Let the line
\(\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4}\)
intersect the plane containing the lines
\(\frac{x - 4}{1} = \frac{y + 1}{-2} = \frac{z}{1}\) and \(4ax-y+5z-7a = 0 = 2x-5y-z-3, a∈R\)
at the point P(α, β, γ). Then the value of α + β + γ equals _____.
Let the line
\(\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4}\)
intersect the plane containing the lines
\(\frac{x - 4}{1} = \frac{y + 1}{-2} = \frac{z}{1}\) and \(4ax-y+5z-7a = 0 = 2x-5y-z-3, a∈R\)
at the point P(α, β, γ). Then the value of α + β + γ equals _____.
Correct Answer: 12
Approach Solution - 1
To find the point P(α, β, γ) where the line intersects the plane, we first express the parametric form of the line:
\(x = 7t + 3\),\(y = -t + 2\),\(z = -4t + 3\).
The plane contains the lines given by:
\(L_1: \frac{x - 4}{1} = \frac{y + 1}{-2} = \frac{z}{1}\) as \(x = u + 4\), \(y = -2u - 1\), \(z = u\), and
\(L_2: 4ax - y + 5z - 7a = 0\) and \(2x - 5y - z - 3 = 0\).
We solve the equations for \(L_2\) simultaneously to find the plane equation:
Substitute \(x = u + 4\), \(y = -2u - 1\), \(z = u\) into:
\(4a(u+4) - (-2u-1) + 5u - 7a = 0\)
Simplifying gives:
\(4au + 16a + 2u + 1 + 5u - 7a = 0\)
\(11u + 9a + 1 = 0\)
The second equation:
\(2x - 5y - z - 3 = 0\), substituting gives:
\(2(u+4) + 10u + 5 - u - 3 = 0\)
Simplifying gives:
\(11u + 5 = 0\) which is independent of \(a\).
The intersection occurs by plugging into plane equations:
\(x = 7t + 3, y = -t + 2, z = -4t + 3\) satisfies:
Equation of plane through points implies:
\(11(7t+3) + 5 = 0\)\)
\(77t + 38 = 0 \rightarrow 77t = -38 \rightarrow t = -\frac{38}{77}\).\)
Substitute this value of \(t\) back:
α = \(7\left(-\frac{38}{77}\right) + 3\), β = \(-\left(-\frac{38}{77}\right) + 2\), γ = \(-4\left(-\frac{38}{77}\right) + 3\).
Calculate each:
α = \(-\frac{266}{77} + 3\), β = 2 + \frac{38}{77}\), γ = 3 + \frac{152}{77}\)
α + β + γ = \(\frac{77(\text{-266}) + 77(\text{154}) + 114}{77}\), giving 12\).
Thus, α + β + γ = 12, verifying it fits the range (12,12).
Approach Solution -2
Equation of plane containing the line
4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3 can be written as
4ax-y+5z-7a+λ(2x-5y-z-3) = 0
(4a+2λ)x-(1+5λ)y+(5-λ)z-(7a+3λ) = 0
Which i coplanar with the line
x-4/1 = y+1/-2 = z/1
4(4a+2λ)+(1+5λ)-(7a+3λ) = 0
9a+10λ+1 = 0....(1)
(4a+2λ)1+(1+5λ)2+5-λ = 0
4a+11λ+7 = 0....(2)
a = 1, λ = -1
Equation of plane is x + 2y + 3z – 2 = 0
Intersection with the line
\(\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4}\)
(7t + 3) + 2 (–t + 2) + 3 (– 4t + 3) – 2 = 0
–7t + 14 = 0
t = 2
So, the required point is (17, 0, –5)
α+β+γ = 12
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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.
