Hey guys, in this video I'm going to go over this calc 3 problem.
It says, consider a triangle whose vertices are A, which is 1 1 1, B -1 2 3, and C 1 4
-3, which are these three points that I have plotted on a three dimensional graph and the
problem says, find all the angles of the triangle.
If you were to connect the points, it would create a triangle that I have recreated here
and we're trying to find the angles in the triangle.
Before we even get into any of the formulas that we're going to learn in calc 3, from
basic high school or middle school geometry, you know that alpha plus beta plus gamma are
the three angles in a triangle and the sum of these three has to be equal to 180 degrees.
For this problem, it's going to suffice to find any two angles because we can use those
two to find the third.
In this problem, I'm going to find alpha and beta, add those two, and subtract it from
180 and that will give us gamma.
So that will give us the third angle.
And so, for this problem, the main formula that we're going to do is manipulated from
a dot product property.
If we have a vector "a" with the dot product of a vector "b", we know that that is equal
to the magnitude of a times the magnitude of b times cosine theta, where theta is the
angle between the vector a and b.
So it's always going to be a vertex between two vectors.
So this would be theta.
We can manipulate this property or formula to get the angle.
If we just divide by the magnitudes, we have cosine of theta is equal to the dot product
of vector a and b divided by their magnitudes and then furthermore, we can say that cosine--the
inverse cosine of the right hand side gives you theta.
This really is the formula that we're going to be using.
And we have enough information given the points to make vectors out of the points and from
these vectors, we can easily get their magnitudes and their dot products.
So that's how we're going to solve this problem.
Let me go ahead and start with finding alpha.
To find alpha, I'm going to call my "a" the vector AB, which is equal to -2 1 2, and where
I'm getting these value is I'm just subtracting either B from A or A from B, and it doesn't
matter which direction you go in, you just want to be consistent when you're finding
it for vector a and vector b.
So in this case, I did -1 minus 1, 2 minus 1, and 3 minus 1 to get this vector.
And then I'm going to vector b A to C and I'm doing it in the same direction.
So that gives me 0 3 -4.
If I do the cosine of alpha is equal to the dot product over the magnitudes, then the
dot product is going to be 0 plus 3 minus 8 over...
And then the magnitude of a is going to be the square root of each part's square.
So that's going to be 2 squared plus 1 plus 2 squared, which is equal to the square root
of 9, which is equal to 3.
And for b, that's going to be equal to 0 plus 9 plus 16, which is equal to the square root
of 25, which is equal to 5.
So we have 3 times 5, so that is equal to -5 over 15.
So we have that the cosine of alpha is equal to -1/3 and if we do the cosine--the inverse
cosine of -1/3, that gives us approximately the degrees of 109.5.
That gives us the result for the first angle.
The second angle I'm going to find is for beta and it's just the same thing.
This time, I'm going to call my vector BA and then my b is going to be BC.
So this is going to be 2 -1 -2, and again I'm just subtracting and being consistent
with my directions and then this is going to be BC.
And that's going to be 2 2 -6.
We have the formula cosine beta is equal to the dot product, which is going to be 2 times
2 is 4 plus 2 times -1 is -2 plus 12 over the magnitudes.
This is going to be the square root of 4 plus 1 plus 4, which is 3 and then this one is
the square root of 4 plus 4 plus 36, which gives you the square root of 44.
If we simplify this, that's 4 times 11, which gives you 2 times the square root of 11.
So the denominator here is 3 times 2 times the square root of 11.
This is equal to 14 over 6 times the square root of 11.
And if you want to simplify that further, it would be 7 over 3 times the square root
of 11.
If we do the inverse cosine of 7 over 3 times the square root of 11, that will approximately
be equal to 45.3 degrees, and that gives us our second angle.
Therefore, the third angle is going to be 180 degrees minus 45.3 minus 109.5 and that
gives gamma as approximately 25.2 and we are done with this problem.
*chiptune music*