- #1

Dukefool

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## Homework Statement

The two lines are in symmetric equations. L

_{1}is (x-1)/2=(y+1)/3=-z. L

_{2}is (x+1)/2=(y-1)/3=3-z. Find the equation of the plane containing the lines.

The point on L

_{1}is (1,-1,0) with slope of (2,3,-1)

The point on L

_{2}is (-1,1,3) with slope of (2,3,-1)

## Homework Equations

a(x-x

_{0})+b(y-y

_{0})+c(z-z

_{0})=0

Normal vector =V(P

_{1}P

_{2})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 x

_{0},y

_{0},z

_{0}correct but why are the a, b, and c values wrong?

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