If the length of the perpendicular drawn from the point P(a, 4, 2), a> 0 on the line
\(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{1}\) is \(2\sqrt6\) units and \(Q(α1, α2, α3)\)
is the image of the point P in this line, then
\(\alpha + \sum_{i=1}^{3} \alpha_i\)
is equal to :
If the length of the perpendicular drawn from the point P(a, 4, 2), a> 0 on the line
\(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{1}\) is \(2\sqrt6\) units and \(Q(α1, α2, α3)\)
is the image of the point P in this line, then
\(\alpha + \sum_{i=1}^{3} \alpha_i\)
is equal to :
- 7
- 8
- 12
- 14
The Correct Option is B
Solution and Explanation
The correct answer is (B) : 8
∵ PR is perpendicular to given line, so
\(2(2λ-1-α)+3(3λ-1)-1(-λ-1)=0\)
\(⇒ α = 7λ-2\)
Now,
\(∵ PR = 2\sqrt6\)
\(⇒ (-5λ+1)^2 + (3λ-1)^2 + (λ+1)^2 = 24\)
\(⇒ 5λ^2 - 2λ-3 = 0\)
\(⇒ λ = 1\ or -\frac{3}{5}\)
\(∵ α>0\ so λ = 1\ and α = 5\)
Now \( \sum_{i=1}^{3} \alpha_i\) = 2(Sum of co-ordinate of R)-(Sum of co-ordinates of P)
= 2(7)-11
= 3
a+\( \sum_{i=1}^{3} \alpha_i\)= 5+3
= 8
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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.
