(x+1)/(3)=(y-2)/(-2)=(z+1)/(-1) লাইনে P(1,0,2) থেকে লম্বের দৈর্ঘ্য হল | ...

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) .

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

The lines (x-2)/(2)=(y)/(-2)=(z-7)/(16) and (x+3)/(4)=(y+2)/(3)=(z+2)/(1) intersect at the point "P" .If the distance of "P" from the line (x+1)/(2)=(y-1)/(3)=(z-1)/(1) is "l" ,then 14l^(2) is equal to

The line (x-2)/(3)=(y+1)/(2)=(z-1)/(-1) intersects the curve x^2+y^2=r^2, z=0 , then

The equation of the plane passing through the poit of intersection of the lines (x-1)/(3)=(y-2)/(1)=(z-3)/(2),(x-3)/(1)=(y-1)/(2)=(z-2)/(3) and perpendicular to the line (x-2)/(2)=(y-3)/(3)=(z-2)/(1) is P = 0. If the distance of the point (1, 1, 3) from P = 0 is k units, then the value of (k^(2))/(2) is equal to

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)

The equation of the plane through the point (-1,2,0) and parallel to the line (x)/(3)=(y+1)/(0)=(z-2)/(-1) and (x)/(3)=(2y+1)/(2)=(2z+1)/(-1) is

The projection of the line (x-1)/(2)=(y+1)/(1)=(z-2)/(3) on a plane P is (x-1)/(1)=(y+1)/(2)=(z-2)/(1) , then the equation of the plane P is

The equation of the plane through the point (-1,2,0) and parallel to the lines (x)/(3)=(y+1)/(0)=(z-2)/(-1) and (x-1)/(1)=(y+1)/(2)=(2z+5)/(-1)