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Question
Find the equation of the plane containing the line `x/(-2) = (y - 1)/3 = (1 - z)/1` and the point (–1, 0, 2).
Solution 1
Let the equation of the plane be a(x – x1) + b(y – y1) + c(z – z1) = 0 is passing through (–1, 0, 2)
`\implies` a(x + 1) + b(y – 0) + c(z – 2) = 0 ...(1)
Given line `(x - 0)/(-2) = (y - 1)/3 = (z - 1)/(-1)` passing through (0, 1, 1) and having d.r. (−2, 3, −1)
Since the plane contains the line and the point
`\implies` a(1) + b(1) + c(1 – 2) = 0
`\implies` a + b – c = 0 ...(2)
Also the line and normal to the plane are perpendicular
`\implies` –2𝑎 + 3b – c = 0 ...(3)
Solving (2) and (3)
`a/2 = b/3 = c/5 = k`
Hence required equation of the plane is
`\implies` 2(x + 1) + 3(y − 0) + 5(z − 2) = 0
`\implies` 2x + 3y + 5z − 8 = 0
Solution 2
Let the equation of the plane be a(x – x1) + b(y – y1) + c(z – z1) = 0 is passing through (–1, 0, 2)
`\implies` a(x + 1) + b(y – 0) + c(z – 2) = 0 ...(1)
Given line `(x - 0)/(-2) = (y - 1)/3 = (z - 1)/(-1)` passing through (0, 1, 1) and having d.r. (−2, 3, −1)
Since the plane contains the line and the point
`\implies` a(1) + b(1) + c(1 – 2) = 0
`\implies` a + b – c = 0 ...(2)
Also the line and normal to the plane are perpendicular
`\implies` –2𝑎 + 3b – c = 0 ...(3)
Hence required equation of the plane is
`|(x + 1, y, z - 2),(1, 1, -1),(-2, 3, -1)| = 0`
`\implies` 2(x + 1) + 3(y − 0) + 5(z − 2) = 0
`\implies` 2x + 3y + 5z − 8 = 0
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