सरल रेखाओ `(x-3)/(3)=(y-8)/(-1)=(z-3)/(1)` और `(x+3)/(-3)=(y+7)/(2)=(z-6)/(4)` के बीच की लघुतम दूरी ज्ञात कीजिए ।

The lines (x)/(1)=(y)/(2)=(z)/(3)and(x-1)/(-2)=(y-2)/(-4)=(z-3)/(-6) are

The shortest distance between line (x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and (x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is

If the shortest distance between the lines (x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and (x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is lambdasqrt(30) unit, then the value of lambda is

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) .

Express the following equation of the lines into vector form : (x - 3)/(3) = (y - 8)/(-1) = (z - 3)/(1) and (x + 3)/(-3) = (y + 7)/(2) = (z - 6)/(4)

(2-3x)/(x)+(2-3y)/(y)+(2-3z)/(z)=0 then (1)/(x)+(1)/(y)+(1)/(z)=

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

if the lines (x)/(1)=(y)/(2)=(z)/(3),(x-1)/(3)=(y-2)/(-1)=(z-3)/(4) and (x+k)/(3)=(y-1)/(2)=(z-2)/(h) are concurrent then