Application Of Linear Algebra To Lines & Planes

 

Direction of normal N to a plane

Basically, we are looking at how lines and planes relate in three dimensions and how linear algebra can be applied to extract information.. For example, a generic line and plan posed in 3-dimensions is shown in the diagram. The vector N here defines the direction of the normal of the plane, that is, the direction perpendicular to its plane. Note that Q is a point in the plane, and P is a point defined along the vector that links up with point P which is not in the plane but is along the vector normal to it. 

A kind of classic problem would ask to find the value of t such that a certain vector (not to be confused with the coordinate!):

X = P + tN, also satisfies:   (X – Q) · N = 0.

Why do we write: (X – Q) · N = 0? On what basis?

Well, it’s because we are looking at the vector dot product. For any two vectors, A and B, the dot product is defined:

A·B = AB cos(Θ) where Θ is the angle between them

Obviously if two vectors are perpendicular to each other then Θ = 90 degrees, and cos(Θ) = 0. Thus, in this case: (X – Q) ·N = 0

Both conditions can actually be integrated into one expression which must be satisfied, to find t:

(P + tN – Q) ·N = 0 or

(P – Q) ·N + tN ·N = 0

So we can solve for t: t = (Q – P) ·N/ N ·N
.
Example Problem  (1) Find the equation of the line in (two-dimensional space) perpendicular to the point A(-5,4) and passing through the point (3,2)

Solution: This example shows that we can apply the preceding formulations for 2-space as well as 3-space. In this case, we can write: (x, y) = P + tA

Which yields: x = 3 – 5t and y = 2 + 4t

So we have the simultaneous equations:

x = 3 – 5t
y = 2 + 4t
------------

Subtracting, we get: 4x + 5y = 22

Which is the equation of the line.

Example  Problem(2):

Show that the lines: 3x – 5y = 1 and 2x + 3y = 5 are not perpendicular

Solution: From the equations, the points in question are: A= (3, -5) and B = (2,3)

As we saw earlier, i.e. the requirement for the dot product to hold:

Cos(Θ) = (A ·B)/ [A][B] = 0

Where [A] = {(3) ² + (-5) ²} ^1/2 = (9 + 25) ^1/2 = (34) ^1/2

And: [B] = {(2) ²  + (3) ² } = (4 + 9) ^½  = (13) ^1/2

And: A · B = {(3) · (2) + (-5)(3)} = {6 – 15) = -9

 cos(Θ) = -9/ {(34) ^½   x (13) ^1/2}   ≠  0

So the lines aren’t perpendicular!

Example Problem (3):

Find the equation of the plane perpendicular to the vector N at (1, -1,3), and passing through the point P= (4, 2, -1).

Solution:  Here, we want:

(x, y, z) = P· N

Using the ordered triples in 3-space, we can write the requirement for the plane perpendicular to N and passing through P in a plane:

x – y + 3z = P ·N

And: P ·N = [(4 ·1) + (-1) · (2) + 2 · (-1)] = -1

Therefore P· N = -1 and:

x – y + 3z = -1

Problems for the Math Whiz:

(1) Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point:

  P= (2, p, -5)


(2)Find the cosine of the angle between the planes:

 x + y + z = 1 and x – y – z = 5