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 + t**N** – Q) **·****N** = 0 or

(P – Q) **·****N** + t**N** **·****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!

**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).

P= (2, p, -5)

2x – y + z = 1 and 3x + y + 2 = 2

x + y + z = 1 and x – y –
z = 5

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