My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Thanks Tree1Thanks
  • 1 Post By Country Boy
Reply
 
LinkBack Thread Tools Display Modes
May 29th, 2017, 07:32 AM   #1
Newbie
 
Joined: May 2017
From: Tanzania

Posts: 8
Thanks: 0

Equations of a plane

I have encountered a question which is kind of different and strange to me; it says,
Find the equations of the planes through (2,-1,3) and contains (x-2)/3 = (y-1)/2 = (1-z)/4 and -x/2 =3-y = (z-3)/2

Now, what is confusing me is those two equations; I really don't know where to start. I need help here.

Last edited by skipjack; June 28th, 2017 at 09:56 AM.
Techgenius is offline  
 
May 29th, 2017, 08:54 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,821
Thanks: 750

Yes, that is a bit peculiar. A plane can be determined by two (non-skew) lines or one point and one line. Unless the given point lies in the plane determined by the two lines, there is no such plane.

The two given lines are (x-2)/3 = (y-1)/2 = (1-z)/4 and -x/2 = 3-y = (z-3)/2. Setting (x-2)/3 = (y-1)/2 = (1-z)/4= t, the first line can be written in parametric equations, as x = 3t + 2, y = 2t + 1, and z = 1 - 4t. Setting -x/2 = 3-y = (z-3)/2 = s, the second line can be written, in parametric equations, as x = 2s, y = -s + 3, z = 2s + 3. The first line is in the same direction as the vector <3, 2, -4> and the second as the vector <2, -1, 2>. Their cross product, $\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 2 \\ 3 & 2 & -4\end{array}\right|= 14\vec{j}+ 7\vec{k}$ is perpendicular to both vectors, so normal to the plane they determine. Taking t = 0, one point on the first line, and so in the plane containing that line, is (2, 1, 1). The equation of the plane with that normal vector and containing that point, is 0(x - 2) + 14(y - 1) + 7(z - 1)= 0 or 14y + 7z = 21.

Now, does the given point, (2, -1, 3)? If not, this is impossible.
Thanks from SenatorArmstrong

Last edited by skipjack; May 29th, 2017 at 11:42 PM.
Country Boy is offline  
May 29th, 2017, 11:01 AM   #3
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,175
Thanks: 90

Any two vectors a,b have a common normal axb. It doesn't mean they are in the same plane.

Vectors along each line are given by a=(3,2,-4) and b=(-2,-1,2) which gives a normal to both lines as n=axb=(0,2,1), which Countryboy also gets.

(2,1,1) is in L1. Plane through L1 with normal n is (r-r0).n=0
2(y-1)+(z-1)=0

(0,3,3) is in L2 but not in above plane.
There is no plane through the two lines.
zylo is offline  
May 29th, 2017, 11:03 AM   #4
Newbie
 
Joined: May 2017
From: Tanzania

Posts: 8
Thanks: 0

Quote:
Originally Posted by Country Boy View Post
Yes, that is a bit peculiar. A plane can be determined by two (non-skew) lines or one point and one line. Unless the given point lies in the plane determined by the two lines, there is no such plane.

The two given lines are (x-2)/3 = (y-1)/2 = (1-z)/4 and -x/2 = 3-y = (z-3)/2. Setting (x-2)/3 = (y-1)/2 = (1-z)/4= t, the first line can be written in parametric equations, as x = 3t + 2, y = 2t + 1, and z = 1 - 4t. Setting -x/2 = 3-y = (z-3)/2 = s, the second line can be written, in parametric equations, as x = 2s, y = -s + 3, z = 2s + 3. The first line is in the same direction as the vector <3, 2, -4> and the second as the vector <2, -1, 2>. Their cross product, $\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 2 \\ 3 & 2 & -4\end{array}\right|= 14\vec{j}+ 7\vec{k}$ is perpendicular to both vectors, so normal to the plane they determine. Taking t = 0, one point on the first line, and so in the plane containing that line, is (2, 1, 1). The equation of the plane with that normal vector and containing that point, is 0(x - 2) + 14(y - 1) + 7(z - 1)= 0 or 14y + 7z = 21.

Now, does the given point, (2, -1, 3)? If not, this is impossible.
He used something called the Cartesian form and got the answer which is 0x+2y+z=1.Your formula seems to be cool though, I don't know where did it get wrong.

Last edited by skipjack; May 29th, 2017 at 11:44 PM.
Techgenius is offline  
June 27th, 2017, 08:51 AM   #5
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,821
Thanks: 750

Well, yes. If you divide each term of my form, 14y+ 7z= 21, by three you get that form: 2y+ z= 1.

The final point was to check if (2, -1, 3), x= 2, y= -1, z= 3, satisfies that. 2(-1)+ 3= 1. So yes, that plane satisfies all of the conditions.
Country Boy is offline  
June 28th, 2017, 08:26 AM   #6
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,175
Thanks: 90

The OP lines are:

$\displaystyle \frac{x-2}{3}=\frac{y-1}{2}=\frac{z-1}{-4}$ and $\displaystyle \frac{x}{-2}=\frac{y-3}{-1}=\frac{z-3}{2}$

These intersect if they have a common solution. The Eqs in y and z are, respectively:
2y+z=3
2y+z=9
which have no solution, so the lines don't intersect and can't have a plane which contains them both, as pointed out in post #3, ie,

There is no plane which satisfies OP.

It is alleged 2y+z=1 is the solution. In that case it must contain the points (2,1,1) and (0,3,3), which are points on the two lines respectively. It doesn't so it isn't.

EDIT: There is another interpretation to OP. Find the planes which go through given point and each of the given lines, in which case find the equation of a plane through a given point and line.

Last edited by zylo; June 28th, 2017 at 08:40 AM.
zylo is offline  
June 28th, 2017, 09:48 AM   #7
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,175
Thanks: 90

Plane Through Line and Point

Quote:
Originally Posted by zylo View Post
The OP lines are:

$\displaystyle \frac{x-2}{3}=\frac{y-1}{2}=\frac{z-1}{-4}$ and $\displaystyle \frac{x}{-2}=\frac{y-3}{-1}=\frac{z-3}{2}$
Given Point: (2,-1,3)

EDIT: There is another interpretation to OP. Find the planes which go through given point and each of the given lines, in which case find the equation of a plane through a given point and line.
Normal to plane: N=(r$\displaystyle _{p}$-r$\displaystyle _{l})$Xl
Equation of plane: (r-r$\displaystyle _{p}$)$\displaystyle \cdot$N=0

r$\displaystyle _{p}$=given point
r$\displaystyle _{l}$=point on line
l=dir vector of line
r=point in plane

Example: Plane through given point and line 1.
r$\displaystyle _{p}$=(2,-1,3)
r$\displaystyle _{l}$=(2,1,1)
l=(3,2,-4)
r=(x,y,z)
zylo is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
equations, plane



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Find all points on surface whose tangent plane is parallel to given plane Addez123 Calculus 4 November 12th, 2016 07:51 AM
2 Points in R3 and 1 point on plane. How to find equation of plane? extreme112 Linear Algebra 2 October 13th, 2015 08:34 AM
How are plane waves, p-forms, and Maxwell's equations related? Ganesh Ujwal Physics 1 January 7th, 2015 04:00 AM
Slove parametric equations for a point of a plane tombfighter Algebra 2 September 11th, 2008 07:15 PM
Slove parametric equations for a point of a plane tombfighter Abstract Algebra 1 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.