BAM Mr. Tarrou. In this math lesson we're going to take a look at an example of
finding the angle between two planes and I've got now quite a large list of math
lessons dealing with the mathematics of vectors I'll put a link to that playlist
in the description if you'd like to check out some more of my other lessons
but we have this diagram here where we have an orange plane and a purple and a
blue plane intersecting at some kind of angle how are we going to find that well
the angle theta between a plane with normal vector m and a plane with a
normal vector n is equal to the angle between the lines in the direction of
those two vectors so there's going to be related as a similarity here between
finding the angle between two lines and the angle between the intersection of
two planes one similarity is we do have these planes that extend in all
directions not like a vector that has an initial point and just goes in one
direction so when we found the angle between the between two planes excuse me
when we found the angle between two vectors we were accepting answers
between 0 and 180 degrees well with these planes and lines extending in
multiple directions you know lines 2 and planes in all kinds of directions we're
really going to be concerned with that acute angle between the intersection not
necessarily the obtuse angle which of course you just subtract from 180 but
that's why they're talking about lines going in the direction of the vectors
and not just measuring the angle between the vectors themselves we're going to
limit those answers between not 0 180 degrees but between 0 and 90 degrees
okay so we have theta is equal to the arc cosine of the absolute value of the
dot product between vector N and vector and a dot product scalar product inner
product whatever your textbook is calling it divided by the magnitude of
vector M times the magnitude of vector n now when we found the angle between two
vectors it was the same looking that was very it
was the same formula basically except there was no absolute value function
here we had that absolute value function there when we were finding the
intersection or finding the angle between two lines why is it there well
in terms of planes how are you going to find these normal
vectors if your planes given to you in a Cartesian format you can just pull the
normal the directional vector that normal vector right out of the equation
of that plane if it's in Cartesian format if you have the equation of your
plane given to you is that original vector format which is vector are
describing position vector for any point along in the plane equal to vector a
position vector for any known point on the plane plus u times maybe alpha +
vector V times a scalar of beta and we had that that addition of the scalar
multiples of two non collinear vectors well you would find that that normal
vector by doing the cross-product between the two non collinear vectors
which are the directional those sort of gradient vectors for that particular
plane well we know that the cross product is not commutative right the
cross product of V and U is well if I have that the cross product of vector U
and vector V is vector M then if I change the order I'm going to get an
opposite vector that's mainly why the absolute value functions are in that
equation given us that's going to give us the equation for theta and looking
here if we were looking for that acute angle between those two vectors or in
this case between those two planes if we take an opposite vector we're just
simply going to get the obtuse angle
like think about your unit circle from way back write the cosine of 30 degrees
is square root of 3 over 2 the cosine of the supplement of this
angle the obtuse part of it would be 150 degrees now on the when we're studying
unit circle we would just highlight how this is in quadrant one and this is in
Quadrant two with that same 30 degree reference angle you would get not the
same answer but opposite answers so the absolute value function is in here to
make sure that the angle that we get is going to be that acute angle that we're
looking for between the intersection of those two planes if you happen to set up
your cross multiplication and get your normal vectors so that you would have
gotten that obtuse angle well that's all that we have here our cute obtuse the
only difference in the answer is the sine so the absolute value function will
take care of that let's take a look at our first example right now for our
example we're going to find the angle between the planes defined by a couple
of equations in different formats we have one here that is in Cartesian
format and we have one here that's in our sort of original standard vector
form for the equation of a plane so we can see how to handle these differently
and this part here are playing X minus 3y plus 2z equals one is given to us in
cartesian format and if we're looking for a normal vector that's awesome
because vector R dotted with vector N equals vector a dotted with vector n
this equation for a plane using a normal vector just comes right out and gives us
an equation in cartesian format so this one is the resultant of a dot product
inner product scalar product of vector a and vector n two vectors with you know
all of our components we know they're all constants given us a numerical value
over here just to really point this out we have vector R a position vector for
every general point on our plane XY and Z the indata ed with
our normal vector our coefficients one negative three and two here right there
is our normal vector one negative three two four one of our planes
I'll drawn our attention over here this format is great if I want a couple of
non collinear parallel vectors to this plane but it is going to cause it's
going to require us to do a little bit of extra work for finding that normal
vector cross product so if this is vector U and vector V the cross product
of vector u with vector V is equal to now let's pull off to the side and give
herself a little bit of guide for scratch work 3 by 3 matrix you've been
seeing me do this now probably I hope if you're watching more than just one video
quite a bit we're going to set up this 3 by 3 matrix and set up with go through
the determinant process by setting up some small 2x2 matrix we're going to
take out that row and column with our I component and we're going to take the
determinant of matrix 2 0 negative 1/2 going across the rows and that's going
to give us a scalar multiple for the I component now we have a positive
determinant minus the next determinate we're going to take out the row and
column with component J negative 3 0 - 2
given us a scalar multiple for our J component and then finally we're going
to add our last determinate so again that positive - positive determinate
taking out our row and column where the K component is to create a coefficient
for our K component vector negative 3 - 2 negative 1 times vector K that's going
to say that give us that the cross product is equal to these components we
have 2 times 2 is 4 - 0 multiply down positive multiply up diagonally that's
going to give us a subtraction value minus negative 3 times 2 is negative 6
and that's zero or J component and then finally plus negative three times
negative one which is three minus two times two which is four for our K vector
and our cross product is going to yield us that normal vector which has the
components of four six negative one I'm pausing because I've see that there's an
error somewhere that error is the fact that I have a negative in my notes and
right here I wrote a positive but you know what let's just roll with it I'll
grab a calculator and finish up my angle when we're done here
so we have this as our normal vector M to this equation vector equation of a
plane which is not what I intended to do but I don't want to reshoot the scene
over just because I forgot to copy a minus sign or negative R angle grab some
maybe orange chalk here theta is going to be equal to the arc cosine of the
absolute value of the dot product between vector n which is 1 negative 3
and 2 dotted with this normal vector here of 4 6 negative 1 absolute value
that again except it's not going to leave me enough room for my magnitudes
the magnitude of vector n is going to come from doing 1 squared plus negative
3 squared plus 2 squared times the square root of or find the magnitude of
vector M or squared plus 6 squared plus negative 1 squared and squeeze that in
right there beautiful so we have B theta is equal to the arc cosine
four plus negative 18 plus negative two over the square roots we've got nine ten
14 times the square root of 36 and 16 which is 36 46 52 53 in with the
absolute value we're looking at theta is equal to the arc cosine of negative 20
negative 16 over the square roots about 14 times the square root of 53 absolute
value which is gonna get rid of that negative and that comes out to be with
the aid of my little calculator here fifty four point oh two nine degrees
hope that helps you understand how to find the angle between the intersection
of two planes say by the way when you're doing your work make sure you copy those
answers or the problem rate to begin with so you get the right answer even
though you need to get all these points marked off just because you have copier
like you might lose a point on the IB exam anyway I missed a true Pam go to
your homework
you