Vector Methods in Three-Space
This handout is not a complete summary about vectors in space. Instead, it summarizes with examples
many of the formulas and concepts used in the manipulation of points, lines, and planes in space.
We generally use three important properties of vectors in three-space to solve many of the problems
which arise with points, lines, angles, and planes. These are:
a) If two vectors and are parallel, then one is a scalar multiple of the other ( = ), and their
).
cross product is equal to zero ( × = 0
b) If two vectors and are orthogonal (perpendicular to each other), then their dot product
equals zero ( ∙ = 0).
c) A vector , which is equal to the cross product of two vectors and , is perpendicular to both and . That is, if = × , then is perpendicular to and perpendicular to .
Find the distance between two points.
Let = , , and = , , be two points in space. Then,
= − + − + − Or,
= 〈 − 〉 →
= ∙ = 〈 − 〉 ∙ 〈 − 〉 →
= 〈 − 〉 ∙ 〈 − 〉
Example: Find the distance between the points = 3, −1,4 and = 5,4,1.
= 5 − 3 + 4 + 1 + 1 − 4 = √4 + 25 + 9 = √38
Or,
= 〈 − 〉 ∙ 〈 − 〉 = 〈2,5, −3〉 ∙ 〈2,5, −3〉 = √4 + 25 + 9 = √38
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Find the equation of the line through a point and parallel to a vector.
Let the point be , , and let the vector be 〈, , 〉. Then the equation of the line in space is
= + $, = + $,
= + $
− − − =
=
parametricform
symmetricform
Example: Find the equation of the line through the point (1, 5, 2) and parallel to the vector 〈2, −2,7〉.
= 1 + 2$, = 5 − 2$, = 2 + 7$
Or,
−1 −5 −2
=
=
2
−2
7
Find the distance from a point to a line in space.
The distance between a point Q and a line in space is
=
× 4
3
3
‖4
‖
where Q is the point, P is a point on the line, and 4
is the direction vector for the line.
Example: Find the distance between the point (3, -1, 4) and the line given by
= −2 + 3$,
= −2$, and
= 1 + 4$
Using the directions numbers from the parametric equations, 4
= 〈3, −2,4〉. To find a point on
the line, let $ = 0 to get = −2,0,1. So,
= 〈5, −1,3〉 →
=
× 4
3 ‖〈5, −1,3〉 × 〈3, −2,4〉‖ √174
3
=
=
= √6
‖〈3, −2,4〉‖
‖4
‖
√29
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Find the distance from a point Q to a plane.
Let the equation of the plane be + + + = 0 and let the point be = , , . Then the
distance is
=
| + + + |
√ + + or =
:;
∙ <;:
‖<‖
where P is a point on the plane, Q is a point not on the plane, and < is a normal vector to the plane.
Example: Find the distance from the point (3, 1, 5) to the plane − 3 + 2 − 5 = 0.
=
|13 − 31 + 25 − 5|
√1 + 3 + 2
=
5
√14
=
5√14
14
= 〈−2,1,5〉, < = 〈1, −3,2〉.
Or, let = 0 and = 0 to obtain the point = 5,0,0. Thus, =
:|〈−2,1,5〉 ∙ 〈1, −3,2〉|:
‖〈1, −3,2〉‖
=
|−2 − 3 + 10|
√1 + 9 + 4
=
5
√14
=
5√14
14
Find the distance between two parallel planes.
All we need to do is to pick a point in one of the planes and use the technique in the previous example.
Example: Find the distance between the parallel planes 10 − 12 + 5 − 8 = 0 and
10 − 12 + 5 + 4 = 0.
= 1, 1, 2 is a point in the first plane. So, using the equation of the second plane,
=
|101 − 121 + 52 + 4|
√10
+
12
+ 5
=
12
√269
Find two parallel planes that contain two skew lines.
Let the equations of the two skew lines be
− − − =
=
and
− − − =
=
=
>
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The vectors 4
= 〈, , 〉 and ? = 〈, =, >〉 are the direction vectors for the lines. The cross product
4
× ? is a vector perpendicular to both 4
and ?, that is, perpendicular to both lines. Since the two
planes will be parallel, both will have the same normal vector. Moreover, since the two lines will be in
the planes (one line in each plane), then the normal vector will also be perpendicular to the lines.
Example: Find the two parallel planes that contain the following two skew lines:
−1 +3 −6
=
=
2
1
−2
and
+1 +1 −1
=
=
4
5
3
So,
4
= 〈2,1, −2〉, ? = 〈4,5,3〉 →
4
× ? = 〈13,14,6〉 thisisthenormalvector
The equations of the two planes are:
13 − 1 − 14 + 3 + 6 − 6 = 0 →
13 + 1 − 14 + 1 + 6 − 1 = 0 →
13 − 14 + 6 − 91 = 0
13 − 14 + 6 − 7 = 0
Find the distance between two skew lines.
First, find the two parallel planes that contain the skew lines. Then, find the distance between these
two planes, which will also equal the distance between the two skew lines.
D
Example: For the two skew lines given above, C0,0, EF is a point on the second plane. Then we have
=
7
G130 + 140 + 6 C6F − 91G
√13
+
14
+ 6
=
84
√401
Find the equation of the line of intersection between two planes.
We have two methods for solving this type of problem, algebraic and vector. Both are illustrated in the
example on the next page.
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Example: Find the intersection line common to the planes + − 3 = 2 and 2 − + = 9.
Algebraic Method: Solve for x and y in terms of z.
+ − 3 = 2
2 − + = 9
Let = $. Then =
H
+
I
H
→ 11 2
+
3
3
5 7
=− +
3 3
3 = 2 + 11 →
3 = 7 − 5
J
H
and = − +
DI
H
=
→
are the parametric equations of the line.
Vector Method: The vectors 4
= 〈1,1, −3〉 and ? = 〈2, −1,1〉 are the normal vectors to the
planes; they are also orthogonal to the common line. Hence, 4
× ? = 〈−2, −7, −3〉 is a
direction vector for the line. If = 0, =
,
H
J
and = − H, then the equation of the line in
parametric form is
11
− 2$
3
5
= − − 7$
3
= −3$
=
Find the equation of a plane through a point and perpendicular to a vector.
Let the point be , , and the vector be 〈, , 〉. Since the point will be in the plane, selecting
another point , , in the plane will give us the vector 〈 − , − , − 〉, which is also in the
plane. Since this vector will be perpendicular to the given vector, their dot product will be zero. That is,
〈 − , − , − 〉 ∙ 〈, , 〉 = 0. This is the equation of the plane.
Example:
Find the equation of the plane through the point (1, -2, 4) and perpendicular to the vector 〈2,4, −4〉.
〈 − 1, + 2, − 4〉 ∙ 〈2,4, −4〉 = 0
2 − 1 + 4 + 2 − 4 − 4 = 0
+ 2 − 2 + 11 = 0
Note that 〈1,2, −2〉 × 〈2,4, −4〉 = 0 shows that the normal vector to the plane is parallel to the given
vector, which means the plane is perpendicular to the given vector.
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Find the equation of a plane through a point and parallel to a vector.
Let the point be , , and let the vector be 〈, , 〉. Since the point will be in the plane, selecting
another point , , in the plane will give us the vector 〈 − , − , − 〉 which is also in the
plane. Since this vector will be parallel to the given vector, their cross product will be zero. That is,
〈 − , − , − 〉 × 〈, , 〉 = 0. This is the equation of the plane.
Example:
Find the equation of the plane through the point 1, −2,4 and parallel to the vector 〈2,4, −4〉.
〈 − 1, + 2, − 4〉 × 〈2,4, −4〉 = 0
8 − 1 − 6 + 2 − 2 − 4 = 0
4 − 3 − = 12
Note that 〈4, −3, −1〉 ∙ 〈2,4, −4〉 = 8 − 12 + 4 = 0 shows that the normal vector to the plane is
perpendicular to the given vector, which means the plane is parallel to the given vector.
Find the equation of a plane containing three nonlinear points.
Let the three points be P, Q, and R. Then 〈 − 〉 and 〈 − K〉 are two vectors in the plane. The vector
? = 〈 − 〉 × 〈 − K〉 is perpendicular to both vectors and is therefore normal to the plane. Using the
vector ? and the point P we can write the equation of the plane.
Example: Find the equation of the plane containing the three points (1, 2, 3), (2, 0, 4), and (3, 3, 1).
Let P = (1, 2, 3), Q = (2, 0, 4), and R = (3, 3, 1). Then,
? = 〈 − 〉 × 〈 − K〉 = 〈−1,2, −1〉 × 〈−2,1,2〉 = 〈5,4,3〉
Using P and ?, the equation of the plane is
5 − 1 + 4 − 2 + 3 − 3 = 0 →
5 + 4 + 3 = 22
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