" fatalize for tear "(x-5)/(7)=(y+2)/(-5)=(z)/(1)" ath "(x)/(1)=(y)/(2)=(z)/(3)" upent "(z)/(3)

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 .

Show that the lines (x-5)/(7) =(y+2)/(-5)=(z)/(1) " and " (x)/(1) =(y)/(2)=(z)/(3) are at right angles .

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

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

The angle between the lines (x-7)/(1)=(y+3)/(-5)=(z)/(3) and (2-x)/(-7)=(y)/(2)=(z+5)/(1) is equal to

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 the angle between the pair of lines (x-5)/7=(y+2)/(-5)=z/1 and x/1=y/2=z/3

The image of the line (x-1)/(3)=(y-3)/(1)=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line (1)(x+3)/(3)=(y-5)/(1)=(z-2)/(-5) (2) (x+3)/(-3)=(y-5)/(-1)=(z+2)/(5) (3) (x-3)/(3)=(y+5)/(1)=(z-2)/(-5) (3) (x-3)/(-3)=(y+5)/(-1)=(z-2)/(5)

Perpendiculars are drawn from points on the line (x+2)/(2)=(y+1)/(-1)=(z)/(3) to the plane x+y+z=3 The feet of perpendiculars lie on the line (a) (x)/(5)=(y-1)/(8)=(z-2)/(-13) (b) (x)/(2)=(y-1)/(3)=(z-2)/(-5)(d)(x)/(4)=(y-1)/(-7)=(z-2)/(5)(d)(x)/(2)=(y-1)/(-7)=(z-2)/(5)