Three Dimensional Vectors, level 2. In this video we will go over slightly more
challenging examples that require the application of various concepts associated with three
dimensional vectors. Alright let's go ahead and jump straight into the first example.
Find vector z, given that vector u equals <1,2,3> vector v equals <2,2,-1> and vector
w equals< 4,0,-4>. Where vector z equals 5u-3v-(1/2)w. Alright in this problem we need to use the
properties of vector addition to find the sum of these vectors. We first need to carry
out a scalar multiplication for vector u, v and w, so let’s go ahead and carry out
the scalar multiplication first. Next it is just a matter of adding or subtracting the
vectors component wise, doing that and simplifying we obtain the final answer equal to negative
3, 4 and 20. Once again notice that vector arithmetic in R cubed is essentially the same
as in R squared. Alright, let’s move along to the next example.
Determine which of the vectors is (are) parallel to vector z. Where vector z equals <3,2,-5>.
Alright recall that two vectors are parallel if one vector is a constant multiple of the
second vector. For this problem we need to find or show that there exists a scalar that
generates vector z when multiplied by the vectors shown below.
For the first vector we are essentially trying to show that vector z equals c times vector
v, we can go ahead and break apart the vector equation into its three components as follows.
then we go ahead and solve for the constant c, notice that in all three components we
obtained the same constant this means that we can multiply vector v by the scalar negative
1/2 and obtain vector z meaning that vector v is parallel to vector z.
For the second vector we do the same and break apart the vector equation into components
and solve for c, doing that we obtain the following values for c. once again notice
that every component has the same value for the scalar this means that vector w is also
parallel to vector z.
Lastly we do the same for vector u and break apart the vector equation into separate components
and solve for the scalar, obtaining the following. Notice that the x and y-components are the
same but the z-component is negative. We need all three scalars to be same for all
components including the sign. Since this scalar does not match the rest of the scalars
on the other components we can conclude that vector u is not parallel to vector z. So vector
z is parallel to vector v and vector w only. Alright let’s move along to the next example.
Use vectors to determine whether the points A(4,-2,7), B(-2,0,3) and C(7,-3,9) are collinear.
Alright let’s go ahead and plot these points so we can get a visual for this problem. We
are asked to show that these 3 points lie along the same line in other words they are
collinear.
We can show this by using vectors by first assuming that they are collinear and break
the line into two pieces or in this case two vectors, making sure that we use a common
point for both vectors. In this case let’s go ahead and find the components of the vector
represented by points A and B and the vector represented by points A and C.
To find the components of the first vector we simply subtract point A from point B and
obtain the following for the components of vector AB, and then we do the same for vector
AC and subtract point A from point C obtaining the following for the components of vector
AC.
Now, if both of this separate vectors lie on the same line they should be parallel to
one another so we need to show that they are a constant multiple of each other so we go
ahead and set up our vector equation and by inspection we see that we can obtain vector
AB by multiplying vector AC by negative 2. Since both of these vectors are parallel and
they pass through the same point, in this case point A we can conclude that they are
collinear. Alright let’s try the next example. Find the magnitude of vector v. Where v equals
-4i+3j+7k. Alright here we have a vector written using
the standard unit vectors, recall that this form is very similar to the component form
of a vector in this case each of the scalars represents the x, y and z-components respectively.
So to find the magnitude we simply use the Pythagorean theorem in three dimensions doing
that and simplifying we obtain the square root of 74 as the final answer. Alright let’s try
the next example.
Determine the value of c that satisfies the equation, the magnitude of scalar c times
vector u equals 4. Where vector u equals negative I plus 2j plus 3k.
Alright we want find the values of scalar c that satisfies the vector equation. We are
given vector u written with unit vectors, so we can go ahead and rewrite this expression
into its component form as follows.
Next let’s go ahead and substitute the component form of vector u into the vector equation.
Then we go ahead and multiply scalar c into each of the components of vector u as follows.
Notice that up to this point the vector equation is essentially stating that the magnitude
of this vector should equal 4. So we go ahead and find the expression for the magnitude of
this vector by using the Pythagorean theorem in three dimensions.
Simplifying the vector equation we obtain the following. Now it is just a matter of
solving for the scalar c, so we go ahead and square both sides, solve for c squared and
then take the square root of both sides doing that we obtain two solutions, positive and
negative the square root of 8 over 7. Let’s go over the final example.
Find a vector v with a magnitude of 3/2 and in the same direction as vector b. Where vector
b equals <2,-2, 1>. Alright here we are asked to generate a vector
that has a magnitude of three halves and points in the same direction as vector b. We first
need to extract the unit vector from vector b this way we can obtain a vector that has
the direction of vector b and will have a magnitude equal to 1 this way, when we multiply
it by a scalar only the length of the vector will change keeping the same direction in
the process.
Let’s start by finding the magnitude of vector b, simplifying the expression we
obtain 3 as the magnitude of vector b. Next let’s go ahead and find the unit vector,
so we take the components of vector b and divide them by the magnitude of vector b doing
that we obtain the following components for the unit vector. Now it is just a matter of
multiplying this unit vector by the magnitude of vector v since the unit vector points in
the same direction as vector b it is just a matter of scaling the vector to the appropriate
magnitude in this case three halves. Doing that and simplifying the expression we obtain
1, -1 and one half for the components of vector v. This vector has a magnitude of 3/2 and
points in the same direction as vector b. Ok in our next and final video on three dimensional
vectors we will go over some more challenging examples.