Three Dimensional Vectors Level 3 In this final video on three dimensional vectors
we will go over 4 examples the last example will be a static equilibrium problem in three
dimensions. Alright let’s go ahead and take a look at the first example.
Sketch vector v and write its component form. Vector v lies in the yz-plane, has a magnitude
of 2, and makes an angle of 30 degrees with the positive y-axis.
Alright since this vector lies on a plane it will have one component equal to 0 in this
case the x-component will be equal to 0.
Now to determine the value of the other components we can go ahead and use right triangle trigonometry
to find the y and z components of the vector. According to the problem vector v makes an
angle of 30 degrees with the positive y-axis this means that it can form 2 separate angles,
positive 30 degrees and negative 30 degrees so we have two solutions that satisfies the
conditions of the problem.
Now that we have identified our vectors it is just a matter of using the 30-60-90 special
right triangle to determine the values of the y and z components, doing that we obtain
the following expression, now for the y-component cosine of positive 30 degrees and cosine of
negative 30 degrees yield the same value and when multiplied by the magnitude it will reduce
to the square root of 3, but for the z-component sine of 30 degrees and sine of negative 30
degrees yield different answers that differ only by a negative sign so taking this into
account and simplifying the expression we obtain positive and negative 1 for the z-component.
So the components of the vectors that satisfy the conditions of the problem are equal to
0, square root of 3 and positive negative 1. Alright let’s try a another similar problem.
Vector v lies in the xz-plane, has a magnitude of 5, and makes an angle of 45 degrees with the
positive z-axis. Similar to the previous example let's go ahead
and make a sketch of the problem. Notice that we can also form 2 separate vectors that make
an angle of 45 degrees with the z-axis so this problem will also have two solutions.
Since the vectors lie in the xz-plane the vectors will have a y-component equal to 0.
At this point it is just a matter of using right triangle trigonometry and find the components
of each vector. If we change our point of view and focus on the xz-plane we would see
the following, now if use the standard conventions for measuring angles the first vector would
make an angle of 45 degrees relative to the positive x-axis and the second vector would
make an angle of 135 degrees relative to the x-axis. Using these values for the angle we
can compute the values for each component by using the ratios of a 45-45-90 special
triangle and then multiply by the magnitude of the vector, doing that we obtain the following
for the first vector, 5 times the square root of 2 over 2, zero and 5 times the square root of 2 over 2
For the second vector the values of each component will be the same but the x or I
hat component will be negative since cosine of 135 degrees yields a negative value. Alright
let’s go over the next example.
Use vectors to find the point that lies two-thirds of the way from point P(4,3,0) to point Q(1,-3,3).
Alright let’s go ahead and plot these points to get a graph that we can work with. We are
asked to determine the coordinate point that is two-thirds of the way from point P to point
Q. We can determine this by first finding the component form of the vector represented
by the points P and Q so we go ahead and subtract the initial point from the terminal point
then we simplify the expression obtaining negative 3, negative 6 and 3 for the components.
Next let’s go ahead and scale this vector by a factor of 2/3 doing that we obtain the
following components. Now this component essentially tells us the difference between point P and
the point that is 2/3 away from point P. So all we need to do is to take point P and add
the components of the scaled vector, doing that we obtain the final coordinate equal
to 2,-1, and 2. Alright let’s end the video with the final example.
Find the tension in each of the supporting cables in the figure below if the weight of
the crate is 500 newtons.
Alright here we have a static equilibrium problem in three dimensional space. The concepts
of static equilibrium are the same the sum of all the forces acting on a body has to
equal zero.
For this particular problem we have a crate that is hanging from 3 ropes, attached at
various parts relative to a three dimensional coordinate system. We are given the coordinates
of points A, B, C and D.
Alright the first step is to find an expression for each of the forces that are acting on
point A. This is going to include 3 tension forces one for each rope and the weight of
the box. Let’s first find an expression for each of the tree tension forces. Let’s
label 3 separate vectors by using the points that are provided to us.
These forces will represent the tension force along point A and point B, along point A and
point C and the tension force along point A and point D.
Next we need to write the component form for each vector; the only problem is that we currently
have no idea what the magnitude of each tension force is, but we are able to at least find
the components for the direction of each of the tension force by using the coordinate
points provided to us.
The component form of the force from Point A to Point B can be found by taking Point
A and subtracting it from point B. The component form of the force from point A to point C
can be found in the similar fashion, and the force from point A to point D will also be
found in the same way, simplifying the expression we obtain the following for the direction
of each tension force.
Now we currently have no idea what the magnitude of the force for each tension is equal to,
but we know that if we were to take the direction of each tension force and multiply it by some
scalar c sub 1, c sub2, and c sub 3 for each of the forces we should be able obtain the
component form for each tension force.
We will use these constants along with the direction of the ropes to temporarily represent
the tension forces. We will also express the weight of the crate as a force with only one
component in this case negative 500 newtons along the negative z axis.
Now we can go ahead and find the sum of all the forces and set it equal to 0 as follows.
Next we need to extract 3 separate equations one for each component, for the x-components
we obtain the following equation, for the y-component we obtain the following, and for
the z-component we get the following expression.
Now we are face to face with a system of equations, so we have 3 unknowns and 3 equations we should
be able to find the solution for each of the constants. Using the first equation we go
ahead and solve for c sub 2 and using the second equation we go ahead and solve for
c sub 1, this allows us to substitute these constants into the third equation reducing
it into an equation with a single variable. From here it is just a matter of collecting
like terms and solving for c sub 3, simplifying the expression we obtain the following value
for c sub 3, now we go ahead and solve for c sub 1 and c sub 2, doing that we obtain
the following values.
The last step is to go back to our expressions representing the tension force for each rope
and substitute these scalars and then multiply it by each rope’s direction vector, this
generates the component form for each tension force, and finally all we need to do is to
find the magnitude of each tension force by using the Pythagorean theorem in three dimensions,
approximating the magnitudes we obtain the following values for each of the forces.
Allright and this ends the video series on three dimensional vectors, in our next series of
videos we will go over a new vector operation called the dot product.