Find an equation of the plane through the point and perpendicular to the given line?

Consider the plane defined by the equation x + y + z = 1, and the point (2,−1, 3). Answer the following?

• (a) Find the equation of the line that passes through this point and which is perpendicular to this plane. (b) Find the point where this line meets the given plane. (c) Find the shortest distance between the given point and the plane. (Do not use any formulas, solve it conceptually using parts a and b.)

• Answer:

  (a) Normal vector to a plane ax + by + cz = d is (a, b, c) So normal vector to plane x + y + z = 1 is (1, 1, 1) This is direction vector of any line that is perpendicular to this plane. Parametric equation of line perpendicular to plane that passes through point (2, -1, 3): x = 2 + t, y = -1 + t, z = 3 + t Converting to regular equation: x - 2 = t, y + 1 = t, z - 3 = t x - 2 = y + 1 = z - 3 (b) x - 2 = y + 1 y = x - 3 x - 2 = z - 3 z = x + 1 Replacing these values of y and z into equation of plane, we get: x + (x - 3) + (x + 1) = 1 3x - 2 = 1 3x = 3 x = 1 y = x - 3 y = -2 z = x + 1 z = 2 Point where line meets plane: (1, -2, 2) (c) Shortest distance between given point (2, -1, 3) and plane is the distance between given point and point where line meets plane (1, -2, 2). This is the perpendicular distance. d = √((2-1)² + (-1+2)² + (3-2)²) = √(1+1+1) = √3