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Find the equations of the two lines through the origin which intersect the line x-32=y-31=z1 at angles of π3 each.

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प्रश्न

Find the equations of the two lines through the origin which intersect the line `(x - 3)/2 = (y - 3)/1 = z/1` at angles of `pi/3` each.

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उत्तर


Any point on the given line is `(x - 3)/2 = (y - 3)/1 = z/1 = lambda`

⇒  x = 2λ + 3, y = λ + 3

And z = λ

Let it be the coordinates of P

∴ Direction ratios of OP are (2λ + 3 – 0), (λ + 3 – 0) and (λ – 0)

⇒ 2λ + 3, λ + 3, λ

But the direction ratios of the line PQ are 2, 1, 1

∴ `cos theta = ("a"_1"a"_2 + "b"_1"b"_2 + "c"_1"c"_2)/(sqrt("a"_1^2 + "b"_1^2 + "c"_1^2)*sqrt("a"_2^2 + "b"_2^2 + "c"_2^2)`

`cos  pi/3 = (2(2lambda + 3) + 1(lambda + 3) + 1.lambda)/(sqrt((2)^2 + (1)^2 + (1)^2) * sqrt((2lambda + 3)^2 + (lambda + 3)^2 + lambda^2)`

⇒ `1/2 = (4lambda + 6 + lambda + 3 + lambda)/(sqrt(6) * sqrt(4lambda^2 + 9 + 12lambda + lambda^2 + 9 + 6lambda + lambda^2)`

⇒ `sqrt(6)/2 = (6lambda + 9)/sqrt(6lambda^2 + 18lambda + 18)`

= `(6lambda + 9)/(sqrt(6)sqrt(lambda^2 + 3lambda + 3)`

⇒ `6/2 = (3(2lambda + 3))/sqrt(lambda^2 + 3lambda + 3)`

⇒ 3 = `(3(2lambda + 3))/sqrt(lambda^2 + 3lambda + 3)`

⇒ 1 = `(2lambda + 3)/sqrt(lambda^2 + 3lambda + 3)`

⇒ `sqrt(lambda^2 + 3lambda + 3) = 2lambda + 3`

⇒ λ2 + 3λ+ 3 = 4λ2 + 9 + 12λ  ......(Squaring both sides)

⇒ 3λ2 + 9λ + 6 = 0

⇒ λ2 + 3λ + 2 = 0

⇒ (λ + 1)(λ + 2) = 0

∴ λ = – 1, λ = – 2

∴ Direction ratios are [2(– 1) + 3, – 1 + 3, – 1]

i.e., 1, 2, – 1

When λ = – 1 and [2(– 2) + 3, – 2 + 3, – 2]

i.e., – 1, 1, – 2

When λ = – 2.

Hence, the required equations are

`x/1 = y/2 = z/(-1)` and `x/(-1) = y/1 = z/(-2)`.

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अध्याय 11: Three Dimensional Geometry - Exercise [पृष्ठ २३६]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise | Q 11 | पृष्ठ २३६

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