For real numbers α and β ≠ 0, if the point of intersection of the straight lines x−α1= y−12= z−13 and x−4β= y−63= z−73,lies on the plane x + 2y − z = 8, then α − β is equal to :

Question

Accepted Answer

Correct answer: D. (d) First line is (ϕ + α, 2ϕ + 1, 3ϕ + 1) and second line is f (qβ + 4, 3q + 6, 3q + 7) For ϕ + α = qβ + 4 …(i) 2ϕ + 1 = 3q + 6 …(i) 3ϕ + 1 = 3q + 7 for (ii) &(iii) ϕ = 1, q = −1 So, from (i) α + β = 3 Now, point of intersection is (α + 1, 3, 4) It lies on the plane. Hence, α = 5&β = −2

$For real numbers α and β \neq 0, if the point of intersection of the straight lines \frac{x - α}{1} = \frac{y - 1}{2} = \frac{z - 1}{3} and \frac{x - 4}{β} = \frac{y - 6}{3} = \frac{z - 7}{3},lies on the plane x + 2y - z = 8, then α - β is equal to :$

Solution

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