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Angles of a Triangle via the Dot (Scalar) 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 angles of the triangle defined by
A
(2, 2, –2),
B
(–4, 4, 2),
C
(2, –5, 4).
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 coordinates are shown in blue, yellow and green.
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Add point B. Then click "Next" to check.
The three coordinates 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 coordinates 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!
How do we proceed?
We can use two sides of the triangle to determine two vectors, such as
AB
and
AC
. By taking their dot product and determining their lengths, we can obtain the cosine of the angle between them. This can be repeated at the other vertices.
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.
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 the vector
AC
to your diagram. Then click "Next" to check.
alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
We can set up the dot product of the vectors in two ways and solve for the cosine of the angle.
State those two ways in terms of the components and the lengths
.
Solve for
.
We get
Equation 1
:
Can we express
in terms of their components, too?
Yes.
State those expressions.
Equation 2
and
Equation 3
How can we get the dot product?
We can determine the components of each vector from the coordinates of the points.
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 dot product of
AB
and
AC
in terms of the literal components
.
Substitute the values.
Simplify.
Equation 4
This takes care of the numerator of Equation 1.
For the denominator, we need to evaluate Equations 2 and 3.
Do that.
We get
Combine these results, Equation 4 and Equation 1 to find
How do we get the angle?
We take the inverse cosine of both sides of the equation.
Do that.
We get
We need to do a similar set of calculations to find the angle at points B and C. Let’s see if you can do those with less help.
For point B, express the two vectors involved.
.
Notice that
what does this tell us about their lengths?
Their lengths are equal.
Find the length of
BC
.
We find that it is 11 units long.
Determine the cosine of the angle.
We get
.
Find the angle.
Now find the angle at C without help.
How can we check these results?
These angles of the triangle should sum to
Do they?
Yes, within a tenth of a degree.
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