Question
What is the answer to a multiplication problem called in mathematical terminology?
Solution
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In mathematical terminology, the answer to a multiplication problem is called the product.
Definition:
If two numbers, called factors, are multiplied together, the result is known as the product:
If two numbers, called factors, are multiplied together, the result is known as the product:
Example:
Here, 5 and 3 are the factors, and 15 is the product.
This terminology applies in various mathematical contexts, including algebra, arithmetic, and higher-level mathematics, where multiplication is used to determine scalar products, vector products, and matrix products.
Exercise- Vector Multiplication
Exercise 14 - Multiplication/Division of fractions/mixed numbers
Exercise 13 - Filling the blanks of the statements
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Exercise- Vector Multiplication
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Hello, in this exercise,
we're given 4 vectors,
A and B, which are 2-dimensional,
and C and D, which are 3-dimensional.
We're asked to find 3 different things.
The first, let's start with A,
is we're supposed to find the scalar multiplication of vectors A and B.
If you recall a scalar multiplication or A vector dot B vector,
remember the dot is equal to A_x,
the x component of A times the x component of B,
B_x plus A_y times B_y.
If we multiply that out,
1 times 1 plus 2 times negative 3,
that's 1 times 1 plus 2 times negative 3.
Our result is 1 minus 6,
which is negative 5.
There you've solved part a.
Of course, there is a second way as well.
We could have found the magnitude of the vectors A and B.
If we knew the angle between the 2 of them,
we could have solved multiplying the magnitude of vector A times
the magnitude of vector B times
the cosine of the angle between them, which we call Alpha.
Of course, in this case,
we didn't have the angle,
and we did have the points.
As soon as you have the points,
it can often be easier to just multiply the x components,
multiply the y components and the z components if there
are z components and find the answer that way.
Keep in mind that negative 5 is only a magnitude,
is a scalar, there's no direction, it's not a vector.
Moving on to part b,
it asks us to do a vector multiplication of the vectors A and B.
Again, we're still in 2-dimensions,
so we can just use our formula.
If you recall the formula A cross B,
which is the vector multiplication of A and B,
is equal to A_x, the x component of A.
Again, A_x times by minus A_y times B_x.
Remember, the order here is important.
This isn't commutative, so A cross B is not the same as B cross A.
In fact, it's equal to the opposite.
If we had B cross A, we would have the opposite result.
If we solve this,
A_x is 1 and by is negative 3.
If we write that out, our first portion is 1 times negative 3.
A_y is 2 and b_x is 1.
Minus 2 times 1,
we get negative 3 minus 2,
and our answer is again minus 5.
This is just a coincidence.
It won't always happen that
your scalar multiplication and your vector multiplication are the same.
This is pure coincidence.
In fact, it's rather unlikely.
Now this is our magnitude,
but if we're looking for the direction,
because this is a vector multiplication,
and it needs both magnitude and direction, length and direction.
We need to remember because we're only given the x and the y coordinates
that the vector will end up going in the direction of the z-axis.
We can write out that the magnitude is negative 5,
and it's in the direction z-hat.
Of course, there's a second way to do this multiplication as well.
We could have taken the magnitude of A times the magnitude of
the vector B and multiply that by the sine of the angle Alpha,
the angle between the 2 vectors.
Of course that's if we're given that angle,
in this case we weren't.
The easier path was what we did earlier.
We would have had the same answer,
negative 5 in the direction of z.
But the way we did it, given our data, was easier.
For part c,
we are given the vector multiplication of C and D. Now these are 3-dimensional vectors.
We can't do the operation we did above.
We need to use the determinant.
Now, if you recall,
if we're doing C cross D,
the way we do a determinant is created
a matrix in 3-dimensions that will have 3 rows and 3 columns.
In the first row, assuming that we're using Cartesian coordinates,
which I can assume because the points here are
given in coordinates that seem to be Cartesian,
I'll write in my first row,
x-hat, y-hat, and z-hat.
In the second row,
remember the order here is important.
I'm going to write my C coordinates,
the x coordinate first,
the y coordinate second,
and the z-coordinate third.
Again, the order is important.
C_x, C_y and C_z.
Again, my first component of the multiplication.
In the bottom row,
D_x, D_y, D_z.
Of course, I have to put my 2 lines bracketing this in,
so I know that there's a determinant.
If we put in our values x,
y, and z stay the same of course.
You have x-hat, y-hat, and z-hat.
For C_x, our value is negative 1.
C_y is 2,
and C_z is negative 2.
D_x is 2,
D_y is 0 and D_z is 1,
and we can close our determinant.
The procedure for determinant is to go component by component first x,
then y, then z, and do a special operation for each one.
What we're going to do is cross out the x, the y,
and the z every time and then take the row,
the column that is of the variable we're working with.
First, we'll do x, and we cross out
that column and do cross multiplications on the diagonals.
We're going to do x, but x-hat in cross out the row and the x column,
and we're left with the y column and the z column. What do we do?
We go on our main diagonal first.
We multiply C_y by D_z,
which gives us 2 times 1.
Then we're going to multiply on the secondary diagonal C_z,
by D_y negative 2 times 0.
In a parenthesis first we'll write x-hat outside of the parenthesis and then within
the parentheses will write 2 times 1 minus negative 2 times 0.
Now I'm going to subtract my y element.
If you recall, we do positive and negative,
than positive and negative, so on and so forth with the determinant.
For my y component,
I'll be subtracting it.
Negative y-hat will be,
remember we're left with,
once we cross out my y row and my y column,
I'm left with D_x and C_x,
and I'm left with C_z and D_z.
First I do my main diagonal,
C_x times D_z,
which is negative 1 times 1 and subtract from that my secondary diagonal,
which is negative 2 times 2.
Now if I move on to my z component,
I'm going to add that because again,
positive, negative, positive, negative.
I put in my z-hat and once again,
I'm going to cross out a different component.
I'm going to cross out the same row,
the x, y, z row,
but this time the z column.
I'm left with D_x and C_x, D_y and C_y.
My main diagonals negative 1 times 0, I subtract from that,
my secondary diagonal negative 2 times 2 and that's the z-hat element.
If I solve here,
my x element will be 2,
2x-hat minus y-hat,
but my element is actually negative 5,
so it's plus 5y-hat and I add my z element,
which actually is negative.
I subtract 4z. Once again,
keep in mind our solution here is a vector, it has direction.
Vector multiplication will always give you a vector,
a scalar multiplication will give you a magnitude or length,
not a vector, it will not have direction.
Now let's assume for a moment that I did want only the magnitude,
only the length of this vector.
We'll call that vector e,
or new vector that we've found is a solution of
C cross D. But if I want the magnitude of that,
signify the magnitude like so.
What I do is take the magnitude of my result here, and how do I do that?
I take the square root of the square values of each of the components.
That means that the magnitude of the vector e will equal the square root of 2^2
plus 5^2 plus 4^2 or negative 4^2.
Of course, my solution will be the same.
The square root of 2^2 is 4 plus 25,
which is 5^2 plus 16,
gives me the square root of 45 and that is the magnitude of the vector e. Of course,
I can always do the second way,
just like we did above in part b. I can find the magnitude of the vector e. Again,
my new vector, the solution,
the same way that I did in the vector in part b. I could take
the magnitude of the vector C multiplied by the magnitude of the vector D,
multiplied by the sine of the angle Theta between these 2.
Now I don't have that angle Theta.
In 3-dimensions, it's rather hard to come by that angle.
I did this in the first method,
but that's always possible.
You've solved parts a, b,
and c, and that's the end of the exercise.
This video explains how to solve three different problems related to vector multiplication. In the first problem, the scalar multiplication of vectors A and B is found by multiplying the x and y components of each vector. The vector multiplication of A and B is also found by using the formula A cross B, which results in a magnitude of negative 5 in the direction of the z-axis. The third problem involves finding the vector multiplication of vectors C and D, which are 3-dimensional. This is done by using a determinant and cross-multiplying the components of each vector. The magnitude of the resulting vector can also be found by taking the square root of the squares of each component.
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