Chapter 1: Vectors, Lines and Planes
Section 1.7: Planes
Example 1.7.2
Obtain an equation for the plane that passes through the point P:5,−3,7, and has for its normal the vector N=3 i−5 j+2 k.
Solution
Mathematical Solution
The equation of the plane is given by the vector equation R−P·N=0, that is, by
(xyz−5−37)·3−52 = 3x−5−5y+3+2z−7=0
which simplifies to 3 x−5 y+2 z=15+15+14=44.
Alternatively, some practitioners would write N·R=d and use point P to determine the value of d. The calculation we go as follows.
3 x−5 y+2 zx=a|f(x)x=5,y=−3,z=7=d ⇒ d=15+15+14=44
Hence, the equation of the plane is 3 x−5 y+2 z=44.
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Enter N as per Table 1.1.1.
Context Panel: Assign to a Name≻N
3,−5,2→assign to a nameN
Write the sequence of the point P and the name N.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation
5,−3,7,N→make plane<< Plane 1 >>→representation3x−5y+2z=44
The traditional vector approach is implemented below.
Define the position vectors P and R
Enter P as per Table 1.1.1.
Context Panel: Assign to a Name≻P
5,−3,7→assign to a nameP
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
x,y,z→assign to a nameR
Obtain the equation of the plane
Write R−P·N=0 Press the Enter key.
R−P·N=0
3x−5y+2z−44=0
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define N and the position vectors P and R.
P,N,R≔5,−3,7,3,−5,2,x,y,z:
Apply the DotProduct command.
DotProductR−P,N=0
3x−44−5y+2z=0
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