Find the coordinates of the point where the line through the points (3,−4,−5) and (2,−3,1) crosses the plane 2x + y + z = 7.

Engineering

Mathematics

Equation of Straight Line in 3D

Distance from a Plane

Question

Find the coordinates of the point where the line through the points (3,−4,−5) and (2,−3,1) crosses the plane 2x + y + z = 7.

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The equation of line passing through the points (3,–4,–5) and (2,–3,1) is

\frac{x - 3}{2 - 3} = \frac{y + 4}{- 3 + 4} = \frac{z + 5}{1 + 5} \Rightarrow \frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6}

......(i)

Let the line (i) crosses the plane 2x + y + z = 7 at point P(α,β,γ) ...(ii)

∵ P lies on line (i), therefore (α,β,γ) must satisfy equation (i)

∴ \frac{α - 3}{- 1} = \frac{β + 4}{1} = \frac{γ + 5}{6} = λ (say)

α = –λ + 3; β = λ – 4 and γ = 6λ – 5

Also P(α,β,γ) lie on plane (ii)
∴ 2α + β + γ = 7
⇒ 2(−λ + 3) + (λ − 4) + (6λ − 5) = 7
⇒ −2λ + 6 + λ − 4 + 6λ − 5 = 7

⇒ 5λ = 10, ⇒ λ = 2

Hence, the coordinate of required point P is (−2 + 3,2 − 4, 6 × 2 − 5)

i.e.(1,−2,7)