सदिश विधि का प्रयोग करते हुए सरल रेखाओ `(x-1)/(-1)=(y+1)/(2)=(z-4)/(-1)` तथा `x=(y-1)/(1)=z-2` के बिच लघुतम दुरी ज्ञात कीजिए !

The line (x+1)/(1)=(y-1)/(2)=(z-2)/(-1),(x-1)/(2)=y/1=(z+1)/(4) are

The lines (x+1)/(1)=(y-1)/(2)=(z-2)/(1),(x-1)/(2)=(y)/(1)=(z+1)/(4) are

The lines (x+1)/(1)=(y-1)/(2)=(z-2)/(1),(x-1)/(2)=(y)/(1)=(z+1)/(4) are

The lines (x-2)/(1)=(y-3)/(2)=(z-4)/(3)and(x-1)/(-5)=(y-2)/(1)=(z-1)/(1) are

2^(x)=4^(y)=8^(z) and (1)/(2x)+(1)/(4y)+(1)/(4z)=4

(x^(-1)-y^(-1))/(z)+(y^(-1)-z^(-1))/(x)+(z^(-1)-x^(-1))/(y)= _______

Show that the lines (x-1)/2=(y-3)/-1=(z)/-1 and (x-4)/3=(y-1)/-2=(z+1)/-1 are coplanar.

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

If xy+yz+zx=1, " then " (x)/( 1+x^(2))+(y)/(1+y^2)+(z)/(1+z^(2))=