The equations of the line in general form are
3x + 2y - z - 4 = 0
and 4x + y - 2z + 3 = 0 (1)
Let l, m and n be the direction cosines of the line. Since the line is common to both the planes, it is perpendicular to the normals to both the planes. Hence,
3l + 2m - n = 0
and
4l + m - 2n = 0
Solving these, we get
Now to find the coordinates of a point on a line, let us find out the point where it meets the plane z = 0.
Putting z = 0 in the equation given by (1), we have
3x + 2y - 4 = 0 and
4x + y + 3 = 0
Solving these, we get x = -2 and y = 5.
So, one point of the line is (-2, 5, 0).
Therefore equation of the line in symmetrical form is
