Question

Let
\(\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z+3}{-1}\)
lie on the plane px – qy + z = 5, for some p, q ∈ ℝ. The shortest distance of the plane from the origin is :

• A. \(\sqrt{\frac{3}{109}}\)
• B. \(\sqrt{\frac{5}{142}}\)
• C. \(\frac{5}{\sqrt{71}}\)
• D. \(\frac{1}{\sqrt{142}}\)

Answer 1

The Correct Option is B

The correct answer is (B) : \(\sqrt{\frac{5}{142}}\)
Given: 
Line L: \(\frac{(x - 2)}{3} = \frac{(y + 1)}{- 2} = \frac{(z + 3)}{- 1}\)
And plane P: px - qy + z = 5
∵ Line L lies on the plane p 
∴ Point (2, -1, -3) will satisfy the equation of plane 
So, 2p + q - 3 = 5 
⇒ 2p + q = 8 .... (i)
The line is also parallel to the plane. 
∴ 3p - 2q - 1 = 0 
3p - 2q =1...(ii) 
By solving equation (i) and equation (ii), 
we get p = 15 ,q = - 22 
Therefore , Equation of plane is 15x + 22y + z - 5 = 0
Now , distance of plane from origin is d \(= | \frac{5 }{\sqrt{(15)²+(22)²+1²}} |\)
\(⇒ d = \frac{5}{\sqrt{710}} = \sqrt{\frac{25}{710}} \)
\(= \sqrt{\frac{5}{142}}\)