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Consider the lines

L 1 L_{1} : x + 1 3 \frac{x+1}{3} = y + 2 1 \frac{y+2}{1} = z + 1 2 \frac{z+1}{2}

&

L 2 L_{2} : x − 2 1 \frac{x-2}{1} = y + 2 2 \frac{y+2}{2} = z − 3 3 \frac{z-3}{3}

(Following Equations Represents Equation of Line In 3 Dimensional Cartesian Coordinate System)

If the unit vector perpendicular to both L 1 L_{1} and L 2 L_{2} can be written as

a i ^ − b j ^ + c k ^ d e \frac{a\hat{i}-b\hat{j}+c\hat{k}}{d\sqrt{e}} ; where i ^ , j ^ , k ^ \hat{i}, \hat{j}, \hat{k} are unit vectors along x-axis, y-axis, z-axis respectively and a,b,c,d,e ∈ \in R

*Now if f ( x ) f(x) a function which is differentiable and *

∫ 0 t 2 x f ( x ) d x \int_0^{t^{2}}xf(x)dx = 2 5 t 5 \frac{2}{5}t^{5}

then f ( e − a c d f(\frac{e-a}{cd} ) = α β \frac{\alpha}{\beta}

where β > α \beta > \alpha

Then value of 1 + ω α + β + ω 2 b 1 + \omega^{\alpha + \beta} + \omega^{2b} =

(Where ω \omega is imaginary cube root of unity)

0 1 - i i i i

Harshil Patel by Harshil Patel , 24, India

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1 solution

Harshil Patel
Feb 11, 2014

From L 1 L_{1} & L 2 L_{2} we get direction ratio's of L 1 L_{1} = (3,1,2) and L 2 L_{2} = (1,2,3)

So, Perpendicular vector to L 1 L_{1} and L 2 L_{2} is ( 3 , 1 , 2 ) × ( 1 , 2 , 3 ) (3,1,2) \times (1,2,3) [Cross Product of 2 Vectors]

We get vector perpendicular to both lines as A ˉ \bar{A} = − 1 i ^ − 7 j ^ + 5 k ^ -1\hat{i}-7\hat{j}+5\hat{k}

We know unit vector of any vector is = A ˉ ∣ A ˉ ∣ \frac{\bar{A}}{|\bar{A}|} [Where ∣ A ˉ ∣ |\bar{A}| is magnitude of A ˉ \bar{A} ]

So, unit vector A ^ \hat{A} = − 1 i ^ − 7 j ^ + 5 k ^ 5 3 \frac{-1\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}

Implies, a=-1, b=7, c=5, d=5, e=3

Now, we have ∫ 0 t 2 x f ( x ) d x = 2 5 t 5 \int_0^{t^{2}} xf(x)dx = \frac{2}{5}t^{5} ,

Differentiating both side by using Newton-Leibnitz's Formula we get t 2 ( f ( t 2 ) [ d d x t 2 ] − 0 = 2 t 4 t^{2}(f(t^{2})[\frac{d}{dx}t^{2}] - 0 = 2t^{4}

Solving Further we are left with f ( t 2 ) = t f(t^{2})=t

e − a c d = 3 − ( − 1 ) 25 = 4 25 = ( 2 5 ) 2 \frac{e-a}{cd} = \frac{3-(-1)}{25} = \frac{4}{25} = (\frac{2}{5})^2

⟹ \implies f ( [ 2 5 ] 2 ) = 2 5 = α β f([\frac{2}{5}]^2) = \frac{2}{5} = \frac{\alpha}{\beta}

⟹ \implies α = 2 , β = 5 \alpha = 2 , \beta = 5

α + β \alpha + \beta = 7

and 2 b 2b = 14

*So, according to given equation 1 + ω α + β + ω 2 b 1+\omega^{\alpha+\beta}+\omega^{2b} = *

* 1 + ω 7 + ω 14 1+\omega^{7}+\omega^{14} and using property of ω \omega *

ω 3 n + 1 = ω , ω 3 n + 2 = ω 2 \omega^{3n+1} = \omega , \omega^{3n+2} = \omega^{2}

* = 1 + ω + ω 2 1+\omega + \omega^{2} *

now , ω = − 1 + 3 i 2 , ω 2 = − 1 − 3 i 2 \omega = \frac{-1+\sqrt{3}i}{2} , \omega^{2} = \frac{-1-\sqrt{3}i}{2}

⟹ \implies ω + ω 2 = − 1 \omega + \omega^{2} = -1

so , 1 + ω + ω 2 1+\omega + \omega^{2} = 1-1 = 0 \boxed{0}

Reference :- Line In 3-D Root of Unity Leibniz Integral Rule

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Your problem seems to have two answers. As f ( t 2 ) = t f(t^2)=t , so from what we get after applying Newton-Leibnitz differentiation, i.e. f ( 4 25 ) = ± 2 5 \displaystyle\,f(\frac{4}{25})=\pm\frac{2}{5} which in turn yields two possible answers as 2 + ω 2 2+\omega^2 & 0 0 .

Nishant Sharma - 7 years, 3 months ago

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