Find the vector and scalar equations of the plane containing the line r=(3,5,1)+t(-2,3,1) and the point (1,2,3)
To find the vector equation of the plane containing a line and a point, we need a vector parallel to the plane and the coordinates of the given point.
Step 1: Find a vector parallel to the plane.
The line given in parametric form of r = (3,5,1) + t(-2,3,1) can be written as r = (3 - 2t, 5 + 3t, 1 + t).
The direction vector of the line is given by (-2,3,1).
This vector is also parallel to the plane we want to find.
Step 2: Determine the coordinates of the given point.
The given point is (1,2,3).
Step 3: Construct the vector equation of the plane.
Using the point-normal form of a plane equation, the vector equation of the plane is:
r = (1,2,3) + s(a,b,c) + t(-2,3,1),
where (a,b,c) is the direction vector of the line.
Therefore, the vector equation of the plane is:
r = (1 + sa - 2t, 2 + sb + 3t, 3 + sc + t).
Step 4: Convert the vector equation to scalar equation.
To find the scalar equation of the plane, we need to eliminate the parameters (s,t) and express the coordinates in terms of variables (x,y,z).
By comparing the corresponding coordinates in the vector equation, we can write the scalar equation:
x = 1 + sa - 2t,
y = 2 + sb + 3t,
z = 3 + sc + t.
Rearranging these equations, we get:
x - sa + 2t = 1,
y - sb - 3t = 2,
z - sc - t = 3.
These are the scalar equations of the plane containing the line r = (3,5,1) + t(-2,3,1) and the point (1,2,3).
We know one direction vector of the plane is (-2,3,1)
We have to find another.
Use the two points (3,5,1) and (1,2,3)
the vector would be (2,3,-2)
I will assume you know how to find the cross-product between these two vectors.
I got (9,2,12)
So the plane has equation
9x + 2y + 12z = c
plug in (1,2,3)
9 + 4 + 36 = c = 49
scalar equation is
9x + 2y + 12z = 49
vector equation is
r = (3,5,1) + t(-2,3,1) + s(2,3,-2)