General Contents
Detailed Contents
Index
Area of a Triangle via the Cross Product
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Find the area of the triangle defined by
A
(1, 1, –1),
B
(–2, 2, 1),
C
(1, 0, 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 clicking on "Next".
The three components are shown in blue, yellow and green.
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Add point B. Then click "Next" to check.
The three components are shown in blue, yellow and green.
alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
Add point C in a similar way and show the triangle.
The three components are shown in blue, yellow and green.
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How do we proceed?
We can determine two vectors, AB, and AC in the plane. By taking their cross product, we can obtain area of the parallelogram with sides AB and AC from the cross product vector. The area of the triangle is half of this.
For the vector
AB
, should the arrow point be at A or B?
B.
Add the vector
AB
to your diagram. Then click "Next" to check.
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Add the vector
AC
to your diagram. Then click "Next" to check.
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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.
How can we get the cross product?
We can determine the components of each vector and use the determinant method.
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?
For the
x
-component of
AB
, we do the subtraction
Substitute values.
We get
Find the other components of this vector.
We get
and
Determine the 3 components of
AC
.
Set up the determinant for the cross product of
AB
and
AC
in terms of the literal components
.
Substitute the values.
Set up the multiplication.
Note the sign before the
j
-component.
Simplify.
Add this vector,
, to the last diagram, with its base at point A.
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Are we done?
No.
What else do we need to do?
We need the magnitude of this vector result.
How do we find the magnitude of a vector?
We take the square root of the sum of the squares of its components.
Set that up in terms of the numerical components we have just found.
The area of the parallelogram with sides AB and AC is
Simplify.
We get
Is this our result?
No, it is the area of the parallelogram. We wanted the area of the triangle.
Determine the area of the triangle.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
General Contents
Detailed Contents
Index
