[(5-x)/(3)=(y+3)/(-2),z=5],[(x)/(1)=(1-y)/(3)=(z-5)/(2)]

Find the angle between the following lines (5-x)/(3)=(y+3)/(-2),z=5 and x/1=(1-y)/(3)=(z-5)/(2)

(5-x)/(3) =(y+3)/(-2) ,z=5 " and " (x-1)/(1)=(1-y)/(3)=(z-5)/(2) Find the angle between the lines.

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 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)

If Q is the image of the point P (2,3,4) under the reflection in the plane x-2y+5z=6 then the equation of the line PQ is a) (x-2)/(-1)= (y-3)/2= (z-4)/5 b) (x-2)/(1)= (y-3)/(-2)= (z-4)/5 c) (x-2)/(-1)= (y-3)/(-2)= (z-4)/5 d) (x-2)/(1)= (y-3)/(2)= (z-4)/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)

The straight lines (x-3)/(2)=(y+5)/(4)=(z-1)/(-13)and(x+1)/(3)=(y-4)/(5)=(z+2)/(2) are

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

The angle between the lines (x+4)/1=(y-3)/2=(z+2)/3 and x/3=(y-1)/(-2)=z/1 is (x-2)/3=(y+1)/(-2),z=2 and (x-1)/1=(2y+3).3=(z+5)/2 is

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