Solving Simple Problems In Linear Algebra (4)

 Basically, in this section, we are looking at how lines and planes relate in three (e.g. x, y, z) dimensions.  This aspect has great relevance for many areas of physics, especially astrophysics and advanced celestial mechanics (i.e. applied to YORP effect in celestial mechanics).   For example, a generic line and plan posed in 3-dimensions is shown in the diagram on the page opposite, with the normal (N) direction to the plane indicated.:

Vector N in relation to plane in x-y-z coordinate system

The vector N here defines the direction of the normal to 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
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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!

Suggested Problems:

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

2) Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point:
  P= (2, p, -5)

3) Find a vector parallel to the line of interception of two planes:

2x – y + z = 1 and 3x + y + 2 = 2

And the cosine of the angle between the planes.

4) Find the cosine of the angle between the planes:

 x + y + z = 1 and x – y – z = 5