Q2b) find the shortest distance between the planes 2x+2y-z-3=0 and 4x+4y-2z+9=o?

Answer 1

2x+2y-z-3=0
Pick a point (0, 0, -3) from the plane.
d = |-2(-3)+9|/6 = 1/2

Answer 2

Q2b) find the shortest distance between the planes
2x + 2y - z - 3 = 0
4x + 4y - 2z + 9 = 0

The two planes are parallel since the normal vectors are non-zero multiples of each other. Since every point in a parallel plane is the same distance from the other plane, pick a point in one plane and measure the distance to the other plane.

Pick the point P(0, 0, -3) in the first plane and measure the distance to the second plane. Use the distance formula.

d = | 4*0 + 4*0 -2*(-3) + 9 | / √[4² + 4² + (-2)²]
d = | 0 + 0 + 6 + 9 | / √(16 + 16 + 4)
d = 15 / √36 = 15/6 = 5/2

Answer 3

let A represent as 2x+2y-z-3=0
let B represent as 4x+4y-2z+9=0
2xA-B=15.
so, the shortest distance between the planes 2x+2y-z-3=0 and 4x+4y-2z+9=0 is 15