question_answer 32)

In the following exercise, find the distance of each of the given points from the corresponding given plane. Points Planes (i) (0, 0, 0) 3x – 4y + 12z = 3 (ii) (3, –2, 1) 2x – y + 2z + 3 = 0 (iii) (2, 3, –5) x + 2y – 2z = 9 (iv) (–6, 0, 0) 2x – 3y + 6z – 2 = 0  

Answer:

As we know that the length of the perpendicular from point P(x1, y1, z1) from the plane a1x+ b1y+ c1z+d1 = 0 is given by (i)     Here point is P(0, 0, 0) and equation fo the plane is 3x ? 4y + 12 z ? 3 = 0  Required distance = length of the perpendicular drawn from point (0, 0, 0) to the plane 3x ? 4y + 12z ? 3 = 0        units. (ii)    Here point is (3, ?1, 1) and equation of the plane is 2x ? y + 2z + 3 = 0.    Required distance = length of the perpendicular drawn from point (3, ?2, 1) to the plane 2x ? y + 2z + 3 = 0 (iii) Here point is (2, 3, ?5) and equation of the plane is       x + 2y ? 2z = 9                     Required distance = length of the perpendicular drawn from point (2, 3, ?5) to the       plane x + 2y ? 2z ? 9 = 0             (iv)  Here point is P(?6, 0, 0) and equation of the plane is       2x ? 3y + 6z ? 2 = 0  Required distance = length of the perpendicular drawn from point (?6, 0, 0) to the plane       2x ? 3y + 6z ? 2 = 0