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3-4-Cartesian-Equation-of-a-Plane

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Ms. Catral • MCV4U • Unit 3: Lines and Planes
Lesson 4: Cartesian Equation of a Plane
Lesson Goal: To be able to represent a plane with a Cartesian equation
Normal: A vector, through the origin, which is perpendicular to a plane
Cartesian Equation of a Plane
The scalar, or Cartesian, equation of a plane with normal, 𝑛⃑ = (𝐴, 𝐡, 𝐢) is of the form
𝐴π‘₯ + 𝐡𝑦 + 𝐢𝑧 + 𝐷 = 0
Note: Since the normal is perpendicular to the plane, 𝑛⃑ = π‘Ž × π‘βƒ‘, where π‘Ž and 𝑏⃑ are any two
noncollinear direction vectors of the plane.
Example 1 – Let 𝑛⃑ be the normal vector 𝑛⃑ = (21,13, −2). Find the Cartesian equation for the plane πœ‹
that has normal vector 𝑛⃑ and passes through (2, 0, 8).
Example 2 – Determine the Cartesian equation of the plane whose equation in vector form is
π‘Ÿ = (1, 2, −1) + 𝑠(1, 0, 2) + 𝑑(−1, 3, 4), 𝑠, 𝑑 ∈ ℝ
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Ms. Catral • MCV4U • Unit 3: Lines and Planes
Parallel and Perpendicular Planes
1. If πœ‹1 and πœ‹2 are two perpendicular planes, with normal 𝑛⃑1 and 𝑛⃑1 , respectively, then their normals
are perpendicular (i.e. 𝑛⃑1 • 𝑛⃑2 = 0)
2. If πœ‹1 and πœ‹2 are two parallel planes, with normal 𝑛⃑1 and 𝑛⃑1 , respectively, then their normals are
parallel (i.e. 𝑛⃑1 = π‘˜π‘›βƒ‘2 , π‘˜ ∈ ℝ)
Example 3 – Show that the planes πœ‹1 : 2π‘₯ − 5𝑦 + 3𝑧 − 1 = 0 and πœ‹2 : 4π‘₯ − 10𝑦 + 6𝑧 + 7 = 0 are
parallel, but not coincident.
Example 4 – Show that the planes πœ‹3 : −4π‘₯ + 8𝑦 + 2𝑧 − 11 = 0 and πœ‹4 : 5π‘₯ + 4𝑦 − 6𝑧 + 1 = 0 are
perpendicular.
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Ms. Catral • MCV4U • Unit 3: Lines and Planes
Angle between Intersecting Planes
The angle, πœƒ, between two planes, πœ‹1 and πœ‹2 , with normal of 𝑛⃑1
and 𝑛⃑1 , respectively, can be calculated using the formula
cos πœƒ =
𝑛⃑1 • 𝑛⃑2
|𝑛⃑1 ||𝑛⃑2 |
Example 4 – What is the angle between the two planes πœ‹1 : π‘₯ − 𝑦 − 2𝑧 + 3 = 0
and πœ‹2 : 2π‘₯ + 𝑦 − 𝑧 + 2 = 0 ?
Homework – pg 468 # 1, 6ab, 7, 10, 13
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