Vectors and Planes Question

Question

There are two planes:

x+4y-3z+12=0 and 2x+8y-6z+4=0

a) Explain how I know these planes are parallel and distanced?

b) How far apart are these lines?

Answer 1

the 2nd equation is just twice the 1st equation, except for the constant term 4 = (1/3)(12)

they would have represented graphically the identical plane, if the 2nd constant term had been 24 not 4.

the planes are parallel

It's similar to 2 dimensions

x + y = 12

2x+2y =24 are the same line

but

x+y =12 and

x+y = 4 are parallel lines

when x=y=0, the lines are 12-4 = 8 units apart, vertically, or horizontally.

but you want the perpendicular distance

distance apart = 8/sqr2 = about 8(/1.414) = 5.67

for the plane when x=y=z=0, a vertical distance between the planes = 12-2=10

but you want the perpendicular distance between the planes

d = |d1-d2|/sqr(a^2 +b^2 +c^2)

=(12-2)/sqr(1^2 +4^2 +3^2) = 10/sqr26 = about 1.96

write the two equations so that all coefficients are the same, with only the constant term different

x +4y -3z -12 = 0

x+4y -3z -2 = 0 after dividing the 2nd equation by 2 (2x+8y-6z-4)/2=0

then a=1, b=4, c=-3, d1=-12, d2 =-2

plug into the formula distance between the parallel planes is the absolute value of the difference in constant terms divided by the square root of the sum of squares of the coefficients of the 3 variables.