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

Answer: (B)

$\sqrt{\frac{5}{142}}$

Answer 2

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 $= | \frac{5 }{\sqrt{(15)²+(22)²+1²}} |$
$⇒ d = \frac{5}{\sqrt{710}} = \sqrt{\frac{25}{710}}$
$= \sqrt{\frac{5}{142}}$