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Vectors

Vectors

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Equations of a plane

We've seen in previous sections that a line is uniquely determined by two points. But there are infinitely many planes which contain those two points. Planes are flat, two-dimensional surfaces extending infinitely in all directions. Our typical Cartesian plane is one of them. You could think of the computer screen you're reading this on as a plane. The point is, in three dimensions, planes are everywhere -- sort of the three dimensional analogue of a line in two dimensions or a point in one.


(picture of planes)

Discussion

How do you think we could uniquely define a plane? That is, if two points don't define one specific plane, what does?

Imagine you’ve already seen that two points, say A and B, only pin down the straight line AB—and you can still “hinge” infinitely many flat sheets through that hinge line. So what extra bit of information kills all those tilts and fixes exactly one sheet?

  • Three non-collinear points

  • • Pick A and B—there’s your hinge line.

  • • Now grab a third point C that doesn’t lie on AB.

  • • That C forces a second direction (from A toward C) as well as the first (from A toward B). • With those two directions nailed down, the flat sheet can no longer swivel—it’s rigidly fixed.

One point plus two sliding directions

  • • Imagine a peg at A, and two straight rails through A, each pointing a different way.

  • • You can “slide” along rail 1 some distance, then along rail 2 some distance.

  • • Every spot on that flat board is reached this way—and you can’t tilt the board without breaking contact with one of the rails.

  • • Thus: a point (A) together with two independent sliding directions uniquely carve out one plane.

All spots at the same “height” from a fixed upright stick

  • • Picture a stick standing straight up through a board, and a little weight hanging on a string from the stick’s top.

  • • Wherever you move on the board, the weight’s hook sits the same distance down the stick.

  • • In other words, every point on the board shares the same “shadow-length” onto that stick.

  • • That common “shadow-length” plus the stick’s direction pick out exactly one flat surface.

You can also think of two non-parallel lines crossing at a point—their union fixes two directions, just like the rails in the second example. All of these are really just three ways to say the same thing: you need enough independent “degrees of freedom” (two directions) plus a base location (one point) so that no more tilting or sliding can occur.

Just as there are several ways of expressing the equation of one line, there can be several ways to express the equation of one plane. One of these is via vectors.

Vector form

A plane in 3D space can be described by a vector equation involving a fixed point and two direction vectors lying in the plane. Planes are often denoted by Π (capital pi).


If the position vector of the fixed point is a, and two non-parallel direction vectors in the plane are b and c, then the plane is represented by:

Π:r=a+λb+μc📖

Here, λ and μ are parameters that can take any real values, allowing r to move freely across the entire surface of the plane.

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Example

Find the vector equation of the plane through points A(1,2,3), B(2,1,4), and C(3,2,1).


First, form direction vectors from one point, e.g., A, to the others:

AB=⎝⎛​2−11−24−3​⎠⎞​=⎝⎛​1−11​⎠⎞​,AC=⎝⎛​3−12−21−3​⎠⎞​=⎝⎛​20−2​⎠⎞​

Thus, a suitable vector equation is:

Π:r=⎝⎛​123​⎠⎞​+λ⎝⎛​1−11​⎠⎞​+μ⎝⎛​20−2​⎠⎞​

Checkpoint

Find the vector equation of the plane through points A(0,−4,21​),B(5,6,1),C(−3,1,0).

Select the correct option

Exercise

The plane Π is defined by the equation

x=y

Find the value(s) of a for which the point ⎝⎛​6−14​⎠⎞​ lies in the plane Π.

Select the correct option

Since the vector equation of a plane can be equivalently expressed in many different ways, it can be difficult to see whether two vector expressions represent the same plane. It's for this reason that we often give the equation of a plane based off of a normal vector to that plane.


A vector which is perpendicular to a given plane is called its normal vector. Since a plane is basically a two dimensional sheet, the normal vector is perpendicular to every line in the plane. We can thereby find the normal vector by taking the vector product of the direction vectors in the form r=a+λb+μc:

n=b×c

The normal is not unique - any vector parallel to this normal will also be normal to the plane. Further, all planes perpendicular to n are parallel to each other, so you can pick out one of those planes by specifying a point on it.

Discussion

Let's say you want to write the equation of a plane. You know the point A is in the plane. You also know the vector n is normal (perpendicular) to the plane.

Can you express the relationship between the point A and the vector n using a formula for the scalar product?


Does this uniquely define a plane?

Suppose P is any point in the plane and A is the given point. The vector from A to P, AP, lies entirely in the plane. Since n is perpendicular to the plane, it must be perpendicular to every such AP. Hence

AP⋅n=0


Next introduce position vectors: let a be the position vector of A and r be that of a generic point P. Then

AP=r−a

so the condition becomes

(r−a)⋅n=0


As long as n=0, this single equation picks out exactly the plane through A with normal n. (Scaling n by any nonzero constant gives the same plane; if n=0 the equation is trivial and does not define a plane.)

This relationship is known as the scalar product form of a plane.

Scalar product form

The scalar product form of a plane uses a vector perpendicular ("normal") to the plane and one known point in the plane. If a point in the plane has position vector a and n is a normal vector, then any other point r lies in the plane if:

Π:r⋅n=a⋅n📖

This equation expresses the idea that the vector from the known point to any other point in the plane is always perpendicular to n.

Powered by Desmos

Example

From the vector equation

Π:r=⎝⎛​123​⎠⎞​+λ⎝⎛​1−11​⎠⎞​+μ⎝⎛​20−2​⎠⎞​

the direction vectors are ⎝⎛​1−11​⎠⎞​ and ⎝⎛​20−2​⎠⎞​.

Their cross product is

n=⎝⎛​1−11​⎠⎞​×⎝⎛​20−2​⎠⎞​=⎝⎛​242​⎠⎞​∥⎝⎛​121​⎠⎞​

Using a=⎝⎛​123​⎠⎞​ as a known point, the plane’s normal form follows from r⋅n=a⋅n. Since

a⋅n=1×1+2×2+3×1=8

the plane’s normal form is

Π:r⋅⎝⎛​121​⎠⎞​=8

Checkpoint

A plane is given by the equation

Π:r=⎝⎛​1−24​⎠⎞​+λ⎝⎛​320​⎠⎞​+μ⎝⎛​−511​⎠⎞​

It is given that the vector n=⎝⎛​011​⎠⎞​ is normal to Π.

Give the scalar product form of Π.

Select the correct option

Exercise

A plane is given by the equation

Π:r=⎝⎛​012​⎠⎞​+λ⎝⎛​−233​⎠⎞​+μ⎝⎛​1−15​⎠⎞​

Find the equation of a vector normal to Π.

Select the correct option

We can also write the equation of a plane in a form that looks more similar to our two-dimensional, Cartesian equations.

Discussion

Let the equation of a plane be given by the scalar product form r⋅n=d, and define n=⎝⎛​n1​n2​n3​​⎠⎞​,r=⎝⎛​xyz​⎠⎞​.

Can you rewrite r⋅n=d in terms of x,y, and z?

We recall that for any two vectors u=(u1​,u2​,u3​) and v=(v1​,v2​,v3​), their dot product is

u⋅v=u1​v1​+u2​v2​+u3​v3​

Here r=(x,y,z) and n=(n1​,n2​,n3​), so

r⋅n=(x,y,z)⋅(n1​,n2​,n3​)=xn1​+yn2​+zn3​

Thus the plane equation r⋅n=d becomes

n1​x+n2​y+n3​z=d

This is known as the Cartesian equation of a plane.

Cartesian equation of a plane

The Cartesian equation of a plane with a normal vector n and containing a point with position vector a is

n1​x+n2​y+n3​z=d📖

where n=⎝⎛​n1​n2​n3​​⎠⎞​,d=a⋅n.

Example

If a plane Π has equation r⋅⎝⎛​121​⎠⎞​=8, then letting r=⎝⎛​xyz​⎠⎞​:

x+2y+z=8

Checkpoint

A plane has normal vector n=⎝⎛​1−52​⎠⎞​ and contains the point A(3,7,0).

Find the equation of the plane in Cartesian form.

Select the correct option

Exercise

A plane is given by the equation

Π:r=⎝⎛​−231​⎠⎞​+λ⎝⎛​111​⎠⎞​+μ⎝⎛​013​⎠⎞​

It is given that the vector n=⎝⎛​2−31​⎠⎞​ is normal to Π.

Give the Cartesian form of Π.

Select the correct option

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