BAM Mr. Tarrou in this math lesson we're going to do one example of finding the
distance from a point to a plane that shortest distance between a given point
in a plane and at its core this is exactly what we did in algebra 1 when we
were finding the distance between a given point and a line so first of all
the shirt just reminding you that when you have an opportunity to vote in your
city state or government national government go vote put some say you know
put some input have some input into the leaders of your local area or country we
have a local city election going on today the distance from a point to a
plane is measured along the perpendicular to the plane that contains
a given point so how do you find the shortest distance between a point and a
given plane well you're going to find the normal vector to the plane you're
going to find basically the perpendicular gradient to that plane
right because when I can necessarily work with them the whole vector just the
direction of that normal vector that perpendicular vector to the plane we're
going to find the equation of a line through the given point perpendicular to
the plane so just like in Algebra one what I need to write the equation of the
line a point and a slope only now our slope is going to be in
three-dimensional space we're going to find the intersection point between the
line which is going through the given point that's perpendicular to the plane
and then now you have the court you know now you have the coordinates of two
points you're just simply going to find the distance between those two points
and because we're talking about a a normal vector we're using that gradient
as our slope we're working with a perpendicular line that's going to
measure the shortest distance between the given point and the plane so let's
see what that example looks like coming up right now
so for example we're going to find the distance from the point 3 negative 1/2
to the plane negative 5x plus 3y minus 4z is equal to negative 1 and this quit
this examples going to go fairly quickly because we are giving you the equation
of this plane in cartesian format which means that we can remember one more time
i would show this on almost all of my videos and involve me in this topic we
have vector our position vector of any general point on the plane times the dot
product of our normal vector is equal to vector a position vector for a known
point on the plane dotted with vector n that's what's
creating this constant of negative 1 and that allows us to here to look and see
that our normal vector is equal to negative 5 or has the direction of
negative 5 3 and negative 4 now if the equation of your plane is given to you
in this vector format vector R is equal to vector a position vector for a known
point Plus vector u times some scalar multiple Plus vector V tons times some
scalar multiple well these are two non collinear vectors which are parallel to
your plane basically acting as a to space gradient and you would have to
find the cross product of course between vector U and vector V to find a normal
vector to this plane but we're being kind of nice here the equation of the
plane is given to us in Cartesian format which allows us to pull that right out
now the vector form for the equation of a line is that vector R is equal to
vector a a known point on the plane excuse me known point on the line
position vector plus some vector U times the scalar multiple which I'm going to
call lambda because we're plum outlines land of sorts with an L
that's the only reason so here we have the equation
of a line which is going to go through this point but perpendicular to the
plane we're going to say that this is equal to position vector for a point on
the line 3 negative 1 2 plus now the normal vector this is perpendicular to
the plane but that's what we want the shortest distance between the point and
the plane so we wanted to use that perpendicular so basically perpendicular
to the plane so the direction of the line the slope of line it basically is
negative 5 3 negative 4 now what we need to do is like we talked about in the
last video where is this line intersecting this plane well we want to
basically substitute some equations from this line into the equation of this
plane in the simplest manner which is going to allow us to find the
intersection point and that means that if we right convert this equation of our
line it's in vector form into parametric form where we have X in terms of lambda
lambda Y in terms of lambda Z in terms of lambda lambda we can plug those
expressions for XY and Z and basically find lambda basically find out you know
how far from this point on the line do we need to move before we get a lambda
that satisfies the equation for the plane so that's basically how far we
have to move we won't know the answer until we get to the you know we'll have
like a scalar multiple for this gradient but it'll give us an idea of how far
away from this known point we need to move to get a point that's on this
that's within this line that's also in this plane so our parametric equations
are going to be X is equal to 3 minus 5 lambda Y is equal to negative 1 plus 3
lambda and Z is equal to 2 minus 4 lambda the coordinates you know out of
this vector form where's the x y&z coming from that's basically
the component form of the coordinate that vector R is describing the position
vector for any general point that's on the plane now we can take and the system
of parametric equations and plug it into our equation of our plane and find that
lambda such that the left-hand side is equal to the right-hand side so we have
negative 5 times our substitution for X is going to be 3 minus 5 lambda plus 3
times y which is going to be the expression represented by the expression
negative 1 plus 3 lambda minus 4 times our expression for Z which is 2 minus 4
lambda we're going to set that equal to and solve it for 1 so it's going to give
us with the distributive property negative 15 plus 25 lambda minus 3 plus
9 lambda minus 8 plus 16 lambda is equal to negative 1 we've got 25 34 44 and 50
lambda we've got negative 15 negative 18 negative 25 no negative 26 we're then
going to add both sides by 26 which is equal to 25 and after we divide by 5 we
see that lambda is equal to 1/2 so the point of intersection therefore is going
to be X is equal to 3 minus 5 times 1/2 which is going to be 6 over 2 minus 5
over 2 is 1/2 we have y is equal to negative 1 plus 3 times 1/2
so we're looking at negative 2 over 2 plus 3 over 2 is also 1/2 and then we've
got Z is 2 minus 4 times 1/2 which is 2 minus 2 which is 0 so the coordinate of
our point is going to be equal to 1/2 1/2 and 0 and just to make sure that I'm
not making some kind of mistake here ok excellent and matching my notes so we're
almost done we have the I might even be able to fit this all up on the board
without having to erase anything we've got a point we've got a point of
intersection where we're looking at that perpendicular distance between this
given point P and the point on the plane and we have to find the distance and
maybe I'm not going to be able to get this to fit so let's erase this
can always back up the tape the video right distance is equal to back up the
tape that's funny the difference between the x-coordinates
so we're looking at 3 minus 1/2 squared so the difference in the x-coordinates
the difference in the y-coordinates now negative 1 minus also 1/2 squared plus
the difference in the Z coordinates 2 minus 0 now we just have a little bit of
cleaning up to do here we have 6 over 2 minus 1/2 is 5 over 2 we've got negative
2 over 2 minus 1 over 2 is negative 3 over 2 plus 2 squared so we're looking
at the square root of 25 over 4 where we're all your proper notation there we
got 9 over 2 so plus 9 over ok let's try to get negative 3 squared is 9 and 2
squared is 4 and now this is 4 over 1 so we're multiplying the top and bottom by
4 and we are looking at your distance it's equal to the square root of 25 and
9 is 34 now this is 16 so 34 plus 10 is 44 and 6 makes 50 over a common
denominator of 4 we've got that that is equal to the square root of 4 is of
course 2 and 50 was reducible that's square root of 25 times 2 so we're
looking at 5 square root of 2 units and double check 5 square root of 2 over 2
units there you go so we have an example of how to find the
distance between a point and a plane i mr true hey go to your homework
you