BAM! Mr. Tarrou In this video we're gonna be taking a look at three points
in three-dimensional space we're going to find the component form of vector a B
and vector AC and then hence or otherwise we're going to find the area
of the triangle of ABC now you can have a basically three ways of finding the
area of this triangle defined by these three vertices we can use the SAS
formula the fact the fact that the area is equal to one-half the magnitude of
side a time's the magnitude of side B times the sine of C the angle between
those in this case B vectors a B and AC you can find the magnitude the distance
basically between the points of a B BC and AC and use Heron's formula we are
going to use the vector formula for finding the area of a parallelogram and
just simply cut that in half okay in the description below you're gonna find a
video - where I sort of talk about more of the geometry and explain where the
formula comes from for finding the area of a triangle or hence a parallelogram
using vectors but we're just going to apply that formula in this video for as
opposed to introduce an it like it in the other video first find the vector a
B okay so we have an initial point and a terminal point and so to find that
vector in component form we work with the terminal point of the vector first
so we have the coordinates of zero four and one movement along the x y&z axis
we're going to subtract that with the components or the x y&z components
excuse me of the initial point so we have zero minus negative 3 4 minus 1 and
1 minus 2 giving us a vector of 3 3 and negative 1 and this is how my my book
that I'm currently teaching out of Wright's vectors in component form in a
stacked format you might see some arrow brackets and these
components moving horizontally you could also write this as 3i plus 3 J minus 1 K
if you're doing sort of like that full written form using the component vectors
I J and K in your notation vector AC again initial point of a terminal
pointing up C we'll start with the terminal point 1 negative 2 and 5 minus
seen with the coordinates of our initial point negative 3 1 and 2 so we're
looking at vector AC has the components of 4 negative 3 and 3 just before I go
on make sure to make some silly copy error now the area of the triangle
defined by these 3 vertices working with vectors okay so the area of a
parallelogram that is going to be equal to the magnitude of the cross product
between vector a and B just have given you that as a formula well a triangle is
simply half of a parallelogram so the formula we're going to use for this
triangle is that the area is equal to 1/2 the magnitude of the cross product
of vector a B and AC so we are going to find that cross product that vector
remember cross product gives you a vector which is perpendicular to the two
vectors that you use to find that cross product with so cross product of a B and
AC we're going to take basically the derivative determinant excuse me we're
going to set up a component row I J and K and then make a row with the
components from vector a and C okay and I talked about in the video I have a
link in the description that the cross product is not commutative if you'd
reverse order you get an opposite answer but
we're going to be ultimately working with the magnitude which is length so
that's not an issue for us here we're going to be looking for the determinant
of the 2x2 matrix taking out the row the row and column
that has this component vector I in it and we're looking at 3 negative 3
negative 1 and 3 and that's going to create a coefficient for the component
vector moving in the direction of the x axis - so we have a plus minus and then
determinate take out the column and row that has the component vector in the
direction of the y axis so 3 4 negative 1 3 add that with a determinate I'm just
writing this in here is really make sure that you realize that that's what I'm
looking for take out the column and row with the
component vector in the direction of the z axis now we're looking at 9 minus 3 so
3 times 3 is 9 and then minus the product of negative 3 and negative 1
that's going to be a coefficient of vector I - we've got 9 minus negative 4
and the determinant of this 2 by 2 matrix one should have a little vector K
over here we look at it we're looking at negative 9 minus 12 vector K so our
cross product is equal to 6 I minus 13 J plus negative 21 vector K or the cross
product is equal to 6 negative
13 and negative 21 and I got to this point after I made this problem up and
realize my numbers are a little bit gross this was really intended to make
to be a non calculator question but yeah that didn't work out and that's okay I'm
just grab our calculators we work through this now simply the area
triangle is one half of magnitude of the specter so the area of triangle ABC is
going to be equal to one half to find the magnitude of a vector in three space
of course we know we just square each of these components and add them up
basically just the distance formula with one extra dimension in it so we're
looking at the square root of 6 squared is 36 13 squared is 169 and negative 21
squared is 441 and now we're looking at one half the square root of 400 500 560
600 6 36 36 637 646 now grab your calculator if need be and that is going
to come out to be approximately twelve point seven one square units and that is
how you find well component form of vectors in 3-space and a quick review
introduction of how to find the area of a triangle using vectors which are
defining two of the three sides of that triangle i mr true yeah go to your
homework
you