The Cross Product Level 3 In this video we will go over various examples
illustrating how to find the cross product of two vectors in space by using both the
geometric and component definition of the cross product.
Let’s jump straight into the first example.
Find the magnitude of vector A crossed with vector B and determine whether A crossed B
is directed into the page or out of the page.
Here we are given two vectors that are located on a common plane.
We are given the magnitude of both vectors and the angle between them.
We can find the magnitude of the cross product by using the geometric definition of the cross
product.
So we go ahead and substitute the magnitude and the angle between the vectors into the
formula.
Computing the product of the magnitudes and sine of 60 degrees we obtain the following.
Simplifying the expression we obtain 25 times the square root of 3 for the magnitude of
the cross product.
Now to determine the direction of the cross product we can go ahead and use the right
hand rule, recall that this vector is going to be perpendicular to vector A and vector
B as well as the common plane containing these two vectors.
Using the right hand rule we place our fingers in the direction of vector A so that our palm
faces vector B and curl them towards vector B, in this case our thumb will be pointing
into the page so we denote this vector with an X if the vector pointed out of the page
then we denote the vector with a dot.
Alright let’s try the next example.
Find the magnitude of vector B crossed with vector A and determine whether B crossed A
is directed into the page or out of the page.
Similar to the previous example we are asked to find the magnitude of vector B crossed
with vector A. We are giving the magnitude of each vector and an angle.
Notice that this angle is not the angle we usually use for computing the magnitude of
the cross product by using the geometric definition since this angle is not between Vector B and
Vector A. We can easily find the angle between the two vectors by shifting vector A so that
its tail aligns with the tail of vector B as follows.
Then it is just a matter of using geometry to find the angle between vector A and vector
B in this case it will be equal to 30 degrees.
Now we can go ahead and find the magnitude of the cross product by using the geometric
definition.
Substituting the magnitudes and the angle between the vectors we obtain the following,
simplifying the expression we obtain 24 as the magnitude of the cross product.
Next we use the right hand rule to determine the direction of vector B crossed with vector
A, so we point our fingers in the direction of vector B with our palms facing vector A
and curl them towards vector A; in this case our thumb will point out of the page so we
use a dot to represent this vector.
Alright let’s try the next example.
The figure below shows vector V in the xy-plane and vector U in the direction of positive
k-hat.
Their lengths are 4 and 5 respectively.
Find the magnitude of vector V crossed with vector U and determine whether the components
of vector V crossed with vector U are positive, negative, or 0.
Alright here we have a diagram of vector V and vector U we are first asked to find the
magnitude of vector V crossed with vector U, we are given the magnitude of each vector,
note that the angle between the vectors can be obtained from the figure.
We know that vector V is located on the xy-plane and vector U has only one component since
it is pointing in the positive k-hat direction.
This means that vector V and vector U are orthogonal to one another so the angle between
these two vectors is 90 degrees or pi over 2.
Next we go ahead and substitute these values into the geometric definition of the cross
product, simplifying the expression we obtain 20 for the magnitude of vector V crossed with
vector U.
Next let’s use the right hand rule to determine the direction of vector V crossed with vector
U.
We point our fingers in the direction of vector V with our palms facing vector U and curl
our fingers towards vector U.
Once again our thumb will point in the direction of vector V crossed with vector U.
In this case it will have an x-component that is positive, a y-component that is negative
and since it lies on the xy-plane it will have a z-component that is equal to 0.
Let’s try the next example.
Find the cross product between vector A and vector B and verify that it is orthogonal
to both vector A and vector B. Alright here we are given vector A and vector
B in component form we are asked to find the cross product between these two vectors specifically
vector A crossed with vector B. Let’s go ahead and use the component definition of
the cross product.
We first need to set up our 3 by 3 determinant by using the standard unit vectors as the
entries on the first row, followed by the components of vector A in the second row and
the components of vector B in the third row as follows.
Next we go ahead and compute the determinant by cofactor expansion.
So we first cover row 1 and column 1 and use the minor of the 2 by 2 matrix this value
will represent the x or i-hat component, next we write a negative sign and then cover row
1 and column 2 and use the minor of the 2 by 2 matrix this value will represent the
y or j-hat component, and finally we cover row 1 and column 3, the minor of the 2 by
2 matrix will represent the z or k-hat component.
Make sure you do not forget the negative sign in the y or j-hat component this is a common
mistake that many students make.
Next we go ahead the find the determinants of the 2 by 2 matrices, then we simplify each
component obtaining the following components for the cross product.
The last thing we need to do is to show that this vector is orthogonal to both vector A
and vector B. The easiest way to show this is by taking the dot product of this vector
with vector A and vector B. The result should equal 0 if the vectors are orthogonal to one
another.
Computing the dot product with vector A gives us 0 as our answer.
In the same manner computing the dot product with vector B also gives us 0 as our answer
hence the vector produced by crossing vector A and vector B is orthogonal to both vector
A and vector B.
Let’s move along to the next example.
Find the cross product between vector A and vector B and verify that it is orthogonal
to both vector A and vector B. Here we are given vector A and vector B written
in standard unit vector form.
Let’s go ahead and rewrite the vectors in component form.
Notice that vector B does not have a value for i-hat so the x-component will be equal
to zero.
Next we go ahead and set up our determinant, the first row will contain the unit standard
vectors, the second row will contain the components of vector A and the third row will contain
the components of vector B. Then we go ahead and find the minors by eliminating the first
row and each of the 3 columns doing that we obtain the following 2 by 2 determinants.
Do not forget to include a negative sign in the j-hat component.
Finding the determinant of each 2 by 2 matrix and simplifying we obtain the following vector
for the cross product between vector A and vector B.
We can verify that this vector is orthogonal to both vector A and vector B by computing
the dot product between these vectors.
Using the component definition of the dot product we obtain 0 for both vector A and
vector B hence vector A crossed B is orthogonal to both vector A and Vector B. Alright in
our next video we will go over more challenging examples.