`(4y+1)/(3)+(2y-1)/(2)-(3y-7)/(5)=(47)/(10)`

Projection of line (x+1)/(2)+(y+1)/(-1)=(z+3)/(4) on the plane x+2y+z=6; has equation x+2y+z-6=0=9x-2y-5z-8bx+2y+z+6=0,9x-2y+5z=4c(x-1)/(4)=(y-3)/(-7)=(z+1)/(10)d.(x+3)/(4)=(y-2)/(7)=(z-7)/(-10)

The distance of point of intersection of lines (x-4)/(1)=(x+3)/(-4)=(z-1)/(7) and (x-1)/(2)=(y+1)/(-3)=(z+10)/(8) from (1,-4,7) is

The equation of the plane which passes through the point of intersection of lines (x-1)/(3)=(y-2)/(1)=(z-3)/(2), and (x-3)/(1)=(y-1)/(2)=(z-2)/(3) and at greatest distance from point (0,0,0) is a.4x+3y+5z=25 b.4x+3y=5z=50c3x+4y+5z=49d.x+7y-5z=2

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) .

Find (x+y)+(x-y), if x=(5)/(4),y=(-1)/(3) (ii) x=(2)/(7),y=(4)/(3)( iii) x=(1)/(4),y=(3)/(2)

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

((y^(3)-3y^(2)+5y-1)/(y-1))

Show that the lines : (i) (x -5)/(7) = (y + 2)/(-5) = (z)/(1) " " and (x)/(1) = (y)/(2) = (z)/(3) (ii) (x - 3)/(2) = (y + 1)/(-3) = (z - 2)/(4) and (x + 2)/(2) = (y - 4)/(4) = (z + 5)/(2) are perpendicular to each other .