1

Lines and Planes in �� using Vectors, Dot and Cross Products

This is a summary of the ways that the dot and cross products are used to solve questions

related to lines and planes in 3ℜ . There is often more than one way to solve a problem,

and in some cases, answers are not unique.

• Preliminary

Let L represent a line in �� passing through a point (��, �, �) and parallel to a vector � = ⟨��, ��, ��⟩, called the direction vector. The coordinates x, y and z for a point on line

L in �� are usually written parametrically as a function of �: � = �� + ��� = � + ��� = � + ���

The equation of a line does not have to be unique. For example, a line passing through

(1,2,3) and parallel to � = ⟨4,−5,6⟩ has the parametric form

� = 1 + 4� = 2 − 5� = 3 + 6�.

However, if � = 1, then a point on that line is (5, −3,9) and the same line could be

written as

� = 5 + 4� = −3 − 5� = 9 + 6�.

These two sets of equations produce the exact same outputs, that is, the exact same line.

All that differs is the “starting point”, where � = 0.

A plane is defined by a point (��, �, �) on the plane, and a vector � = ⟨ , !, "⟩ normal

(orthogonal) to the plane. The equation of a plane is given by:

� + ! + " = #

where # = �� + !� + "�.

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There are many questions that can be asked regarding the interactions between lines with

other lines, lines with planes, and planes with other planes.

• Show that two lines are parallel.

Show that the two direction vectors are scalar multiples of one another, while ensuring

the two lines are not identical.

Example: Let $� and $� be two lines, defined below by

$� = % � = 1 + 2� = 2 − 3� = −1 + 4�& and $� = %� = 3 + 4� = 1 − 6� = 2 + 8�

&.

The directional vectors are �� = ⟨2,−3,4⟩ and �� = ⟨4,−6,8⟩. Therefore, �( = 2��, so at

the very least, the vectors �� and �� are parallel. Now we need to show that these are two

different lines. We know that (1,2, −1) is a point on $�. We then show whether $�

includes this point; that is, we show whether or not there exists a t-value that when

substituted into $�, we generate (1,2, −1).

Using the equation � = 3 + 4� from $�, we substitute 1 for �, and solve for �. We get � = −1 2⁄ . Now, using the equation = 1 − 6� from $�, we substitute � = −1 2⁄ and

find that = 4. Thus, $� does not pass through (1,2, −1), so we conclude that $� and $�

are two separate parallel lines.

Now consider $� and $� given below.

$� = %� = 1 + � = 4� = 2 − �& and $� = % � = −5 + 2� = −24 + 8� = 8 − 2�

&.

You should be able to show that they are the same line. On the other hand, consider

$� = % � = 6 − � = 2 + 7� = 1 − 3�& and $� = % � = 5 − � = 4 + 7� = 2 + 2�

&.

These are clearly not the same line. They are not parallel. This summarily removes the

possibility of them being the same line.

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• Find the parameterized equation of a line through two points.

Recall that to find the parameterized equation of a line, we need any point on that line

and a direction vector �. Suppose the two points are represented as A and B. We find the

direction vector � from A to B and from this, we can build the equations for the line.

Example: Let + = (1,3,4) and , = (−3,2,6). The direction vector from A to B is � = ⟨−4, −1,2⟩. Therefore, the line can be written as

$ = %� = 1 − 4� = 3 − � = 4 + 2�&.

Note that when � = 0, we generate + = (1,3,4), and when � = 1, we generate , =(−3,2,6). This is a good way to check for correctness.

This same line could also start at B and be traced toward A. The line would be written as

$ = %� = −3 + 4� = 2 + � = 6 − 2�&.

Note that the coefficients of � are negated, and that � = 0 gives B while � = 1 gives A.

• Show that two lines intersect or are skew.

Suppose two lines are given, $� with � as its parameter variable, and $� with - as its

parameter variable. We set the respective �, and equations equal, forming three

equations in two variables (- and �). We then solve two of the equations for - and �. Assuming we find values for - and �, we then substitute those values into the third

equation and check for the statement’s truth. If true, then the two lines intersect. If not,

the two lines do not intersect. If the lines are not parallel and do not intersect, then the

lines are skew.

Example: Let $� and $� be two lines, defined below by

$� = % � = 1 + 2� = 2 − 3� = −1 + 4�& and $� = % � = 1 + 4- = 8 − 12- = 9 − 2-

&.

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We form a system of two equations in two variables (- and �) by equating the equations

for � and for :

1 + 2� = 1 + 4-2 − 3� = 8 − 12- → −2- + � = 012- − 3� = 6 → - = 1� = 2.

We then substitute 1 for - and 2 for � into the equations defining the z-coordinates in $�

and $�:

−1 + 4� = 9 − 2- −1 + 4(2) = 9 − 2(1) 7 = 7.

This is a true statement, so the lines intersect. We should note they cannot possibly be the

same line since they are not parallel. In any case, we can now find the point of

intersection. We evaluate - = 1 or � = 2 into $� or $�, respectively. In both cases, we

find that the lines intersect at (5, −4,7).

If at any step during this process we come to a false statement, then the lines do not

intersect. For example, it is easy to show that

$� = % � = 1 + 2� = 2 − 3� = −1 + 4�& and $� = % � = 1 + 4- = 8 − 12- = 9 − 4-

&

are skew lines.

• Find the shortest distance from a point to a line.

Let Q represent the point that is not on the line. We pick any point P on the line L, and

create a vector 0 from P to Q. Remember, line L is defined (partially) by its direction

vector �. We then find the projection of 0 onto �, then determine the orthogonal

component. The distance from Q to L is the magnitude of this orthogonal component.

Example: Let $ = % � = 1 + 2� = 2 − 3� = −1 + 4�& and let Q = (4,8,3), which is not on line $. The

direction vector of L is � = ⟨2,−3,4⟩, found by reading off the coefficients of �.

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We now choose any point P on L. We choose 1 = (1,2, −1). Therefore, the vector 0

from P to Q is

0 = 12333334 = ⟨4 − 1, 8 − 2, 3 − (−1)⟩ = ⟨3,6,4⟩.

The projection of w onto v is:

proj�0 = 9�∙0�∙�; � = <�=� = <

�= ⟨2, −3,4⟩ = ⟨ >�= , − ���= , �?�=⟩.

The normal vector to proj�0 is found by subtraction, and is called norm�0:

norm�0 = 0 − proj�0 = ⟨3,6,4⟩ − ⟨ >�= , − ���= , �?�=⟩ = ⟨B=�= , �>?�= , ����= ⟩.

The magnitude of this orthogonal vector is the distance from the point Q to the line L:

Distance = |norm�0| = D(79/29)� + (186/29)� + (100/29)� ≈ 7.78 units.

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• Find the intersection of a line with a plane.

Substitute the parameterized equations from the line for �, and into the equation of

the plane, and solve for �. Then substitute this � into the equations for the parameterized

form of the line to find the values for �, and .

In the case when the coefficients of � simplify to 0, but a non-zero constant remains, then

there is no solution, and the line is parallel to the plane. If the coefficients of � simplify to

0 and so do the constants, then we have the identity 0 = 0, which indicates that the line is

a subset of (contained within) the plane.

Example: Let $ = % � = 1 + 2� = 2 − 3� = −1 + 4�& and the plane be given by 2� − + 7 = 11.

We substitute the equations from L into the plane, and simplify:

2(1 + 2�) − (2 − 3�) + 7(−1 + 4�) = 11 2 + 4� − 2 + 3� − 7 + 28� = 11 −7 + 35� = 11 35� = 18 � = �>�G.

Therefore, the point of intersection is found by substituting 18 35⁄ for �:

� = 1 + 2 9�>�G; = B��G, = 2 − 3 9�>�G; = �?�G, = −1 + 4 9�>�G; = �B�G.

Suppose we want to find the point at which the line L above intersects the plane � + 2 + = 1. Substituting as above, we have

(1 + 2�) + 2(2 − 3�) + (−1 + 4�) = 1 1 + 2� + 4 − 6� − 1 + 4� = 1 4 = 1.

The terms containing � summed to zero, and the remaining statement, 4 = 1, is false.

Thus, the line L does not intersect the plane � + 2 + = 1. However, the line L lies

within the plane � + 2 + = 4. You can check this.

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• Find the equation of a plane given three non-collinear points.

From the three points, we form two vectors which necessarily must lie in the same plane

as the three points. Then we find the cross product of these two vectors, which results in a

vector that is orthogonal to the plane. If we call this vector � = ⟨ , !, "⟩, then a, b and c

are the coefficients of x, y and z in the equation of the plane. To find the constant d, we

evaluate at any of the three given points, and check by evaluating the other given points.

Example: Let + = (1,3,4), , = (−3,2,6) and H = (1,0, −6). We form vectors I = +,333334 and � = +H333334: I = +,333334 = ⟨−4, −1,2⟩ and � = +H333334 = ⟨0, −3, −10⟩

We then find I × �:

� = I × � = ⟨16,−40,12⟩.

(Before proceeding, be sure to check that this cross product is correct by showing it is

orthogonal to I as well as � by showing � ∙ I = 0 and � ∙ � = 0.) After ensuring this is

correct, we move on.

Thus, the equation of the plane has the form 16� − 40 + 12 = #. To find #, we

evaluate any one of the original points into �, and . Using point + = (1,3,4), we have

16(1) − 40(3) + 12(4) = −56.

The equation of the plane is 16� − 40 + 12 = −56, or simplified, it is

4� − 10 + 3 = −14.

To check, let’s evaluate this equation at , = (−3,2,6). We have

4(−3) − 10(2) + 3(6) = −14 −12 − 20 + 18 = −14 −32 + 18 = −14 −14 = −14, �KLM

A similar check shows that H = (1,0, −6) also gives a true statement.

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• Find the planar (solid) angle between two non-parallel planes.

The angle between the two planes is the same as the angle between the two normal

vectors of the planes. Planar angles are always acute, or at most 90 degrees in the case of

orthogonal planes. We use the dot product formula with the cosine to find the angle.

Example: Let the planes be � + 3 − 2 = 5 and 4� − + 5 = −2. Their normal

vectors are �� = ⟨1,3, −2⟩ and �� = ⟨4,−1,5⟩. The angle between these two planes is:

N = cosQ� 9 �R∙�S|�R||�S|; = cosQ� 9− =√�<∙<�; ≈ 111.79 degrees.

However, planes always intersect at an acute angle, so the preferred answer is the

supplement: 180 – 111.79 = 68.21 degrees.

• Find the equation of a line of intersection between two planes

For the equation of a line in R3, recall from earlier in this section that we need a direction

vector �, and any point on the line.

From the two planes, we can determine their normal vectors �� and ��. Thus, their cross

product, � = �� × ��, provides a vector parallel to the line of intersection of the two

planes. To find any point on the line of intersection, we usually set one of the variables

equal to 0, then solve the remaining system for the values of the other two variables.

From this, we can build the parametric form of the equation of the line.

Example: To find the equation of the line of intersection between the two planes � − 3 + 2 = 6 and 2� + − 7 = 1, we determine their respective normal vectors.

From the first plane, we have �� = ⟨1,−3,2⟩, and from the second plane, we have �� = ⟨2,1, −7⟩. Next, we find their cross product:

� = �� × �� = ⟨19,11,7⟩.

Now we need any point on this line of intersection. We set � = 0. This results in two

equations in variables y and z:

−3 + 2 = 6 − 7 = −1

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Solving this equation, we find that = − <��= and = − �

�=. We have the necessary

ingredients to assemble the parametric form of the equation of the line. We have the

direction vector � = ⟨19,11,7⟩ and a point on the line, 90,− <��= , − �

�=;. Thus, the equation

of the line of intersection of the planes � − 3 + 2 = 6 and 2� + − 7 = 1 is

$ = % � = 19� = −40 19⁄ + 11� = −3 19⁄ + 7� .&

• Find the shortest distance from a point to a plane.

Let Q represent the point that is not in the plane. Determine the normal vector � from the

equation of the plane. Then choose any point P on the plane and form a vector from P to

Q, � = 12333334. We then project � onto �, and find the magnitude of this projection. This

gives the distance from point Q to the plane.

Example: Let � − 3 + 2 = 6 be a plane, and let Q = (1,4,3) be a point not on the

plane. Note that the normal vector to the plane is � = ⟨1,−3,2⟩.

We now need any point in the plane. This can be easily achieved by choosing arbitrary

values for two of the variables and solving for the third. In this example, we let � = 0 and = 0, which means that = 3. Thus, we have a point 1 = (0,0,3).

The vector from P to Q is � = ⟨1,4,0⟩. Projecting � onto �, we have

proj�� = �∙��∙�� = − ��

�< ⟨1, −3,2⟩.

The distance from Q = (1,4,3) to the plane � − 3 + 2 = 6 is the magnitude of this

projection: ���<D1� + (−3)� + 2� = ��

�<√14 ≈ 2.94.

On the following page is a visual explanation of the process shown above.

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• Find the distance between two parallel planes.

Since the planes are parallel, their normal vectors �� and �� are parallel, and thus scalar

multiples of one another. In other words, it suffices to choose one of the normal vectors

for this process. We then form a vector � from any point in one plane to any point in the

other plane. Using the same technique as in the previous example, we then project � onto � and find the magnitude of this projection.

Example: Let � − 3 + 2 = 6 and −2� + 6 − 4 = 1 be parallel planes. Their

normal vectors are (respectively) �� = ⟨1,−3,2⟩ and �� = ⟨−2,6, −4⟩. Note that each is

a scalar multiple of the other. In this example, we’ll use � = ⟨1, −3,2⟩.

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We then find a point in one plane and a point in the other plane. From � − 3 + 2 = 6,

we obtain (0,0,3), and from −2� + 6 − 4 = 1 we obtain 9− �� , 0, 0;. Thus, a vector

that connects the two points is � = ⟨�� , 0, 3⟩. Projecting � onto �, we have

proj�� = �∙��∙�� = ��

�> ⟨1, −3,2⟩.

The magnitude of this vector is ���>D1� + (−3)� + 2� ≈ 1.737.

• Find the shortest distance between two skew lines.

Since two skew lines can always be placed into two parallel planes, this problem reduces

to finding the distance between the two planes. From the two lines, we can determine the

direction vector of each, �� and ��. Their cross product will produce an orthogonal vector � to both lines. We then choose one point from each line and form a vector � between

these two points. We then project � onto � and find the magnitude of this projection.

Example. Find the shortest distance between lines $� and $�, defined below by

$� = % � = 1 + 2� = 2 − 3� = −1 + 4�& and $� = % � = 5 + � = 3� = −1 + 6�

&.

The lines are skew since they are not parallel, nor do they intersect. From each line, we

determine their direction vectors as �� = ⟨2,−3,4⟩ and �� = ⟨1,3,6⟩. Then we find �, a

vector orthogonal to both lines.

� = �� × �� = ⟨−30,−8,9⟩.

Now we need a vector from one line to another. From the first line, we have a point 1 = (1,2, −1) and from the second line, we have point 2 = (5,0,−1). Thus, a vector

connecting P to Q is � = ⟨4,−2,0⟩. Projecting � onto �, we have

proj�� = �∙��∙�� = − ��<

��<G ⟨−30,−8,9⟩.

Therefore, the distance between the two skew lines $� and $� is the magnitude of this

projection, ��<��<GD(−30)� + (−8)� + 9� ≈ 3.217.

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• Practice

Let + = (1,4, −2), , = (2,−3,0), H = (−5,4,1) and U = (0,7,2) be four points in ��.

Use these four points to answer Exercises 1-8.

1. Find the parameterized equation of the line passing through B and C such that � = 0 gives B and � = 1 gives C.

2. Find the angle between vectors +,333334 and +U333334 at point A.

3. Find the shortest distance from C to the line passing through A and D.

4. Find the equation of the plane passing through A, B and C.

5. Find the shortest distance from point D to the plane ABC.

6. Find the parameterized equation of the line normal to the plane BCD and passing

through A.

7. Find the point at which the line passing through A and normal to plane BCD

(question 6) actually passes through the plane BCD.

8. Find the planar angle formed by planes ABC and BCD.

9. Find the distance between the planes � − + 2 = 1 and 2� − 2 + 4 = 0.

10. Consider lines $� = % � = 2� = 1 − � = 3�& and $� = % � = 4� = 3 + 2� = 1 − 4�

&. Find their point of

intersection. If they do not intersect, then find the shortest distance between the

two lines. Hint: use parameter variable - for one of the lines.

11. Consider lines $� = %� = 3 + 8� = 2 + 3� = 4 − �& and $� = %� = −1 + 3� = 1 + � = 23 − 5�

&. Find their point of

intersection. If they do not intersect, then find the shortest distance between the

two lines.

Answers:

1. $ = % � = 2 − 7� = −3 + 7� = �&.

2. 111.94 degrees.

3.

4.

5.

6.

7.

13

8.

9.

10. 11. Intersect at (11,5,3).

Note: I have not worked these all out.