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Finding the Equation of a Plane with Given Point and Line - Prof. Sean F. Ellermeyer, Quizzes of Advanced Calculus

A solution to finding the equation of a plane that contains a given point (p0) and a line with parametric equations. How to find the normal vector of the plane and derive the equation using vector algebra. It also checks the correctness of the answer by verifying that the point and the line are indeed in the plane.

Typology: Quizzes

2010/2011

Uploaded on 06/03/2011

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MATH 2203 โ€“Quiz 2
January 25, 2006
NAME____________________________________
Find an equation for the plane that contains the point P0(๎˜€9;0;2) and
also contains the line, L, with parametric equations
x= 2 ๎˜€3t
y= 5
z= 2 ๎˜€t.
After you have found an equation for this plane, show how you check to see
that your answer is correct.
Solution: By setting t= 0 and t= 1, we see that the points P1(2;5;2)
and P2(๎˜€1;5;1) are on the line L. Since ๎˜€๎˜€!
P0P1=h11;5;0iand ๎˜€๎˜€!
P0P2=
h8;5;๎˜€1i, a vector that is perpendicular to the plane in question is
n=๎˜€๎˜€!
P0P1๎˜‚๎˜€๎˜€!
P0P2=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
i j k
11 5 0
8 5 ๎˜€1
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜€5i+ 11j+ 15k.
If P(x; y; z)is any point in the plane in question, then n๎˜๎˜€๎˜€!
P0P= 0. Thus,
an equation for this plane is
h๎˜€5;11;15i๎˜hx+ 9; y ; z ๎˜€2i= 0
or
๎˜€5 (x+ 9) + 11y+ 15 (z๎˜€2) = 0
or
๎˜€5x๎˜€45 + 11y+ 15z๎˜€30 = 0
or
๎˜€5x+ 11y+ 15z= 75.
Let us check to see that our answer is correct:
๎˜€5 (๎˜€9) + 11 (0) + 15 (2) = 75
shows that the point P0is in this plane.
๎˜€5 (2 ๎˜€3t) + 11 (5) + 15 (2 ๎˜€t) = ๎˜€10 + 15t+ 55 + 30 ๎˜€15t= 75
shows that the line Lis contained in this plane.
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MATH 2203 รฑQuiz 2 January 25, 2006

NAME____________________________________ Find an equation for the plane that contains the point P 0 ( 9 ; 0 ; 2) and also contains the line, L, with parametric equations

x = 2 3 t y = 5 z = 2 t.

After you have found an equation for this plane, show how you check to see that your answer is correct. Solution: By setting t = 0 and t = 1, we see that the points P 1 (2; 5 ; 2)

and P 2 ( 1 ; 5 ; 1) are on the line L. Since

P 0 P 1 = h 11 ; 5 ; 0 i and

P 0 P 2 =

h 8 ; 5 ; 1 i, a vector that is perpendicular to the plane in question is

n =

P 0 P 1 

P 0 P 2 =

i j k 11 5 0 8 5 1

= 5 i + 11j + 15k.

If P (x; y; z) is any point in the plane in question, then n 

P 0 P = 0. Thus, an equation for this plane is

h 5 ; 11 ; 15 i  hx + 9; y; z 2 i = 0

or 5 (x + 9) + 11y + 15 (z 2) = 0

or 5 x 45 + 11y + 15z 30 = 0

or 5 x + 11y + 15z = 75.

Let us check to see that our answer is correct:

5 (9) + 11 (0) + 15 (2) = 75

shows that the point P 0 is in this plane.

5 (2 3 t) + 11 (5) + 15 (2 t) = 10 + 15t + 55 + 30 15 t = 75

shows that the line L is contained in this plane.