सदिश विधि से रेखाओ `(x-1)/(2)=(y+3)/(-1)=(z+1)/(1)` तथा `(x+3)/(1)=(y-2)/(-1)=(z+1)/(2)` के मध्य का कोण ज्ञात कीजिए!

(x+1)/(3)=(y+2)/(1)=(z+1)/(2) and (x-2)/(1)=(y+2)/(2)=(z-3)/(3)

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

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

(2-3x)/(x)+(2-3y)/(y)+(2-3z)/(z)=0 then (1)/(x)+(1)/(y)+(1)/(z)=

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) .

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

The lines x/2=y/1=z/3 and (x-2)/(2)=(y+1)/(1)=(3-z)/(-3) are ….

Solve : (1)/(2x)+(1)/(4y)-(1)/(3z)=(1)/(4),(1)/(x)=(1)/(3y),(1)/(x)-(1)/(5y)+(4)/(z)=2 (2)/(15)