OpenStax-CNX module: m47413
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Show that Three Vectors are
Coplanar
∗
John Taylor
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 4.0
Abstract
Demonstrates that three vectors are coplanar by forming the scalar triple product to see if the volume
of the related parallelopiped is zero. The determinant method is used.
1 Three Vectors Coplanar
(The small steps below can be used for a self test. To do so, Scroll in small increments.)
(Updated 5/19/14)
Determine whether the vectors specied by the following four points are coplanar:
A (6, 0, 2) , B (2, 0, 4) , C (6, 6, 1) , and D (2, 6, 3).
To visualize the problem, let's draw a diagram. On paper, see if you can plot the point A. Use an x -axis
out of the plane of the diagram. Then check your graph by scrolling down.
The three components are shown in blue, yellow and green.
∗ Version
1.4: May 20, 2014 4:04 pm -0500
† http://creativecommons.org/licenses/by/4.0/
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Figure 1
Add point B . Then scroll down to check.
The three components are shown in blue, yellow and green.
Figure 2
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Add point C in a similar way.
The three components are shown in blue, yellow and green.
Figure 3
Add point D.
The three components are shown in blue, yellow and green.
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Figure 4
How do we proceed?
Since any three points, such as A, B, and C, determine a plane, we can determine two vectors,
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AB and AC, in that plane. By taking their cross product, we can obtain a vector perpendicular to
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their plane. Then we can determine the dot product of that vector and the vector AD.
How will that help?
That dot product is the volume of the parallelepiped determined by the three vectors.
What should we look for?
If that dot product is zero, the volume is zero.
And then?
We can conclude that the three vectors lie in the same plane.
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For the vector AB, should the arrow point be at A or B ?
B.
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Add the vector AB to your diagram. Then scroll down to check.
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Figure 5
Add the vectors
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AC
and
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→
AD
to your diagram. Then scroll down to check.
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Figure 6
Can we use the magnitudes of these vectors and the angle between them to get the cross product?
No.
Why?
We don't know the angle between them.
Then how can we get the cross product?
We can determine the components of each vector and use the determinant method.
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How do we get the x -component of AB?
We simply subtract the x -coordinates of the two points.
What is the order of the subtraction?
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For the x -component of AB, we do the subtraction ABx = Bx − Ax .
Substitute values.
We get ABx = Bx − Ax = 2 − 6 = −4.
Find the other components of this vector.
We get ABy = By − Ay = 0 − 0 = 0 and ABz = Bz − Az = 4 − 2 = 2
→
Determine the 3 components of AC.
ACx = Cx − Ax = 6 − 6 = 0
ACy = Cy − Ay = 6 − 0 = 6
ACz = Cz − Az = 1 − 2 = −1
Set up the determinant for the cross product of
ABx , ABy , ABz , ACx , ACy , and ACz .
→
→
→
i
→
j
→
k
AB × AC=| ABx
ABy
ABz |
ACx
ACy
ACz
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→
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AB and AC in terms of the literal components
OpenStax-CNX module: m47413
Substitute the values.
→
→
→
i
→
AB × AC=| −4
0
j
7
→
k
0
2
6
−1
|
Set up the multiplication.
→
→
→
AB × AC= i ∗ [0 ∗ (−1) − 6 ∗ 2]
+
→
∗ [(−4) ∗ (−1) − 0 ∗ 2]
j
→
∗ [(−4) ∗ 6 − 0 ∗ 0]
k
Note the sign before the
Simplify.
→
→
→
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j -component.
→
AB × AC= i ∗ (−12) + j ∗ (−4) + k ∗ (−24)
→
→
→
To get an idea of the direction of this vector, scale it down by a factor of 4 to − 3 i − j −6 k so that
it will t better on the diagram. Add this vector to the last diagram, with its base at point A.
Figure 7
Notice that the colored lines showing the components of this vector start at point A. They correspond
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to the vector we found: − 3 i − j −6 k = 0.25∗ AB × AC
Hence the blue line is 3 units in the negative x-direction, the yellow line is one unit in the negative
y-direction, and the green line is 6 units in the negative z-direction.
Are we done?
No.
What else do we need to do?
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→
We need to determine dot product of this vector with AD.
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Determine the 3 components of AD.
ADx = Dx − Ax = 2 − 6 = −4
ADy = Dy − Ay = 6 − 0 = 6
ADz = Dz − Az = 3 − 2 = 1
Now set up the dot product.
→
→
→
We need AD • AB × AC
Substitute
the vectors.
→
→
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AD • AB × AC
→
→
→
→ →
→
= −4 i +6 j +1 k • −12 i −4 j −24 k
Set upthe products
of the components.
→
→
→
AD • AB × AC
Simplify.
= (−4) ∗ (−12) + 6 ∗ (−4) + 1 ∗ (−24)
→
→
We get AD • AB × AC = 48 − 24 − 24 = 0
→
What is our conclusion?
The volume is zero and the three vectors are coplanar.
If you found this helpful and would recommend that I create more pages like this one, please let me know
using the email link at the top of the page.
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