfying 𝑢=𝑐𝑣.

To check whether the two direction vectors are parallel, we can substitute our direction vectors into 𝑢 and 𝑣 and check for possible values for 𝑐. We get (2,1,2)=𝑐(5,4,7)=(5𝑐,4𝑐,7𝑐).

This leads to the three equations

2=5𝑐,1=4𝑐,2=7𝑐.(3)

The first equation gives 𝑐=25. But substituting this value into the second and the third equations yields 1=4×25=85,2=7×25=145.??

Since both equations are false, we conclude that there is no solution 𝑐 for system (3). So, the direction vectors of the two lines are not parallel, which means that the two lines are not parallel.

Let us check whether the two lines are perpendicular. We recall that two lines are perpendicular if they intersect and their direction vectors are perpendicular. The direction vectors are perpendicular if their dot product is equal to zero. Taking the dot product yields (2,1,2)(5,4,7)=2×5+1×4+(2)×7=0.

Hence, the direction vectors must be perpendicular. So, it remains to show that the lines intersect at a point. To find the intersection, we can write the equations of the lines in the parametric form. The equation of the first line is already given in the parametric form, so let us convert the second equation into the parametric form: 𝑥=6+5𝑡,𝑦=7+4𝑡,𝑧=7𝑡.

If the two lines share a common point, there must be parameter values 𝑡 and 𝑡 satisfying 4+2𝑡=6+5𝑡,6+𝑡=7+4𝑡,22𝑡=7𝑡.

We can use the first two equations to solve this system by elimination. We begin by multiplying the second equation by 2: 12+2𝑡=14+8𝑡.

Then, we subtract the first equation from this one to get 12+2𝑡=14+8𝑡4+2𝑡=6+5𝑡8=8+3𝑡

Solving this equation for 𝑡, we get 𝑡=0. Substituting this into 22𝑡=7𝑡22𝑡=0,gives which leads to 𝑡=1. So, we have obtained 𝑡=1 and 𝑡=0. Let us see whether these parameter values produce a common point in both lines. First, we substitute 𝑡=1 in the parametric form of the equation of the first line: 𝑥=4+2𝑡=4+2×1=6,𝑦=6+𝑡=6+1=7,𝑧=22𝑡=22×1=0.

This leads to the point (6,7,0). Next, we substitute 𝑡=0 into the parametric form of the second line: 𝑥=6+5𝑡=6+5×0=6,𝑦=7+4𝑡=7+4×0=7,𝑧=7𝑡=7×0=0, which leads to the same point (6,7,0). Hence, the two lines share a common point (6,7,0).

Since the direction vectors of the two lines are perpendicular and the two lines intersect at a point, the lines are perpendicular.

In the next example, we will find the line which is perpendicular to a given line which also passes through the origin.

Example 6: Finding the Equation of the Line Passing through the Origin and Intersecting Another Straight Line Orthogonally

Find the equation of the line through the origin that intersects the line 𝑟=(1,2,3)+𝑡(3,5,1) orthogonally.

Answer

We are given that our line passes through the origin, so, if we can identify a direction vector 𝑑 of the line, then the vector form of the equation would be 𝑟=𝑡𝑑.

Since our line must intersect orthogonally with the given line, the direction vector 𝑑 should be perpendicular to the direction vector of the given line, (3,5,1). We recall that two vectors are perpendicular if their dot product is equal to zero. Thus, we must have 𝑑(3,5,1)=0.

Say that our line intersects with the given line at point 𝑃. Then, we see from the figure below that our line has a direction vector 𝑃𝑂 since it is mentioned in the stem that the line passes through the origin.

So, we need to identify the point 𝑃 lying on the given line such that

𝑂𝑃(3,5,1)=0.(4)

Since 𝑃 is a point on the given line, its coordinates must be given by the equation of the line: 𝑟=(1,2,3)+𝑡(3,5,1)=(1+3𝑡,25𝑡,3+𝑡).

This means that there is a parameter value 𝑡 giving the coordinates 𝑃(1+3𝑡,25𝑡,3+𝑡).

Then, the left-hand side of equation (4) is 𝑂𝑃(3,5,1)=(1+3𝑡,25𝑡,3+𝑡)(3,5,1)=3(1+3𝑡)5(25𝑡)+(3+𝑡)=3+9𝑡10+25𝑡+3+𝑡=10+35𝑡.

This expression is equal to zero if the lines intersect orthogonally. Setting the resulting expression equal to zero will give us the parameter value 𝑡 corresponding to point 𝑃 of intersection: 10+35𝑡=0,𝑡=1035=27.whichleadsto

Substituting the parameter value 𝑡=27 into the coordinates of 𝑃: 𝑃(1+3𝑡,25𝑡,3+𝑡)=𝑃1+3×27,25×27,3+27=𝑃77+67,147107,217+27=𝑃17,47,237.

Thus, our line and the given line intersect orthogonally at the point 𝑃17,47,237. This tells us that our line is parallel to the vector 𝑂𝑃=17,47,237.

We can multiply this vector by the scalar 7 to obtain a simpler direction vector parallel to 𝑂𝑃: 𝑑=717,47,237=(1,4,23).

Using this direction vector, the equation of our line in the vector form is 𝑟=𝑡(1,4,23).

So far, we have discussed parallel lines, intersecting lines, and perpendicular lines in three dimensions. We will now discuss a few exceptional classes, which are coincident lines and skew lines in three dimensions.

If two parallel lines intersect at a point, then they share a point and a direction vector; hence, their equations would be identical. In this case, we say that the lines are coincident. On the other hand, if two parallel lines do not overlap entirely, then they do not share any point in common. So, two parallel lines either overlap entirely or do not intersect at all.

Let us consider an example of coincident lines. Consider the following two lines: 𝐿𝑥12=𝑦24=𝑧36,𝐿𝑥21=𝑦2=𝑧63.

The lines 𝐿 and 𝐿 have direction vectors (2,4,6) and (1,2,3) respectively. We note that (2,4,6)=2(1,2,3), which tells us that the direction vectors are parallel. Thus, the two lines 𝐿 and 𝐿 are parallel. In addition, we note that the line 𝐿 passes through the point (1,2,3). If these coordinates satisfy the equation of 𝐿, then we know that this point also lies in 𝐿. Substituting this point into the Cartesian form of the equation of 𝐿, we obtain 121=22=363.

We note that the equation above is true; hence the coordinates of (1,2,3) also satisfy the equation of 𝐿. So, the lines 𝐿 and 𝐿 are parallel and also intersect at point (1,2,3). This tells us that these lines are coincident.

We have discussed that two distinct parallel lines do not share any point. In three dimensions, it is also possible for two nonparallel lines to lack a point of intersection. In this case, we say that these lines are skew. We state the following definition.

Definition: Skew Lines in Three Dimensions

Two lines in three dimensions are skew if they are not parallel and do not intersect.

Below, we see an example of two skew lines.

The red line is coincident with the 𝑥-axis, so it has the direction vector (1,0,0). The blue line passes through (0,1,0) and (0,0,1), so it has the direction vector (0,1,0)(0,0,1)=(0,1,1).

We recall that two vectors are parallel if one is a scalar multiple of the other. If the vector (0,1,1) were a scalar multiple of (1,0,0), the scalar would have to be zero when considering their 𝑥-components. However, this would lead to (0,1,2)=0×(1,0,0),? which is clearly false. So, (0,1,1) is not a scalar multiple of (1,0,0), and the two direction vectors, and hence the two lines, are not parallel. It is clear from the figure that the blue line never intersects the 𝑥-axis. Therefore, these two lines do not intersect and are not parallel. So, the two lines are skew.

In our final example, we will determine whether two lines represented by their equations are parallel, coincident, intersecting, or skew.

Example 7: Determining Whether Two Straight Lines are Parallel, Coincident, Intersecting, or Skew

Consider the two lines 𝐿𝑟=(3,4,1)+𝑡(1,1,1) and 𝐿𝑟=(2,0,5)+𝑡(2,3,0). Choose the correct statement about the two lines.

  1. 𝐿 and 𝐿 are parallel but not coincident.
  2. 𝐿 and 𝐿 are coincident lines.
  3. 𝐿 and 𝐿 are intersecting at a point.
  4. 𝐿 and 𝐿 are skew lines.

Answer

We recall that two lines are parallel if their direction vectors are parallel. From the vector form of the equations of 𝐿 and 𝐿, we obtain the direction vectors (1,1,1) for 𝐿 and (2,3,0) for 𝐿.

Recall that two vectors 𝑢 and 𝑣 are parallel if there is a scalar 𝑐 satisfying 𝑢=𝑐𝑣.

Substituting the direction vectors of 𝐿 and 𝐿 in place of 𝑢 and 𝑣, we get (1,1,1)=𝑐(2,3,0)=(2𝑐,3𝑐,0).

We note that the 𝑧-component of the vector on the rightmost side of the equation is zero, regardless of the value 𝑐. Since the corresponding component of the vector on the leftmost side is nonzero, this equation does not have any solution. This means that (1,1,1) is not a scalar multiple of (2,3,0). So, the lines 𝐿 and 𝐿 are not parallel.

Next, let us examine whether or not the lines intersect. To examine the intersection between two lines, we can use the parametric form for the equations of the lines. From the given vector form, we can write 𝐿𝑟=(3,4,1)+𝑡(1,1,1)=(3+𝑡,4𝑡,1+𝑡).

Then, the equation of 𝐿 in the parametric form is

𝐿𝑥=3+𝑡,𝑦=4𝑡,𝑧=1+𝑡.(5)

Likewise, for 𝐿, 𝐿𝑟=(2,0,5)+𝑡(2,3,0)=(22𝑡,3𝑡,5), which means that the vector form of this equation is given by

𝐿𝑥=22𝑡,𝑦=3𝑡,𝑧=5.(6)

If the two lines have a point in common, there must be parameter values 𝑡 and 𝑡 which yield the coordinates of this point which satisfy the equations (5) and (6). We substitute 𝑡=𝑡 in equation (5) and 𝑡=𝑡 in equation (6) to create an expression for the coordinates in 𝐿 and 𝐿. Setting each coordinate from 𝐿 and 𝐿 equal to each other, we get 3+𝑡=22𝑡,4𝑡=3𝑡,1+𝑡=5.

The third equation immediately leads to 𝑡=4. Substituting this into the second coordinate equation above gives us 44=3𝑡,𝑡=0.whichleadsto

Hence, we have obtained 𝑡=4 and 𝑡=0. This means that if there is a shared set of coordinates between the lines 𝐿 and 𝐿, then these coordinates should be given by the parameter value 𝑡=4 for 𝐿 and 𝑡=0 for 𝐿.

Let us see whether these parameter values produce a common set of coordinates from 𝐿 and 𝐿. First, we substitute 𝑡=4 into the parametric equations in (5): 𝑥=3+𝑡=3+4=7,𝑦=4𝑡=44=0,𝑧=1+𝑡=1+4=5.

This leads to the coordinates (7,0,5) from 𝐿 corresponding to the parameter value 4. Next, we substitute 𝑡=0 into the parametric equations in (6): 𝑥=22𝑡=22×0=2,𝑦=3𝑡=3×0=0,𝑧=5.

This gives us the coordinates (2,0,5) from 𝐿 corresponding to the parameter value 0. We note that this is different from the coordinates (7,0,5) from 𝐿. This means that there is no shared set of coordinates. In other words, the lines 𝐿 and 𝐿 do not intersect.

Since the lines are not parallel and also do not intersect, 𝐿 and 𝐿 are skew lines.

Let us summarize a few important concepts from this explainer.

Key Points

  • Two lines are parallel if they share a common direction vector. Alternatively, two lines are parallel if their direction vectors are parallel.
  • To find the point of intersection between two lines, we find the parameter values 𝑡 and 𝑡 that produce the same set of coordinates in the parametric form of the equation of either line.
  • Two lines are perpendicular if they intersect at a point and their direction vectors are perpendicular.
  • If two parallel lines intersect at a point, they overlap entirely. In this case, the two lines are coincident.
  • Two lines are skew if they are not parallel and also they do not intersect at any point.

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