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Question
Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
Solution
Given point is (3, 0, 1) and the equation of planes are
x + 2y = 0 ......(i)
And 3y – z = 0 ......(ii)
Equation of any line l passing through (3, 0, 1) is
`l : (x - 3)/"a" = (y - 0)/"b" = (z - 1)/"c"`
Direction ratios of the normal to plane (i) and plane (ii) are (1, 2, 0) and (0, 3, – 1)
Since the line is parallel to both the planes.
∴ 1.a + 2.b + 0.c = 0
⇒ a + 2b + 0c = 0
And 0.a + 3.b – 1.c = 0
⇒ 0.a + 3b – c = 0
So `"a"/(-2 - 0) = (-"b")/(-1 - 0) = "c"/(3 - 0) = lambda`
∴ a = – 2λ, b = λ, c = 3λ
So, the equation of line is `(x - 3)/(-2lambda) = y/lambda = (z - 1)/(3lambda)`
Hence, the required equation is `(x - 3)/(-2) = y/1 = (z - 1)/3`
or in vector form is `(x - 3)hat"i" + yhat"j" + (z - 1)hat"k" = lambda(-2hat"i" + hat"j" + 3hat"k")`
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