Find the equation of the plane that passes through the point (3,7,-1) and is perpendicular to the line of intersection of the planes x-y-2z+3=0 and 3x-2y+z+5=0
The line of intersection of the planes is in the direction of the cross-product of the two normals:
{1,-1,-2}x{3,-2,1} = {-5,-7,1}
Now we have a point and a normal vector.
The equation of the plane is thus
-5(x-3) - 7(y-7) + 1(z-1) = 0
-5x + 15 - 7y + 49 + z - 1 = 0
5x+7y-z = 63
To find the equation of a plane passing through a given point and perpendicular to a line of intersection between two planes, we need to follow a few steps:
Step 1: Find the direction vector of the line of intersection between the two planes.
Step 2: Determine a normal vector for the plane by taking the cross product of the direction vectors of the line of intersection and a vector from the point on the plane to any other point.
Step 3: Write the equation of the plane in the form ax + by + cz = d using the normal vector and the coordinates of the given point.
Let's go through these steps one by one:
Step 1: Finding the direction vector of the line of intersection
To find the direction vector of the line of intersection, we equate the normal vectors of the two planes:
The normal vector of the first plane (denoted by N1) is <1, -1, -2>.
The normal vector of the second plane (denoted by N2) is <3, -2, 1>.
Taking the cross product of N1 and N2 will give us the direction vector of the line of intersection:
N1 x N2 = <1, -1, -2> x <3, -2, 1> = <-3, -1, 1>.
So, the direction vector of the line of intersection is <-3, -1, 1>.
Step 2: Determining a normal vector for the plane
Now that we have the direction vector of the line of intersection, we need another vector that goes from the given point (3, 7, -1) on the plane to any other point. Let's take the vector pointing from (3, 7, -1) to (2, 9, 4), for example:
Vector P = <2 - 3, 9 - 7, 4 - (-1)> = <-1, 2, 5>.
We can now take the cross product of the direction vector of the line of intersection and vector P to get a normal vector for the plane:
N = <-3, -1, 1> x <-1, 2, 5> = <-7, 2, -5>.
So, the normal vector for the plane is <-7, 2, -5>.
Step 3: Writing the equation of the plane
Using the normal vector <a, b, c> and the coordinates of the given point (x0, y0, z0) = (3, 7, -1), we can write the equation in the form:
a(x - x0) + b(y - y0) + c(z - z0) = 0.
Substituting the values, we have:
-7(x - 3) + 2(y - 7) - 5(z - (-1)) = 0.
Simplifying:
-7x + 21 + 2y - 14 - 5z + 5 = 0.
-7x + 2y - 5z + 12 = 0.
Therefore, the equation of the plane that passes through the point (3, 7, -1) and is perpendicular to the line of intersection of the planes x - y - 2z + 3 = 0 and 3x - 2y + z + 5 = 0 is -7x + 2y - 5z + 12 = 0.