Find the possible value of k for which the distance between two plane 6x + 3y - 2z + k = 0 and 3x + 1.5y - z + 2 = 0 is 5 ?

Concept:

The distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz +d2 = 0 is given by: \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)

Calculation:

Given: 6x + 3y - 2z + k = 0 and 3x + 1.5y - z + 2 = 0 are two planes and the shortest distance between them is 5.

The equation of plane 3x + 1.5y - z + 2 = 0 can be re-written as: 6x + 3y - 2z + 4 = 0 by multiplying both sides of the equation 3x + 1.5y - z + 2 = 0 by 2.

As we can see that, the given planes 6x + 3y - 2z + k = 0 and 6x + 3y - 2z + 4 = 0 are parallel.

As we know that, the distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz +d2 = 0 is given by: \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)

Here, d₁ = k, d₂ = 4, a = 6, b = 3, c = - 2 and D = 5.

​⇒ \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)​

⇒ \(5= \left | \frac{k-4}{\sqrt{6^2+3^2+(-2)^2}} \right |\)

⇒ ​\(5= \left | \frac{k-4}{\sqrt{36+9+4}} \right |\)

⇒ 5 x 7 = |k - 4|

Case -1: If k ≥ 4 then k - 4 = 35 ⇒ k = 39

Case - 2: If k < 4 then -k + 4 = 35 ⇒ k = - 31

So, k = 39, - 31

Hence, option 4 is correct.