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Planes in 3-dimensional Space

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To find an equation for a plane in 3-dimensional space we need to know two things. We need to find a vector normal to the plane and also a point that lies on the plane. Then we can use the formula below to find an equation for the plane.

a(x – x1) + b(y – y1) + c(z – z1) = 0

Note: <a,b,c> is the vector normal to the plane and (x1,y1,z1) is a point that lies on the plane.

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Application Problem:

Find an equation for the plane normal to the vector <2,1,1> and passing through the point (1,2,1).

Solution:

Our normal vector is <2,1,1> and therefore a = 2, b = 1, and c = 1 in our formula. We are given that (1,2,1) is a point on the plane, which implies we can use x1 = 1, x2 = 2, and x3 = 1 in our formula. Then by plugging the corresponding values into our formula we get that 2(x – 1) + 1(y – 2) + 1(z – 1) = 0. Then by foiling the left side of our equation we get

2x – 2 + y – 2 + z – 1 = 0. Simplifying the left side yields 2x + y + z – 5 = 0 and finally adding 5 to both sides gives 2x + y + z = 5. Thus 2x + y + z = 5 is an equation for the plane passing through the point (1,2,1) and normal to the vector <2,1,1>.

Calculus and Analytic Geometry:

• Basic Differentiation Rules (Power rule, product rule, quotient rule)
• Basic Differentiation Drill Problems
• Differentiating Trigonometric Functions
• Differentiating Logarithmic Functions
• Mapping illustrating the logarithmic properties (**New**)
• Finding the angle between 2 vectors: Application of the dot product
• Basic Integration Techniques
• Advanced Integration Techniques
• Finding the tangent line of a point in relation to a graph
• Equations of planes in 3-dimension
• Equations of lines in 3-dimension
• Significance of the cross product
• Introduction to polar coordinates
• Implicit Differentiation
• Logarithmic Differentiation
• Partial Derivatives (3D)