Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Equations of
planes
Equations of Planes
In the first section of this chapter we saw a couple of equations of planes
...
We would like a more general equation for planes
...
Let’s also
suppose that we have a vector that is orthogonal (perpendicular) to the plane, n = a,b, c
...
Now, assume that P = ( x, y, z ) is any point in the plane
...
Here is a sketch of all these vectors
...
Also notice that
we put the normal vector on the plane, but there is actually no reason to expect this to be the case
...
It is completely possible that the normal vector does not
touch the plane in any way
...
In particular it’s orthogonal to r − r0
...
In other words,
(
)
n i r − r0 = 0
⇒
This is called the vector equation of the plane
...
Start with the first form of the vector
equation and write down a vector for the difference
...
Often this will be written as,
ax + by + cz = d
where d = ax0 + by0 + cz0
...
Notice that if we are given the
equation of a plane in this form we can quickly get a normal vector for the plane
...
Example 1 Determine the equation of the plane that contains the points P = (1, −2, 0 ) ,
Q = ( 3,1, 4 ) and R = ( 0, −1, 2 )
...
We need to find a normal vector
...
We can form the following two vectors from the given points
...
Notice as well that there are many possible vectors to use here, we just chose two of
the possibilities
...
Since both of these are in the plane any vector that is orthogonal to both of these will also be
orthogonal to the plane
...
i
j k i
j
n = PQ × PR = 2 3 4 2 3 = 2i − 8 j + 5k
−1 1 2 −1 1
The equation of the plane is then,
2 ( x − 1) − 8 ( y + 2 ) + 5 ( z − 0 ) = 0
2 x − 8 y + 5 z = 18
We used P for the point, but could have used any of the three points
...
Solution
This is not as difficult a problem as it may at first appear to be
...
This is n = −1, 0, 2
...
This is v = 0, −1, 4
...
If you think
about it this makes some sense
...
So, if the two vectors are parallel the line and plane will be orthogonal
...
i
j k i
j
n × v = −1 0 2 −1 0 = 2i + 4 j + k ≠ 0
0 −1 4 0 −1
So, the vectors aren’t parallel and so the plane and the line are not orthogonal
...
If the line is parallel to the plane then
any vector parallel to the line will be orthogonal to the normal vector of the plane
...
Let’s check this
...
So, the line and the plane are neither orthogonal nor parallel