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.

Answer 1

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")`