# How to find an equation for two parallel lines

## Homework Statement

The two lines are in symmetric equations. L1 is (x-1)/2=(y+1)/3=-z. L2 is (x+1)/2=(y-1)/3=3-z. Find the equation of the plane containing the lines.
The point on L1 is (1,-1,0) with slope of (2,3,-1)
The point on L2 is (-1,1,3) with slope of (2,3,-1)

## Homework Equations

a(x-x0)+b(y-y0)+c(z-z0)=0
Normal vector =V(P1P2)XSlope

## The Attempt at a Solution

With same slope, you know the lines are parallel.
So the vector from P1 to P2 is (-2,2,3)
Cross product of vector P1 to P2 and the slope is (-11,-8,4)
My thought was that the normal vector has the value of (a,b,c) in the equation for plane. I plugged in the values from the normal vector along with P1 to get:
-11(x-1)-8(y+1)+4z=0.
According to the solutions sheet, the answer is 11(x-1)-4(y+1)+10z=0.
I got the x0,y0,z0 correct but why are the a, b, and c values wrong?

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## Homework Statement

The two lines are in symmetric equations. L1 is x-1/2=y+1/3=-z. L2 is x+1/2=y-1/3=3-z. Find the equation of the plane containing the lines.
The point on L1 is (1,-1,0) with slope of (2,3,0)
To start with, neither of these is true. 1- 1/2= 1/2 while -1+1/3= -2/3 and neither of those is equal to -0. (1, -1, 0) is NOT a point on this line.

If x= 1, then 1-1/2= 1/2= y+ 1/3 so y= 1/2- 1/3= 1/6 and -z= 1/2 so z= -1/2. The point (1, 1/6, -1/2) is on L1.

If x= 2, then 2- 1/2= 3/2= y+ 1/3 so y= 3/2- 1/3= 7/6 and -z= 3/2 so z= -3/2. The point (2, 7/6, -3/2) is also on the line and the vector from the first point to the second, and so in the direction of the line is <2-1, 7/6- 1/6, -1/2-(-3/2)>= <1, 1, 1>

The point on L2 is (-1,1,3) with slope of (2,3,0)
Again, not true. -1+1/2= -1/2 which is not equal to 1-1/3= -2/3 and neither is equal to 3-3= 0 so (-1, 1, 3) is NOT a point on this line. If x= -1, then x+1/2= -1/2 so y-1/3= -1/2 gives y= 1/3- 1/2=1/6 and 3- z= -1/2 so z=3+ 1/2=7/2. A point on L2 is (-1, 1/6, 7/2). if x= 1, x+1/2= 3/2 so y- 1/3= 3/2 and y= 1/3+ 3/2= 11/6. 3- z= 3/2 gives z= 3- 3/2= 3/2. Another point on L2 is (1, 11/6, 3/2). the vector from the first to the seocnd, and so in the direction of the line is <-1- 1, 11/6- 1/6, 3/2- 7/2>= <-2, 5/3, -2>. Since this is NOT a multiple of the previous vector, <1, 1, 1>, the two lines are NOT parallel.

## Homework Equations

a(x-x0)+b(y-y0)+c(z-z0)=0
Normal vector =V(P1P2)XSlope
I don't know what this means. The normal vector is the cross product of the previous two vectors and you can use any point on either line, in particular, any of the four points already calculated.

## The Attempt at a Solution

With same slope, you know the lines are parallel.
So the vector from P1 to P2 is (-2,2,3)
Cross product of vector P1 to P2 and the slope is (-11,-8,4)
My thought was that the normal vector has the value of (a,b,c) in the equation for plane. I plugged in the values from the normal vector along with P1 to get:
-11(x-1)-8(y+1)+4z=0.
According to the solutions sheet, the answer is 11(x-1)-4(y+1)+10z=0.
I got the x0,y0,z0 correct but why are the a, b, and c values wrong?

My apologizes as I have forgotten to add in the parenthesis to clarify the equations of the lines.