Welcome to an Introduction to Vectors in Space or in 3D.
The goals of this video are to determine the component form of a vector in space, determine the magnitude of a vector in space,
and also to determine a unit vector in space.
I think a good place to start is to take a look at the graph of a vector in 3D.
So here we have the XYZ coordinate system.
And in blue here we have the graph of a vector where it's initial point is at the origin, 0,0,0, and the terminal point is at 3,1,5.
So we rotate the coordinate system you can get a better idea of the location of this blue vector. And this is an example of a vector in space.
And of course as we change the terminal point of this vector, its location moves, and now I can see the vector is below the XY plane as we see here.
This is also called the position vector because it's initial point is at the origin 000 . A lot of time if we are asked to graph a vector in space on our paper,
it's helpful to draw a box to give it a better perspective, in 3D. Okay, so for a vector in space with initial point x1 y1 z1 and terminal point x2 y2 z2,
The unique position vector V, can be found by subtracting the x coordinates, y coordinates and Z coordinates.
Resulting in what's called the position vector ABC.
Noticed the special notation that we use here, there pointed brackets to represent a position vector. So this means that the initial point,
is at the origin 000 and the terminal point would be ABC. One more thing I want to mention here, to note a vector you can either use,
a lower case letter in bold or if you don't have the option of using bold print, you can use the letter V with little vector or little arrow above it.
The magnitude or lengh of a vector in spacer or R3, is very similar to the magnitue of a vector in R2,
except now we have to take the square root of the sum of the squares of the X,Y and Z. components.
And then in order to find a unit vector in the direction of vector V in space, we need to divide the original vector V
by the magnitude or multiply the vector by one over the magnitude. That'll result in a vector with the same direction that has that length or magnitude of 1.
Let's talk a little bit more about unit vectors. I, is unit vector 1,0,0 or the unit vector along the positive x-axis.
Vector J, 0,1,0 ,is a unit vector along the y-axis. And vector K 0,0,1 is a unit vector along the positive z-axis.
And with these vectors, I,J and K, we can then rewrite any vector as a sum or difference of these unit vectors.
So for example if we have the position vector ABC, we rewrite it as a sum or difference of I,J, and K, as A times vector I, plus be B times vector J,
plus C times vector K. Let's go ahead and take a look at some sample problems.
Let's start by plotting the position vector 2,3,4 on the XYZ coordinate system.
Remember this is the x-axis let's go an extended in the negative direction. This would be the y-axis.
This is the positive y-axis this would be the negative y-axis. And this is the z-axis and this would be the negative z-axis.
So this vector has initial point at the origin and its terminal point at the point 2,3,4. Let's go ahead and plot the point 2,3,4.
So It will move out 2 units in the positive direction along the x-axis, 3 units in the positive direction along the y-axis.
And from here will construct a parallelogram in the XY plane, so it will look something like this.
And from here we have to go up four units along the z-axis. So let's go ahead and go to the z-axis and go up 4 units.
And then from here we're gonna go ahead and construct a box to give it a better perspective.
So let's go ahead and draw a edge here that is parallel to the x-axis. And then go ahead and connected it down. And then we'll draw anothr edge here that is parallel
to the y-axis.
And down. And now we can go ahead and complete this box by completing the top of the box, something like this.
So the point 2,3,4 is this point here, at that corner of the box. And so the vector want to sketch starts at the origin, and terminates at that point.
So it looks something like this. Now It's a little cramped but you can kinda see that this does give it a perspective of 3 dimensions.
Now we're given a vector in space with initial point 5, negative 1,1, and terminal point negative 1,1,4. We first want to determine the position vector.
Remember the position vector is the vector in the same direction with the same magnitude, but has the initial point at the origin.
So to determine the position vector we'll subtract the initial coordinate from the terminal coordinate.
So we'll have negative 1 minus 5, comma, 1 minus negative 1,comma, 4 minus 1.
let's go ahead and simplify this, we'll have negative 6,2, and 3.
This is the position vector. Again it's the vector has the same magnitude and the same direction as the given vector,
but has the initial point at the origin and terminal point at negative 6,2,3.
now that we have the position vector we can rewrite this vector as a linear combination of vectors I,J and K.
So vector V could be written as negative 6i, plus 2j, plus 3k. The same vectors written as a linear combination of the unit vectors IJK.
Now we want to determine the magnitude of vector v. That's gona equal to the square root of the sum of the squares of each component.
So we'll have negative 6 squared, plus 2 squared, plus 3 squared. Let's go ahead and simplify this. Look like we'll have 36 plus 4, plus 9.
So the magnitude of vector V is equals to the square root of 49, which is equal to 7.
So to find the unit vector of vector V, for the vector that has a magnitude of 1 but has the same direction as vector V,
That gona be be equal to 1 over the magnitude of vector V, times vector V. So we'll have 1/7 times vector V. Negative 6,2,3.
We'll go ahead and preform this scalar multiplication. So the unit vector for vector V will be negative 6/7,2/7 and 3/7.
We'll go ahead and finish by talking about vector operations in space or R3. And this is very similar to vectors in R2.
If we want to add or subtract vectors in space, we just need to add or subtract the corresponding components.
And scalr multiplication if we want to find k times vector U, we just multiply k times each component of vector U as we see here.
And since we performed similar procedures in R2, I'm not going to take the time to show examples for this in R3.
I hope we found this video helpful thank you for watching