Question

What's the Distance between

2x-2y+z-9=0 and (-1,2,0)?​

Answer 1

Let's check up, whether planes are parallel, for this purpose we will multiply the equation of the second plane on 2

2x + 4y - 4z + 18 = 0

As planes are parallel than for calculation distance between planes we use the formula:

d =   |18 - (-6)|   =   |24|   =   24   = 4

√22 + 42 + (-4)2  √36  6

Answer: distance from plane to plane is equal to 4.

Answer 2

Answer:

$D = \dfrac{11}{3}$

Step-by-step explanation:

Given equation-

2x - 2y + z - 9 = 0

Comparing with ax + by + cz + d = 0

Here a = 2, b = -2, c = 1 and d = -9

Distance between 2x - 2y + z - 9 = 0 and points (-1, 2, 0) = ?

Using formula-

Distance between ax + by + cz + d = 0 and point (p, q, r) is-

$D = \dfrac{ap + bq + cr + d}{\sqrt{a^{2}+b^{2}+c^{2}   } }$

So, D = $\dfrac{(-2 \times-1)+(-2 \times2)+(1 \times0)-9}{\sqrt{(2)^{2}+(-2)^{2}+(1)^{2}   } }$

               = $\dfrac{2 -4 -9}{\sqrt{4+4+1} }$

               = $\dfrac{2-13}{\sqrt{9} }$

D = $-\dfrac{11}{3}$

Neglecting -ive sign because D is distance.

So, $D = \dfrac{11}{3}$