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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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