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.
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
thanks!
To determine the vector equation of the plane that contains the two given lines, we need two vectors that lie in the plane. One way to obtain these vectors is by taking the direction vectors of the given lines.
Let's call the direction vector of line l1 as vector v1 and the direction vector of line l2 as vector v2.
v1 = [2, 0, 3]
v2 = [5, 1, -1]
Now, we need a point on the plane. We can choose any point that lies on both lines, such as the initial points of the lines.
Point A = [4, -3, 5]
Using the point A and the direction vectors v1 and v2, we can write the vector equation of the plane as follows:
vector r = A + s * v1 + t * v2
r = [4, -3, 5] + s[2, 0, 3] + t[5, 1, -1]
This is the vector equation of the plane that contains the given lines.
To determine the corresponding Cartesian equation, we can expand the vector equation. Let's assume the components of vector r as x, y, and z.
x = 4 + 2s + 5t
y = -3 + t
z = 5 + 3s - t
Now, we can rearrange these equations to obtain the Cartesian equation.
x - 2s - 5t = 4
y - t = -3
z - 3s + t = 5
Thus, the corresponding Cartesian equation of the plane is:
x - 2s - 5t = 4
y - t = -3
z - 3s + t = 5