The Cross Product Level 6 In this video we will focus on the geometric
properties of the cross product between two vectors in space.
Let’s start with the first geometric property.
Let vector U and vector V be nonzero vectors in space, and let theta be the angle between
vector U and vector V. Then vector U crossed with vector V is orthogonal to both vector
U and vector V, this property was illustrated in the first video of this series.
Now we want to formally prove this geometric property.
In order to show that vector U crossed with vector V is orthogonal to vector U and vector
V, we will make use of the dot product, if both vectors are orthogonal to one another,
then the dot product should equal zero.
So we first denote the vectors in component form as follows.
Now we go ahead and find the cross product between vector U and vector V doing that we
obtain the following vector.
Next we go ahead and dot this vector with vector U and Vector V separately.
Then we go ahead and distribute the components of vector U and vector V for each expression.
From here notice that every single value cancels out reducing both expressions to zero.
Therefore vector U crossed with vector V is orthogonal to both vector U and vector V,
and this ends the proof.
Let’s move along to the next geometric property.
This property was also introduced in the first video of this series.
Recall that we used the right hand rule to determine the direction of the vector produced
from the cross product of two vectors.
This property is used to calculate the length or magnitude of the vector produced when crossing
two vectors.
So the magnitude of the cross product between vector U and vector V is equal to the magnitude
of vector U times the magnitude of vector V times sine of theta, recall that theta is
an angle between 0 and pi.
We can prove this property by starting with the component definition of the cross product
and finding an expression for the magnitude of the cross product.
We are going to work with the square of the magnitude so this expression will simplify
to the following expression.
Let’s go ahead and expand each of the terms as follows.
Next we notice that the magnitude of vector U times the magnitude of vector V squared
can be represented by using the components of the vectors as follows, distributing the
quantities we obtain the following expression.
In a similar fashion the dot product between vector U and vector V squared is written as
follows.
Expanding the expression we obtain the following.
Notice that if we take the expression that represents the product of the magnitudes squared
and subtract it from the expression representing the dot product squared we obtain the same
expression for the magnitude of the cross product squared.
So this long expression can be written as the difference between these two quantities.
This expression is referred to as Lagrange’s identity which was introduced in the previous
video.
Now that we have rewritten the right hand side using these expressions let’s go ahead
and use the geometric definition of the dot product and write vector U dotted with vector
V as the magnitude of vector U times the magnitude of vector V times cosine of theta.
Then we use the property of exponents to rewrite this expression, next we factor out the product
of the magnitudes squared, from here we replace 1 minus cosine of theta squared with sin of
theta squared by making use of the Pythagorean identity sin squared plus cosine squared equals 1.
The last step is to take the square root of both sides and since theta is an angle between
0 and pi, the square root of Sine of theta will always be positive, and this ends the
proof.
The third geometric property is vector U crossed with vector V is equal to the zero vector
if and only if vector U and vector V are scalar multiples of each other.
In other words two nonzero vectors U and V are parallel if and only if vector U crossed
with vector V is equal to the zero vector.
We can prove this by noticing that if vector U is a scalar multiple of vector V then vector
U can be replaced with scalar c times vector V, then we apply an algebraic property of
the cross product which rewrites the expression to scalar c times the cross product of vector
V with itself, recall that a vector crossed with itself is just equal to the zero vector.
So this entire expression simplifies to the zero vector.
Alternatively vector U and vector V are parallel if and only if theta equals 0 or pi.
Since sine of theta at these angles reduces to zero and therefore the magnitude of vector
U crossed with vector V reduces to the zero vector, and this ends the proof.
The final geometric property was also introduced in the first video of this series.
The magnitude of vector U crossed with vector V is equal to the area of a parallelogram
having vector U and vector V as adjacent sides.
To prove this property we can use a parallelogram with vector U and vector V as adjacent sides
we can find an expression for the area of the parallelogram given the magnitude of vector
U and vector V as well as the angle between both vectors.
In this case the base of the parallelogram will be equal to the magnitude of vector U,
and the altitude of the parallelogram will be equal to the magnitude of vector V times
sine of theta, this expression is nothing more than the magnitude of vector U crossed
with vector V, and this ends the proof.
Now let’s go over a couple of examples that make use of the geometric properties of the
cross product.
Show that the quadrilateral with vertices A, B, C and D is a parallelogram, and find
its area.
Determine if the parallelogram is a rectangle.
Ok here we are given the coordinates of 4 distinct points.
We need to show that these points form a parallelogram.
We do this by showing that both pairs of opposite sides are parallel.
Let’s go ahead and find the vector representation of each side.
Vector AB, AD, CD and CB will have following values for their components.
Notice that vector AB is equal to the negative of vector CD and vector AD is equal to the
negative of vector CB, this means that vector AB is parallel to vector CD and vector AD
is parallel to vector CB since the vectors are a constant multiple of one another.
Now that we have shown that the vertices form a parallelogram we go ahead and find the area
by finding the magnitude of the cross product between two adjacent vectors.
Let’s first find the cross product between vector AB and vector AD.
Setting up the determinant and finding the minors we obtain the following,
simplifying the expression we obtain the following components for the cross product.
Next we go ahead and find the magnitude of this vector, doing that we obtain the following
for the area of the parallelogram.
Lastly we can quickly check to see if the parallelogram is a rectangle by computing
the dot product between the same vectors, if the dot product is equal to 0 then the
vectors are orthogonal and therefore represent the sides of a rectangle.
Computing the dot product we obtain negative 2 since the dot product is not equal to zero
this parallelogram is not a rectangle.
Alright let’s try the final example.
Find the area of a triangle that contains the points P, Q and R.
This problem looks very familiar.
The points in this problem are exactly the same from an example in the previous video.
In this case we are now asked to find the area of the triangle as oppose to a vector
that is orthogonal to the plane.
We can find the area by making use of the geometric definition of the cross product.
So let’s first find the area of a parallelogram by crossing two adjacent vectors we will then
divide that area in half to obtain the area of the triangle.
Let’s find the component form of vector PQ and vector PR, then we go ahead and cross
these two vectors with one another, simplifying the expression we obtain the following components.
Next it is just a matter of finding the magnitude of this vector, computing the magnitude and
simplifying we obtain the following value.
Since this value represents the area of a parallelogram we need to divide this value
in half to obtain the area of the triangle doing that we obtain the final answer equal
to 2 times the square root of 6.
Alright in our next video we go over our first applications that make use of the cross product.