Anyone who’s taken a course in analytic geometry knows how much fun it can be. However, one can explore aspects of simple lines and planes, equations of intersections etc. just by the use of some basic linear algebra – which we’ll use in this installment.

Basically, we are looking at how lines and planes relate in three dimensions. 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

Now, question for those who’ve already done some vector analysis: 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(theta) where theta is the angle between them

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

Now 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

Let’s look at some problem applications!

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)

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, to get: 4x + 5y = 22

Which is the equation of the line.

Example (2)

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

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

O be perpendicular (as we saw earlier as the requirement for the dot product to hold:

Cos(theta) = (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)^1/2 = (13)^1/2

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

And cos(theta) = -9/ {(34)^1/2 x (13)^1/2} not equal 0

So the lines aren’t perpendicular

Example (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).

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

Additional problems to work on your own:

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

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

2x – y + z = 1 and 3x + y + 2 = 2

(3)Find the cosine of the angle between the planes: x + y + z = 1 and x – y – z = 5

The solutions to these will be given in the next installment along with the solutions to the matrix groups problems!

Basically, we are looking at how lines and planes relate in three dimensions. 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

Now, question for those who’ve already done some vector analysis: 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(theta) where theta is the angle between them

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

Now 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

Let’s look at some problem applications!

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)

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, to get: 4x + 5y = 22

Which is the equation of the line.

Example (2)

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

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

O be perpendicular (as we saw earlier as the requirement for the dot product to hold:

Cos(theta) = (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)^1/2 = (13)^1/2

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

And cos(theta) = -9/ {(34)^1/2 x (13)^1/2} not equal 0

So the lines aren’t perpendicular

Example (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).

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

Additional problems to work on your own:

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

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

2x – y + z = 1 and 3x + y + 2 = 2

(3)Find the cosine of the angle between the planes: x + y + z = 1 and x – y – z = 5

The solutions to these will be given in the next installment along with the solutions to the matrix groups problems!

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