Question

find the scalar equation of the plane that passes through point p(−1, 1, 4) and is perpendicular to the line of intersection of planes x y − z − 2 = 0 and 2x − y+3z − 1 = 0.

Answer 1

The scalar equation of the plane passing through point P(-1, 1, 4) and perpendicular to the line of intersection of the given planes is -4(x + 1) - 5(y - 1) - 3(z - 4) = 0.

Step-by-step explanation:

To find the scalar equation of the plane passing through point P(-1, 1, 4) and perpendicular to the line of intersection of the planes x + y - z - 2 = 0 and 2x - y + 3z - 1 = 0, we can first find the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes. The normal vector of the first plane is (1, 1, -1) and the normal vector of the second plane is (2, -1, 3). Taking the cross product of these two vectors, we get the direction vector of the line of intersection as (-4, -5, -3).

Since the plane we are looking for is perpendicular to this line, the normal vector of the plane will be the same as the direction vector of the line. So, the equation of the plane is given by:

-4(x + 1) - 5(y - 1) - 3(z - 4) = 0