The image of the point (3,-1,11) w.r.t the line `x/2=(Y-2)/3=(z - 3)/4` is
(2,5,7)
(1,11,3)
(0,0,0)
(0,2,3)

Find the image of the point (0,2,3) in the line (x+3)/(5)= (y-1)/(2) = (z+4)/(3) .

The image of the point A (1,0,0) in the line (x - 1)/(2) = (x + 1)/(-3) = (z + 10)/(8) is :

Find image (reflection )of the point (7,4,-3) in the line (x)/(1)=(y-1)/(2)=(z-2)/(3)

Locus of image of the point (2,3) in the line (2x-3y+4)+k(x-2y+3)=0 ,k in R is

The value of a for which the image of the point (a, a-1) w.r.t the line mirror 3x+y=6a is the point (a^2 + 1, a) is (A) 0 (B) 1 (C) 2 (D) none of these

(a) Find the foot of the perpendicular from the point (i) (2, -1,5) on the line : (x -11)/(10) = (y + 2)/(-5) = (z + 8)/(11) (ii) (0,2,3) on the line (x +3)/(5) = (y-1)/(2) = (z + 4)/(3) . (b) Also, find the length of perpendicular in part (ii).

The image of the point (1, 3, 4) in the plane 2x-y+z+3=0 is (3,\ 5,\ 2) b. (-3,5,\ 2) c. (3,\ 5,\ -2) d. (3,\ -5,\ 2)

The image line of 2x-y-1=0 w.r.t 3x-2y+4=0 is

The distance,of the point (7,-2,11) from the line (x-6)/(1)=(y-4)/(0)=(z-8)/(3) along the line (x-5)/2=(y-1)/(-3)=(z-5)/(6) ,is :