Given
L1: r= [4,-3,5] + t[2,0,3], t∈R
L2: r= [4,-3,5] + s[5,1,-1], s∈R
We need to find the plane that contains both L1 & L2. (changed to uppercase to avoid confusion with 1).
Notice that P0=[4,-3,5] lies in the given plane, in fact, it is the intersection point of L1 and L2 (when t=s=0).
The normal vector of a plane is the cross product of two (non-parallel) vectors that lie in the plane.
Two such vector have been given as:
V1=<2,0,3> and V2=<5,1,-1> corresponding to the direction vector of each line.
the cross-product V1xV2 is the vector N obtained by expanding the following determinant:
i j k
2 0 3
5 1 -1
which gives N=<-3, 17, 2>
Using the fact that P0 lies in the plane, we have the vector equation
N.(R-P0)=0 where R=(x,y,z) is a general point in the plane with coordinates (x,y,z).
For diagrams and explanations, see for example:
http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
The Cartesian equation is obtained by expanding the vector equation:
<-3, 17, 2>.(<x,y,z>-<4,-3,5>)=0
into
<-3, 17, 2>.<x-4, y+3, z-5>=0
=>
2(z-5)+17(y+3)-3(x-4)=0
=>
-3x+17y+2z+53=0
A) Determine the vector equation of the plane that contains the following two lines:
l1: vector r= [4,-3,5] + t[2,0,3], tER
l2: vector r= [4,-3,5] + s[5,1,-1], sER
b)Determine the corresponding Cartesian equation.
2 answers
thanks!
3 answers