Question

find the distance between the planes x+2y-3z=8 and 3x+6y-9z=10

Answer 1

Distance between two parallel planes , ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is $\bold{\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}}$

Here given two planes x + 2y - 3z = 8
and 3x + 6y - 9z = 10 ⇒3(x + 2y - 3z) = 10
⇒x + 2y - 3z = 10/3

Now, you can both the given planes are parallel
So, distance between it is |8 - 10/3|/√{1² + (2)² + (-3)²} = |14/3|/√14 = √14/3 unit

Answer 2

Hi ,

First , we must rearrange the equations in

standard form

x + 2y - 3z - 8 = 0 -----( 1 )

3x + 6y - 9z - 10 = 0 ---( 2 )

1 ) check the ratios of coefficients ,

i ) ratio of x coefficients : 1/3

ii ) ratio of y coefficients : 2/6 = 1/3

iii ) ratio of z coefficients : ( - 3 ) / ( - 9 ) = 1/3

Therefore ,

ratios are equal .

given planes are parallel.

2 ) Finding a point on a first plane

assume y = 0 , z = 0 in the

equation x + 2y - 3z - 8 = 0

we get ,

x - 8 = 0

x = 8

P( x1 , y1 , z1 ) = ( 8 , 0 , 0 )

3 ) finding the distance between two

planes :

distance from the point P to second plane

D = | ax1 + by1 + cz1 + d | / √( a² + b² + c² )

= | 3 × 8 - 10 | / √ ( 3² + 6² + 9² )

= 14 / √ ( 126 )

I hope this helps you.

: )