# Finding the parametric equation of a plane

• rafehi
And, for what b is (0, 1, b) perpendicular to (7, 1, 4)? We must have 1+ 4b= ...? So you are looking for the point (1, 4, 5) plus a combination of (1, 0, ...)(4, 0, ...), and (0, 1, ...)(0, 4, ...). You could go through the same process for part b).In summary, the conversation discusses finding the equation of planes in cartesian and (vector) parametric form. For part a), the cartesian form is found using the equation a(x-x0) + b(y-y0)
rafehi

## Homework Statement

Find the equation of the following planes in cartesian and (vector) parametric form:
a) the plane through the point (1,4,5) and perpendicular to the vector (7,1,4)
b) the plane through the origin and the points (1,1,1) and (1,2,3)

## The Attempt at a Solution

For part a)
I found the cartesian form of the equation using a(x-x0) + b(y-y0) + c(z-z0), to give:
7(x-1) + 1(y-4) + 4(z-5) = 0, which expands to:
7x + y + 4z = 31, which is correct.

However, I've no idea how to find the parametric form. The answer is given as r = (1,4,5) + s(4,0,-7) + t(0,4-1).

For part b)
To find the cartesian form, I found the vectors from the origin to (1, 1, 1,) and (1, 2, 3), using V1->V2 = V2 - V1, giving me V0->V1= (1,1,1) and V0->V2 = (1,2,3). Using the cross product to get the orthogonal, then dotting that with (0,0,0) gives the correct equation, x-2y+z=0, which is correct.

Again, I've got no idea how to find the parametric equation.

Any help would be greatly appreciated. These aren't assignment questions or anything like that - just revision for my L.A exam next week.

Let's pluck a plane out of the air: x + 3y - 7z = 20.

To find the parametric equations for x, y and z is just to solve the equation as you would simultaneous equations, remember that?

One equation, three unknowns => 3-1=2 parameters. Let z=t, y = s, so x = 20 - 3y + 7z = 20 - 3s + 7t. Hence (x, y, z) = (20 - 3s + 7t, s, t) = (20, 0, 0) + (-3, 1, 0)s + (7, 0 , 1)t.

For a, for the parametric form you want, you have one point in the plane, Q(1, 4, 5). You need to find two additional points in the plane, call them R and S.

The points R and S should be such that QR is not parallel to QS.

Then any point in the plane satisfies P = OQ + r*QR + s*QS.

Here's what's going on geometrically. To get from the origin to any point in the plane, I have to get to some point on the plane (OQ does that), and from there, some linear combination of QR and QS gets me to the exact point. The parameters of this parametric form are r and s.

Unco said:
Let's pluck a plane out of the air: x + 3y - 7z = 20.

To find the parametric equations for x, y and z is just to solve the equation as you would simultaneous equations, remember that?

One equation, three unknowns => 3-1=2 parameters. Let z=t, y = s, so x = 20 - 3y + 7z = 20 - 3s + 7t. Hence (x, y, z) = (20 - 3s + 7t, s, t) = (20, 0, 0) + (-3, 1, 0)s + (7, 0 , 1)t.

Applying this method to a), I get (x,y,z)=(31-4t-7s,s,t)= (31,0,0)+(-7,1,0)+(-4,0,1), which differs from the answer given (shown in the OP). How would we answer the question without having to use a different position (as seems to be the expectation in all the questions)?

Mark44 said:
For a, for the parametric form you want, you have one point in the plane, Q(1, 4, 5). You need to find two additional points in the plane, call them R and S.

The points R and S should be such that QR is not parallel to QS.

Then any point in the plane satisfies P = OQ + r*QR + s*QS.

Here's what's going on geometrically. To get from the origin to any point in the plane, I have to get to some point on the plane (OQ does that), and from there, some linear combination of QR and QS gets me to the exact point. The parameters of this parametric form are r and s.

OK, I get that much. But how would be find the points QR and QS?

I'm guessing one possible method would be to use the dot product, such that QR . (7,1,4) and QS . (7,1,4) = 0. Is there an easier approach to the question (and one that doesn't give me infinite possible solutions, which I'm guessing would make life pretty hard for the examiner)?

Thank you both for your help.

rafehi said:
Applying this method to a), I get (x,y,z)=(31-4t-7s,s,t)= (31,0,0)+(-7,1,0)+(-4,0,1), which differs from the answer given (shown in the OP). How would we answer the question without having to use a different position (as seems to be the expectation in all the questions)?
Your answer is incorrect because taking s= t= 0 gives (31, 0, 0) which does NOT satisfy 7x- y+ 4z= 31. Did you forget to divide by 7? Solving the cartesian equation for x we have x= (y- 4z+ 31)/7 and, taking y and z as the parameters, y= s, z= t, x= (s- 4t+ 31)/7 so
r= ((s- 4t+ 31)/7, s, t)= (31/7, 0, 0)+ (1/7, 1, 0)s+ (-4/7, 0, 1)t.

There exist an infinite number of parametric equations for anything because the choice of parameter is arbitrary. I, because I dislike fractions, might have chosen to solve for y rather than x: y= x+ 4z- 31 so taking x= s, z= t, r= (s, s+ 4t- 31, t)= (0, -31, 0)+ (1, 1, 0)s+ (0, 4, 1)t.

In the answer your book gives, r = (1,4,5) + s(4,0,-7) + t(0,4,-1), they have not solved the cartesian equation but worked directly from the given information. Any point in the plane can be written as a single point in the plane (and (1, 4, 5) is given) plus a linear combination of two independent vectors in the plane. Any vector on the plane must be perpendicular to (7, 1, 4) and we can be sure two vectors are independent by taking one (1, 0, a) and the other (0, 1, b). For what a is (1, 0, a) perpendicular to (7, 1, 4)? We must have 7+ 4a= 0 or a= -7/4. (1, 0, -7/4) will do, or since I dislike fractions, (4, 0, -7). A vector of the form (0, 1, b)is perpendicular to (7, 1, 4) if 1+ 4b= 0 so b= -1/4. Such a vector is (0, 1, -1/4) as well as (0, 4, -1). We can write any point on the plane as the intial point, (1, 4, 5) plus a linear combination of those two vectors in the plane: (1, 4, 5)+ (4, 0, -7)s+ (0, 4, -1)t.

HallsofIvy said:
Your answer is incorrect because taking s= t= 0 gives (31, 0, 0) which does NOT satisfy 7x- y+ 4z= 31. Did you forget to divide by 7? Solving the cartesian equation for x we have x= (y- 4z+ 31)/7 and, taking y and z as the parameters, y= s, z= t, x= (s- 4t+ 31)/7 so
r= ((s- 4t+ 31)/7, s, t)= (31/7, 0, 0)+ (1/7, 1, 0)s+ (-4/7, 0, 1)t.

There exist an infinite number of parametric equations for anything because the choice of parameter is arbitrary. I, because I dislike fractions, might have chosen to solve for y rather than x: y= x+ 4z- 31 so taking x= s, z= t, r= (s, s+ 4t- 31, t)= (0, -31, 0)+ (1, 1, 0)s+ (0, 4, 1)t.

In the answer your book gives, r = (1,4,5) + s(4,0,-7) + t(0,4,-1), they have not solved the cartesian equation but worked directly from the given information. Any point in the plane can be written as a single point in the plane (and (1, 4, 5) is given) plus a linear combination of two independent vectors in the plane. Any vector on the plane must be perpendicular to (7, 1, 4) and we can be sure two vectors are independent by taking one (1, 0, a) and the other (0, 1, b). For what a is (1, 0, a) perpendicular to (7, 1, 4)? We must have 7+ 4a= 0 or a= -7/4. (1, 0, -7/4) will do, or since I dislike fractions, (4, 0, -7). A vector of the form (0, 1, b)is perpendicular to (7, 1, 4) if 1+ 4b= 0 so b= -1/4. Such a vector is (0, 1, -1/4) as well as (0, 4, -1). We can write any point on the plane as the intial point, (1, 4, 5) plus a linear combination of those two vectors in the plane: (1, 4, 5)+ (4, 0, -7)s+ (0, 4, -1)t.

Thanks heaps for clearing that up, HoI. On your first paragraph, I did solve for y and let x and z equal arbitrary constants, but I apparently wrote the order as (y,x,z) instead of (x,y,z).

Alright - one more question.

If we're given a point and a vector parallel to the plane, how would we find the vector equation?

The question is: Find the vector equation of the plane through the point (6, 5, -2) and parallel to the plane x + y - z + 1 = 0.

How I went about solving it was find a vector perpendicular to the plane using the dot product:
(a,b,c).(1,1,-1)=0.
(a,b,c)=(1,0,1) satisfies the above, thus it is perpendicular to the plane.

Finding two vectors perpendicular to this (or one, as we already have a parallel vector) would allow us to make a linear combination along the plane and thus define the plane.
So:
(a,b,c).(1,0,1)=0
Therefore, (a,b,c)=(1,0,-1), with only the one possible independent vector, using the parallel vector given in the question:
(6,5,-2) + s(1,0,-1) + t(1,1,-1) (equation of the plane)

(6,5,-2) + s(1,0,1) + t(0,1,1)
Now both the vectors used in the equation are perpendicular to the plane, so how is it possible that the above is an equation for the plane? Unless I'm missing something, I'm guessing it's just an error on their behalf?

rafehi said:
Alright - one more question.

If we're given a point and a vector parallel to the plane, how would we find the vector equation?

The question is: Find the vector equation of the plane through the point (6, 5, -2) and parallel to the plane x + y - z + 1 = 0.
If two planes are parallel, they have the same normal vector. Find the plane containing (6, 5, -2) with normal vector <1, 1, -1>. You seem to be forgetting that for a plane written as Ax+ By+ Cz= constant, <A, B, C> is the normal vector.

(x- 6)+ (y- 5)- (z+ 2)= 0 so x+ y- z= 13 is the "Cartesian" equation. Solving form z, z= x+ y- 13 so x= s, y= t, z= s+ t- 13 are parametric equations for the plane and <0, 0, -13>+ <1, 0, 1>s+ <0, 1, 1>t is a vector equation.

How I went about solving it was find a vector perpendicular to the plane using the dot product:
(a,b,c).(1,1,-1)=0.
(a,b,c)=(1,0,1) satisfies the above, thus it is perpendicular to the plane.
No. (1, 1, -1) itself is perpendicular to the given plane.

Finding two vectors perpendicular to this (or one, as we already have a parallel vector) would allow us to make a linear combination along the plane and thus define the plane.
So:
(a,b,c).(1,0,1)=0
Therefore, (a,b,c)=(1,0,-1), with only the one possible independent vector, using the parallel vector given in the question:
(6,5,-2) + s(1,0,-1) + t(1,1,-1) (equation of the plane)

(6,5,-2) + s(1,0,1) + t(0,1,1)
Now both the vectors used in the equation are perpendicular to the plane, so how is it possible that the above is an equation for the plane? Unless I'm missing something, I'm guessing it's just an error on their behalf?
Again, (1, 1, -1) is perpendicular to the given plane so (x, y, z) is parallel to the plane if and only if x+ y- z= 0. z= x+ y in that case. Taking x= 1, y= 0, z= 1 so (1, 0, 1) is parallel to the plane. Taking x= 0, y= 1, z= 1 so (0, 1, 1) is parallel to the plane. Since (6, 5, -2) is a point in the plane, (6,5, -2)+ s(1, 0, 1) is in the plane for all s and (6, 5, -2)+ t(0,1,1) is in the plane: since the plane is two dimensional all points in the plane are of the form (6, 5, -2)+ s(1, 0, 1)+ t(0, 1, 1) for all s and t.

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You really are a legend HoI. The section is barely covered in the textbook and we skimmed over it in the lectures. Thanks heaps for your help. *insert thumbs up smilie here*

## What is a parametric equation of a plane?

A parametric equation of a plane is a set of equations that describe the location of every point on a plane in terms of one or more parameters. It allows for the representation of a plane in three-dimensional space.

## Why is it important to find the parametric equation of a plane?

Finding the parametric equation of a plane is important because it allows for a more precise and efficient way to describe and analyze the behavior of a plane in three-dimensional space. It also enables the use of vector calculus techniques in solving problems involving planes.

## How do you find the parametric equation of a plane?

To find the parametric equation of a plane, you need to know the coordinates of three non-collinear points on the plane. These points can then be used to determine the direction vectors of the plane, which are used in the parametric equations. The parametric equations for a plane are of the form x = a + bt + cu, y = d + et + fu, and z = g + ht + iu, where t and u are the parameters and a, b, c, d, e, f, g, h, and i are constants.

## What are some applications of the parametric equation of a plane?

The parametric equation of a plane has many practical applications, including in computer graphics, engineering, and physics. In computer graphics, it is used to create 3D models and animations. In engineering, it is used to analyze the behavior of planes and solve problems involving intersections and intersections of planes. In physics, it is used to describe the motion of objects in three-dimensional space.

## Can the parametric equation of a plane be written in different forms?

Yes, the parametric equation of a plane can be written in different forms depending on the specific problem being solved. For example, it can be written using different parameters or constants, or it can be written as a vector equation. However, the general form of the parametric equation remains the same.

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