How do you find the parametric equation through the point (6, 0, 10) and parallel...

hornejada1c

hornejada1c

Answered

2022-07-06

How do you find the parametric equation through the point (6, 0, 10) and parallel to the plane 3 x + 2 y 4 z = 0 ?

Answer & Explanation

Alexis Fields

Alexis Fields

Expert

2022-07-07Added 14 answers

Step 1
Any plane parallel to the plane - 3 x + 2 y - 4 z = 0 will have an equation of the same form but equal to a different constant:
- 3 x + 2 y - 4 z = C [1]
To force the point ( 6 , 0 , 10 ) to be in the plane, substitute the point into equation [1] and solve for C:
- 3 ( 6 ) + 2 ( 0 ) - 4 ( 10 ) = C
C = - 58
The scalar equation of the plane parallel to the given plane is:
- 3 x + 2 y - 4 z = - 58 [2]
The vector equation of the plane is
( x , y , z ) = ( 6 , 0 , 10 ) + s ( u ¯ ) + t ( v ¯ )
Where u ¯ and v ¯ are any two vectors in the plane.
Let's use equation [2] to find two other points in the plane:
Let x = 6 and y = 2
- 3 ( 6 ) + 2 ( 2 ) - 4 z = - 58
- 4 z = - 44
z = 11
The vector from the point ( 6 , 0 , 10 ) to the point ( 6 , 2 , 11 ) is:
u ¯ = 2 j ^ + k ^
Let x = 10 and y = 0
- 3 ( 10 ) + 2 ( 0 ) - 4 z = - 58
- 4 z = - 28
z = 7
The vector from point ( 6 , 0 , 10 ) to the point ( 6 , 0 , 7 ) is:
v ¯ = 4 i ^ - 3 k ^
A vector equation of the plane is:
( x , y , z ) = ( 6 , 0 , 10 ) + s ( 2 j ^ + k ^ ) + t ( 4 i ^ - 3 k ^ )
From the above equation, we can write a set of parametric equations:
x = 4 t + 6
y = 2 s
z = s - 3 t + 10
NOTE: These are not unique equations, because we could have chosen other points and obtained different vectors.

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