Find the scalar, vector, and parametric equations of a plane

In summary, the scalar equation for the plane is 3x-4y+6z+11=0, and the vector equation is y=3/4x+3/2z+11/4. The parametric equations are x=9-9s-7t, y=2+3/4s+15/4t, and z=-5+5s+6t. These equations were verified by plugging in the given point P(9,2,-5) and getting a "true" statement.
  • #1
Physics345
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Homework Statement


Find the scalar, vector, and parametric equations of a plane
that has a normal vector n→=(3,−4,6) and passes through the point P(9, 2, –5).

Homework Equations


Ax+By+Cz+D=0
(x,y,z)=(x0,y0,z0)+s(a1,a2,a3)+t(b1,b2,b3)
x=x0+sa1+tb1
y=y0+sa2+tb2
z=z0+sa3+tb3

The Attempt at a Solution


Scalar Equation:
3x-4y+6z+D=0
3(9)-4(2)+6(-5)+D=0
D=11
3x- 4y +6z+11=0
Vector Equation:
4y=3x+6z+11
y=3/4 x+6/4 z+11
y=3/4 x+3/2 z+11
P(9,2,-5)
Let x=0 and z=0
y=11
Q(0,11,0)
let x=2 and z=1
y=3/4(2)+3/2(1)+11
y=14
R(2,14,1)
(PQ) ⃗ =Q-P
(PQ) ⃗ =(0,11,0)-(9,2,-5)=(-9,9,5)
(PR) ⃗ =R-P
(PR) ⃗ =(2,14,1)-(9,2,-5)=(-7,12,6)
(x,y,z)=(9,2,-5)+s(-9,9,5)+t(-7,12,6)
Parametric Equations:
x=9-9s-7t
y=2+9s+12t
z=-5+5s+6t
Did I do this correctly? And is there anyway to confirm my answers through math? Thanks in advance =)
 
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  • #2
Physics345 said:

Homework Statement


Find the scalar, vector, and parametric equations of a plane
that has a normal vector n→=(3,−4,6) and passes through the point P(9, 2, –5).

Homework Equations


Ax+By+Cz+D=0
(x,y,z)=(x0,y0,z0)+s(a1,a2,a3)+t(b1,b2,b3)
x=x0+sa1+tb1
y=y0+sa2+tb2
z=z0+sa3+tb3

The Attempt at a Solution

Scalar Equation: is OK.
Vector Equation:
4y=3x+6z+11
y=3/4 x+6/4 z+11
You did not divide 11 by 4 .
y=3/4 x+3/2 z+11
P(9,2,-5)
Let x=0 and z=0
y=11
Q(0,11,0)
That point is not in the plane.
let x=2 and z=1
y=3/4(2)+3/2(1)+11
y=14
R(2,14,1)
(PQ) ⃗ =Q-P
(PQ) ⃗ =(0,11,0)-(9,2,-5)=(-9,9,5)
(PR) ⃗ =R-P
(PR) ⃗ =(2,14,1)-(9,2,-5)=(-7,12,6)
(x,y,z)=(9,2,-5)+s(-9,9,5)+t(-7,12,6)
Parametric Equations:
x=9-9s-7t
y=2+9s+12t
z=-5+5s+6t
Did I do this correctly? And is there anyway to confirm my answers through math? Thanks in advance =)
The parametric version is also incorrect.
 
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  • #3
SammyS said:
Scalar Equation: is OK.You did not divide 11 by 4 .That point is not in the plane.
The parametric version is also incorrect.
Which point are you referring to from PQR? Is it because of the division mistake?
 
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  • #4
Physics345 said:
Which point are you referring to from PQR? Is it because of the division mistake?
I am referring to the point Q at (0, 11, 0) .

That does appear to me to be the result of your division mistake.
 
  • #5
Okay according, to your comments I made some adjustments, how are you able to tell by the way?

y=3/4 x+3/2 z+11/4
P(9,2,-5)
Let x=0 and z=0
y=11/4
Q(0,11/4,0)
let x=2 and z=1
y=3/4(2)+3/2(1)+11/4
y=23/4
R(2,23/4,1)
(PQ) ⃗=Q-P
(PQ) ⃗=(0,11/4,0)-(9,2,-5)=(-9, 3/4,5)
(PR) ⃗=R-P
(PR) ⃗=(2,23/4,1)-(9,2,-5)=(-7,15/4,6)
(x,y,z)=(9,2,-5)+s(-9, 3/4,5) +t(-7,15/4,6)
Parametric Equations:
x=9-9s-7t
y=2+3/4 s+15/4 t
z=-5+5s+6t
 
  • #6
Physics345 said:
Okay according, to your comments I made some adjustments, how are you able to tell by the way?
What is the thing that you are asking about? I made four different statements in my post #2 of this thread.
 
  • #7
SammyS said:
What is the thing that you are asking about? I made four different statements in my post #2 of this thread.
That the points were wrong. Also is it right now?
 
  • #8
Physics345 said:
Okay according, to your comments I made some adjustments, how are you able to tell by the way?
...

(x,y,z)=(9,2,-5)+s(-9, 3/4,5) +t(-7,15/4,6)

Parametric Equations:
x=9-9s-7t
y=2+3/4 s+15/4 t
z=-5+5s+6t
Physics345 said:
That the points were wrong. Also is it right now?
Your results are now correct.

As far as checking your results:
I knew that your scalar equation for the plane was correct, by verifying the steps you took.

Using the scalar equation, it's easy to check that a given point is in the plane. Plug-in x, y, and z .
 
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  • #9
P(9,2,-5)
3x- 4y +6z+11=0
3(9)-4(2)+6(-5)+11=0
=0
How did it verify? because it equals zero?
 
  • #10
Physics345 said:
P(9,2,-5)
3x- 4y +6z+11=0
3(9)-4(2)+6(-5)+11=0
=0
How did it verify? because it equals zero?
Yes. Plugging into the equation gave a "true" statement.
 
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  • #11
SammyS said:
Yes. Plugging into the equation gave a "true" statement.
Thank you, I understand now. Your help was very much appreciated.
 
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Related to Find the scalar, vector, and parametric equations of a plane

1. What is a scalar equation of a plane?

A scalar equation of a plane is a linear equation that represents the relationship between three variables in a plane. It can be written in the form ax + by + cz = d, where a, b, and c are constants and x, y, and z are variables.

2. How do you find the vector equation of a plane?

The vector equation of a plane can be found using the normal vector of the plane and a point on the plane. The equation is written as r = a + tb, where r is a position vector, a is a point on the plane, and b is the normal vector.

3. What is a parametric equation of a plane?

A parametric equation of a plane represents the coordinates of any point on the plane in terms of two independent parameters, usually denoted by u and v. It can be written as r(u,v) = a + ub + vc, where a, b, and c are constant vectors and u and v are parameters.

4. How do you convert a scalar equation to a vector equation?

To convert a scalar equation of a plane to a vector equation, you can use the formula r = a + tb, where a is a point on the plane and b is the normal vector. The coefficients of the variables in the scalar equation become the components of the normal vector in the vector equation.

5. In what situations would you use a parametric equation of a plane?

Parametric equations of a plane are often used in situations where there are two independent variables that can be used to describe the coordinates of points on the plane. This can be useful in applications such as computer graphics, where curves and surfaces need to be represented using parametric equations.

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