My Math Forum Equations of a plane

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 May 29th, 2017, 06:32 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 08:56 AM.
 May 29th, 2017, 07:54 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 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 10:42 PM.
 May 29th, 2017, 10:01 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 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.
May 29th, 2017, 10:03 AM   #4
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Quote:
 Originally Posted by Country Boy 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 10:44 PM.

 June 27th, 2017, 07:51 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 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.
 June 28th, 2017, 07:26 AM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 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 07:40 AM.
June 28th, 2017, 08:48 AM   #7
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Plane Through Line and Point

Quote:
 Originally Posted by zylo 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)

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