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Three Dimensional Geometry

Question

Let the line $fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y minus 1 over denominator negative 5 end fraction space equals space fraction numerator straight z plus 2 over denominator 2 end fraction$ lie in the plane x + 3y – αz + β = 0. Then (α, β) equals

(6, – 17)

(–6, 7)

(5, –15)

(–5, 5)

Solution

(–6, 7)

2 + 3 × 1 – α (–2) + β = 0
2α + β = –5 ... (i)
3 – 15 – 2α = 0
2α = –12
B = –5 + 12 = 7
(α, β) ≡ (–6, 7)