Question

Let `(x - 2)/3 = (y + 1)/(-2) = (z + 3)/(-1)` lie on the plane px – qy + z = 5, for p, q ∈ R. The shortest distance of the plane from the origin is ______.

Options

• `sqrt(3/109)`

• `sqrt(5/142)`

• `5/sqrt(71)`

• `1/sqrt(142)`

Answer 1

Let `(x - 2)/3 = (y + 1)/(-2) = (z + 3)/(-1)` lie on the plane px – qy + z = 5, for p, q ∈ R. The shortest distance of the plane from the origin is `underlinebb(sqrt(5/142))`.

Explanation:

Given: Line L: `(x - 2)/3 = (y + 1)/(-2) = (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)

On solving equation (i) and equation (ii), we get

p = 15, q = –22

∴ Equation of plane is 15x + 22y + z – 5 = 0

Now, distance of plane from origin is d = `|5/sqrt((15)^2 + (22)^2 + 1^2)|`

⇒ d = `5/sqrt(710) = sqrt(25/710) = sqrt(5/142)`