InterviewSolution
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InterviewSolution
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The distance of the point `P(3, 8, 2)` from the line `(x-1)/2 = (y-3)/4=(z-2)/3` measured parallel to the plane `3x+2y-2z+17=0` is |
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Answer» plane given is `3x+2y - 2z + 17=0` `3x+2y-2z+c=0-> (3,8,2)` `9+16-4 + c =0` `c=-21` so eqn of plane will be `3x + 2y-2z-21=0` for coordinates lets the points are equal to `lamda` so, `x=2lamda +1` `y= 4 lamda+3` `z= 3 lamda + 2` putting it in eqn we get `3(2 lamda+1) + 2(4 lamda+3) - 2( 3lamda+2) - 21=0` `6 lamda + 3 + 8 lamda+ 6 - 6 lamda - 4 -21=0` `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get so,`lamda = 2` so points will be like `x= 5` `y=11` `z=8` now we have to find the distance from point `(3,8,2)` of point (5,11,8) `` `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get `d= sqrt(4+9+36) =7` option 1 is correct |
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