Find parametric equation at point, and parallel to planes?

In summary, the parametric equation of the line through the point (4,0,-4) that is parallel to the planes x-8y+7z=0 and 4x+3y-z+4=0 can be obtained by taking the cross product of the normal vectors of the two planes and using it as the direction vector in the equation ##L = \lbrack 4,0,-4 \rbrack + t\lbrack a,b,c \rbrack ##, where a, b, and c are the components of the cross product.
  • #1
Destroxia
204
7

Homework Statement



Find parametric equation of the line through the point ##(4,0,-4)## that is parallel to the planes ##x-8y+7z=0## and ##4x+3y-z+4=0##.

Homework Equations



## \vec r = \vec r_0 + t\vec v ##
(for orthogonal vectors) ## v \bullet w = 0 ##[/B]

The Attempt at a Solution



So I started out develeping the equation of the line through the point and came up with

## L = \lbrack 4,0,-4 \rbrack + t\lbrack a,b,c \rbrack ##

Now I needed to find the direction vector a, b, and c.

Since the line is parallel to the 2 planes, the direction vector should be orthogonal to both of the planes' normal vectors.

## \lbrack a,b,c \rbrack \bullet \lbrack 1,-8,7 \rbrack = 0##

## \lbrack a,b,c \rbrack \bullet \lbrack 4,3,-1 \rbrack = 0 ##

So, now I have 2 equations with the 3 variables, and I went to solve them in terms of c.

## a - 8b + 7c = 0 ## and ## 4a + 3b - c = 0 ##

## b = \frac {29c} {35} ##
## a = \frac {-216c} {35} ##

I wasn't really sure what to do at this point, as these answers seemed pretty obscure, and didn't know what to plug in for C, besides maybe c = 35. But when I do that, it doesn't match any of the multiple choice responses.

But the problem comes with pretty defined answers to pick from, multiple choice:

##a) x = 4 + t, y = -8t, z = -4 + 7t ##
##b) x = 4 - t, y = 8t, z = -4 - 7t ##
##c) x = 4 - 13t, y = 29t, z = -4 + 35t ##
##d) x = 4 + 13t, y = -29t, z = -4 - 35 ##
##e) x = 4 + 4t, y = 3t, z = -4 - t ##
 
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  • #2
RyanTAsher said:

Homework Statement



Find parametric equation of the line through the point ##(4,0,-4)## that is parallel to the planes ##x-8y+7z=0## and ##4x+3y-z+4=0##.

Homework Equations



## \vec r = \vec r_0 + t\vec v ##
(for orthogonal vectors) ## v \bullet w = 0 ##[/B]

The Attempt at a Solution



So I started out develeping the equation of the line through the point and came up with

## L = \lbrack 4,0,-4 \rbrack + t\lbrack a,b,c \rbrack ##

Now I needed to find the direction vector a, b, and c.

Since the line is parallel to the 2 planes, the direction vector should be orthogonal to both of the planes' normal vectors.

## \lbrack a,b,c \rbrack \bullet \lbrack 1,-8,7 \rbrack = 0##

## \lbrack a,b,c \rbrack \bullet \lbrack 4,3,-1 \rbrack = 0 ##

So, now I have 2 equations with the 3 variables, and I went to solve them in terms of c.

I didn't check your arithmetic, but you can probably pick c arbitrarily since many different length vectors have the same direction. But wouldn't it be easier to just note the direction you want is in the direction of the cross product of your two normals?
 
  • #3
RyanTAsher said:

Homework Statement



Find parametric equation of the line through the point ##(4,0,-4)## that is parallel to the planes ##x-8y+7z=0## and ##4x+3y-z+4=0##.

Homework Equations



## \vec r = \vec r_0 + t\vec v ##
(for orthogonal vectors) ## v \bullet w = 0 ##[/B]

The Attempt at a Solution



So I started out develeping the equation of the line through the point and came up with

## L = \lbrack 4,0,-4 \rbrack + t\lbrack a,b,c \rbrack ##

Now I needed to find the direction vector a, b, and c.

Since the line is parallel to the 2 planes, the direction vector should be orthogonal to both of the planes' normal vectors.

## \lbrack a,b,c \rbrack \bullet \lbrack 1,-8,7 \rbrack = 0##

## \lbrack a,b,c \rbrack \bullet \lbrack 4,3,-1 \rbrack = 0 ##

So, now I have 2 equations with the 3 variables, and I went to solve them in terms of c.

## a - 8b + 7c = 0 ## and ## 4a + 3b - c = 0 ##

## b = \frac {29c} {35} ##
## a = \frac {-216c} {35} ##
Your value for a is incorrect, which you can see by substituting back into your equations.

I found it much easier just to take the cross product of the normals to the two planes.
RyanTAsher said:
I wasn't really sure what to do at this point, as these answers seemed pretty obscure, and didn't know what to plug in for C, besides maybe c = 35. But when I do that, it doesn't match any of the multiple choice responses.

But the problem comes with pretty defined answers to pick from, multiple choice:

##a) x = 4 + t, y = -8t, z = -4 + 7t ##
##b) x = 4 - t, y = 8t, z = -4 - 7t ##
##c) x = 4 - 13t, y = 29t, z = -4 + 35t ##
##d) x = 4 + 13t, y = -29t, z = -4 - 35 ##
##e) x = 4 + 4t, y = 3t, z = -4 - t ##
 
  • #4
LCKurtz said:
I didn't check your arithmetic, but you can probably pick c arbitrarily since many different length vectors have the same direction. But wouldn't it be easier to just note the direction you want is in the direction of the cross product of your two normals?

Mark44 said:
Your value for a is incorrect, which you can see by substituting back into your equations.

I found it much easier just to take the cross product of the normals to the two planes.

How do you mean? Do you just mean literally taking the cross product of the normals and doing it that way? How would you make sure it goes through the point then?
 
  • #5
RyanTAsher said:
How do you mean? Do you just mean literally taking the cross product of the normals and doing it that way? How would you make sure it goes through the point then?

You have the point given. You just need the direction vector to finish the equation of the line. That's what the cross product gives.
 
  • #6
LCKurtz said:
You have the point given. You just need the direction vector to finish the equation of the line. That's what the cross product gives.

Okay, I get it! Thank you!
 

FAQ: Find parametric equation at point, and parallel to planes?

1. What is a parametric equation at a given point?

A parametric equation is a set of equations that represent a curve or surface in terms of one or more parameters, such as time or distance. At a given point, the parametric equations will provide the x, y, and z coordinates of that point on the curve or surface.

2. How do you find a parametric equation at a given point?

To find a parametric equation at a given point, you will need to know the coordinates of the point as well as the direction of the curve or surface at that point. This can be determined by finding the tangent vector at the point. Then, the parametric equations can be written in terms of the tangent vector and the coordinates of the point.

3. What is the significance of finding a parametric equation at a given point?

Finding a parametric equation at a given point allows us to describe the behavior of a curve or surface at that specific point. This can be useful in many applications, such as in physics, engineering, and computer graphics.

4. How do you find a parametric equation parallel to a given plane?

To find a parametric equation parallel to a given plane, we can use the normal vector of the plane to determine the direction of the parametric equations. The coordinates of any point on the plane can be used as the starting point for the parametric equations. Then, we can write the parametric equations in terms of the normal vector and the coordinates of the starting point.

5. Why is it important to find a parametric equation parallel to a given plane?

Finding a parametric equation parallel to a given plane allows us to describe a curve or surface that lies in the same direction as the plane. This can be useful in visualizing and analyzing geometric shapes, as well as in solving various mathematical problems.

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