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 :
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 :
\(\sqrt{\frac{3}{109}}\)
\(\sqrt{\frac{5}{142}}\)
\(\frac{5}{\sqrt{71}}\)
\(\frac{1}{\sqrt{142}}\)
The Correct Option is B
Solution and Explanation
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 \(= | \frac{5 }{\sqrt{(15)²+(22)²+1²}} |\)
\(⇒ d = \frac{5}{\sqrt{710}} = \sqrt{\frac{25}{710}} \)
\(= \sqrt{\frac{5}{142}}\)
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Concepts Used:
Three Dimensional Geometry
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
