Question

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 _____.

Answer 1

Correct Answer: 12

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).

Answer 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