MCV4U1-UNIT EIGHT-LESSON FOUR
Lesson Four: Vector and Parametric Equations of a Plane
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Let P(x,y,z) be any point on the plane.
Let
P1 ( x1 , y1 , z1 ) be a specific point on the plane.
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a  a1 , a2 , a3 and b  b1 , b2 , b3  be vectors in the plane.
Let
(They are called direction vectors for the plane.)
Using the triangle law of addition of vectors….
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OP  OP1  P1 P
Replacing these vectors with their components….
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x, y, z  x1 , y1 , z1   P1P
and since
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P1 P, a, and b
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a and b .
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 P1P  sa  tb ,
are coplanar (dependent),
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P1 P can be written as a linear combination of
where s and t are parameters (real numbers).
Therefore, the vector equation of the plane is….
x, y, z   x1 , y1 , z1   sa1 , a2 , a3   t b1 , b2 , b3 
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alternate notation: r  r1  sa  tb
The parametric equations of the plane are….
MCV4U1-UNIT EIGHT-LESSON FOUR
x  x1  sa1  tb1
y  y1  sa 2  tb2
z  z1  sa 3  tb3
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(No symmetric equations.)
Example 1:
Find a vector equation of the plane that contains the points
A1,2,1, B 5,3,0 , and C 2,1,1 .
SOLUTION:
We need ONE point and TWO direction vectors to write the equation of a plane.
P.
We can use ANY of the points for 1
To find the direction vectors, you may use the components of any two of the following vectors…..
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AB, AC, BC, BA, CA, CB,...
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Suppose we use AB and AC.
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AB  4,1,1 and AC  1,3,2
Then a vector equation of the plane is…..
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r  1,2,1  s4,1,1  t1,3,2
Example 2:
a) Write the parametric equations of the plane
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r  0,2,5  s1,.1,4  t4,2,1
SOLUTION:
x  s  4t
y  2  s  2t
z  5  4s  t
b)
Find two points that lie on the plane.
SOLUTION:
One point that lies on the plane is
Let s=2 and t=1, then …..
0,2,5 . To find another point, we sub in values for s and t.
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r  0,2,5  21,1,4   14,2,1
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r  0,2,5  2,2,8  4,2,1
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r  6,2,14
MCV4U1-UNIT EIGHT-LESSON FOUR
Therefore
6,2,14 is another point on the plane.
Example 3:
Find a vector equation of the plane that contains the two intersecting lines
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r1  1,3,2  t 9,2,1 and r2  1,0,2  u5,0,1.
SOLUTION:
Since the plane contains the lines, the direction vectors for the lines are direction vectors for the plane. And
the plane contains the same points as the lines.
Therefore, a vector equation of the plane is….
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r  1,3,2  s9,2,1  t 5,0,1