Question

find the point of intersection of the line

1) Find the equation of the line passing through the points A(1,-5,-3)and B(2,-4,8) (3 marks) b) Find the equation of the plane perpendicular to the line in part (a) given that C(1,-9,6) is a point on the plane. (3 marks) c) Find the point of intersection of the line and the plane in parts (a) and (b) above respectively. (3 marks)

Answer 1

We know that: the equation of the line passing through the points (x_1, y_1, z_1) and (x_2, y_2, z_2) is implies x - x_1/x_2 - x_1 = y - y_1/y_2 - y_1 = z - z_1/z_2 - z_1 therefore The Required eqn of the line passing through the points A(1, -5, -3) and B(2, -4, 8) is implies x - 1/2 - 1 = y + 5/-4 + 5 = z + 3/8 + 3 x - 1/1 = y + 5/1 = z + 3/11 (Answer) def: the equation of the plane passing through e(1, -9, 6) is implies A(x - 1) + B(y + 9) + C(z - 6) = 0 The plane is perpendicular to the line x - 1/1 = y + 5/1 = z + 3/11 therefore Its Normal is parallel to the st line therefore A/1 = B/1 = C/11 = lambda(Def) A = lambda: B = lambda: C = 11 lambda therefore the eqn of the plane is implies