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Planes in 3-space are defined by one of the following
1. one point and a normal vector |
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For calculations, it is most convenient to have one point and a normal vector. You can get these from these other two cases. |
2. one point and 2 non-parallel vectors |
If you have one point and 2 non-parallel vectors that lie in the plane, you can take the cross product of the two vectors to get a vector that is perpendicular to the original 2 vectors and perpendicular to the plane (a normal vector). See the practice problems below for examples. |
3. three points that are not collinear |
If we have 3 points that are not collinear, we use these points to get two vectors. Then, by choosing one of the points (any of the three will work), we have the previous case of one point and 2 non-parallel vectors. We then use that procedure to get a normal vector. See the practice problems below for examples. |
There may be times that you are given a point and a normal vector and you need either two vectors in the plane or three points in the plane. See the practice problems below for examples on how to extract that information.
You can determine an incredible amount of information about a plane and it's relationship to other planes just knowing a point and a normal vector. Examples include the angle between two planes including whether two planes are perpendicular or parallel, the distance between a point or line and a plane and other equations to define a plane. Most of these calculations are covered on this page. For now, we continue our discussion on how to define a plane.
Here is a good video clip explaining this idea again.
video by PatrickJMT |
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General and Standard Forms (Equations of Planes)
The (simplified) general form of the equation of a plane is \( Ax + By + Cz + D = 0 \). The cool thing about this form is that the vector formed from the coefficients of the \(x,y,z\) components, i.e. \( \langle A,B,C \rangle \) is a normal vector to the plane. To get one point on the plane, we just choose values for \(x\) and \(y\) and solve for \(z\).
Alternatively, if we are given one point \((x_0, y_0, z_0)\) and a normal vector \(\langle A,B,C \rangle\), we can derive the general form of the equation of the plane using the standard equation \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\). By multiplying out, we can calculate d and end up with an equation in the form \( Ax + By + Cz + D = 0 \). In the simplified general form, the constants \( A,B,C,D \) are all integers and have no common factor.
Theorem: Standard Equation of a Plane |
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Given a point \( P = ( x_0, y_0, z_0 ) \) in a plane and a normal vector \( \vec{n} = \langle A,B,C \rangle \) to the plane, the standard equation of the plane is |
Proof - - Let the point \( Q = (x,y,z) \) be any point in the plane. We can form a vector in the plane \( \overrightarrow{PQ} = \langle x-x_0, y-y_0, z-z_0 \rangle \).
Since the normal vector is orthogonal to vector \( \overrightarrow{PQ} \) in the plane, we know that the dot product is zero. So we can write
\(\begin{array}{rcl}
\vec{n} \cdot \overrightarrow{PQ} & = & 0 \\
\langle A,B,C \rangle \cdot \langle x-x_0, y-y_0, z-z_0 \rangle & = & 0 \\
A(x-x_0) + B(y-y_0) + C(z-z_0) & = & 0 ~~~ \text{[qed]}
\end{array}\)
Parametric Equations
We can define a plane parametrically using two parameters. For this discussion, we will use the greek letters \( \lambda \) and \( \mu \).
Given one point \( \langle x_0, y_0, z_0 \rangle \) and two non-parallel vectors in the plane, we can get a set of parametric equations similar to how we did above for lines. Let's call the two vectors \(\vec{a} = \langle a_x,a_y,a_z \rangle \) and \(\vec{b} = \langle b_x,b_y,b_z \rangle\).
Using the two parameters \( \lambda \) and \( \mu \), any point on the plane can be written as \( \langle x,y,z \rangle = \langle x_0, y_0, z_0 \rangle + \lambda \vec{a} + \mu \vec{b} \). Equating each component, we get the three parametric equations
\( x = x_0 + \lambda a_x + \mu b_x \);
\( y = y_0 + \lambda a_y + \mu b_y \);
\( z = z_0 + \lambda a_z + \mu b_z \)
We can emphasize that these parametric equations are dependent on the parameters \(\lambda\) and \(\mu\) by writing them as
\( x(\lambda, \mu) = x_0 + \lambda a_x + \mu b_x \);
\( y(\lambda, \mu) = y_0 + \lambda a_y + \mu b_y \);
\( z(\lambda, \mu) = z_0 + \lambda a_z + \mu b_z \)
Vector Function
We can use the above parametric equations to build a vector function to define a plane.
\( \vec{W}(\lambda, \mu) = (x_0 + \lambda a_x + \mu b_x)\hat{i} + (y_0 + \lambda a_y + \mu b_y)\hat{j} + (z_0 + \lambda a_z + \mu b_z)\hat{k} \)
More simply, we can write \( \vec{W}(\lambda, \mu) = x(\lambda, \mu)\hat{i} + y(\lambda, \mu)\hat{j} + z(\lambda, \mu)\hat{k} \)
Calculating The Angle Between Two Planes
Using the idea of the dot product, we can calculate the angle between two planes. All we need are two normal vectors, one for each plane. The angle between the normal vectors is the same as the angle between the planes.
Remember that one equation to calculate the dot product is
\( \vec{u} \cdot \vec{v} = \norm{\vec{u}} \norm{\vec{v}} \cos \theta \) where \(\theta\) is the angle between vectors \(\vec{u}\) and \(\vec{v}\).
In our case, if we have two normal vectors \(\vec{n}_1\) and \(\vec{n}_2\), the dot product is
\( \vec{n}_1 \cdot \vec{n}_2 = \norm{\vec{n}_1} \norm{\vec{n}_2} \cos \theta \). Solving for \(\theta\) we get
\[ \displaystyle{ \cos(\theta) = \frac{| \vec{n}_1 \cdot \vec{n}_2 |}{\| \vec{n}_1 \| ~ \| \vec{n}_2 \|} }\]
Of course, then we would need to take the inverse cosine of each side to solve for \(\theta\).
Note: Notice that in the final equation for \(\cos\theta\) we take the absolute value of the dot product. This will always give us a positive angle \(\theta\) and we are only interested in the magnitude of the angle.
Okay, time for some practice problems.
Practice
Unless otherwise instructed, find the equation of the plane using the given information.
point: \( (2,3,-1) \), normal vector: \( \vec{n} = \langle 3,5,-2 \rangle \)
Problem Statement |
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Find the equation of the plane containing the point \( (2,3,-1) \) with normal vector \( \vec{n} = \langle 3,5,-2 \rangle \).
Final Answer |
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\( 3x + 5y - 2z = 23 \)
Problem Statement |
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Find the equation of the plane containing the point \( (2,3,-1) \) with normal vector \( \vec{n} = \langle 3,5,-2 \rangle \).
Solution |
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From the general equation of a plane, we have
\( 3(x - 2) + 5(y - 3) - 2(z + 1) = 0 \) |
\( 3x - 6 + 5y - 15 -2z -2 = 0 \) |
\( 3x + 5y - 2z = 23 \) |
Final Answer |
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\( 3x + 5y - 2z = 23 \)
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point: \( (-1,4,-2) \), normal vector \( \vec{n} = \langle 2,-9,8 \rangle \)
Problem Statement |
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Find the scalar equation of the plane with normal vector \( \vec{n} = \langle 2,-9,8 \rangle \) through the point \( (-1,4,-2) \).
Solution |
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3 points: \( (3,-1,2) \), \( (8,2,4) \), \( (-1,-2,-3) \)
Problem Statement |
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Find the equation of the plane that goes through the points \( (3,-1,2) \), \( (8,2,4) \) and \( (-1,-2,-3) \).
Solution |
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3 points: \( P(1,0,2) \), \( Q(-1,1,2) \), \( R(5,0,3) \)
Problem Statement |
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Find the equation of the plane that passes through the points \( P(1,0,2) \), \( Q(-1,1,2) \) and \( R(5,0,3) \).
Solution |
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3 points: \( P(0,1,0) \), \( Q(2,-1,4) \), \( R(-3,5,-4) \)
Problem Statement |
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Find the scalar equation of the plane that goes through the points \( P(0,1,0) \), \( Q(2,-1,4) \), \( R(-3,5,-4) \).
Solution |
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Really UNDERSTAND Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, find the equation of the plane using the given information.