Find the projection of the line 3x – y +2z – 1 = 0, x +2y – z = 2 on the plane 3x + 2y +z = 0.
probably the intersection of the first two planes.
Find the projection of the line 3x – y +2z – 1 = 0, x +2y – z = 2 on the plane 3x + 2y +z = 0
To find the projection of a line onto a plane, we can use the formula for the projection vector.
First, let's find the direction vector of the line. In this case, the direction vector is the coefficients of x, y, and z in the equation of the line. Therefore, the direction vector of the line is <3, -1, 2>.
Next, we need to find a normal vector of the plane. The normal vector of a plane is the coefficients of x, y, and z in the equation of the plane. Therefore, the normal vector of the plane is <3, 2, 1>.
The projection vector can be found by projecting the direction vector of the line onto the normal vector of the plane. The formula for the projection vector is:
proj = (dot product of the direction vector and the normal vector) / (magnitude of the normal vector)^2 * normal vector
Let's calculate the projection vector:
dot product of the direction vector and the normal vector = (3)(3) + (-1)(2) + (2)(1) = 9 - 2 + 2 = 9
magnitude of the normal vector = sqrt(3^2 + 2^2 + 1^2) = sqrt(14)
proj = 9 / (14) * <3, 2, 1> = <27/14, 18/14, 9/14>
Therefore, the projection of the line 3x - y + 2z - 1 = 0, x + 2y - z = 2 on the plane 3x + 2y + z = 0 is given by the equation:
x = 27/14
y = 18/14
z = 9/14