रेखा `(x-2)/(3) = (y+1)/(-1) = (z-3)/(2)` और समतल `3x + 4y + z + 5 = 0` के मध्य कोण ज्ञात कीजिए।

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) .

The (x+4)/(3) = (y+6)/(5) = (z-1)/(-2) and 3x - 2y + z+5 =0 = 2x + 3y + 4z -k are coplanar for k is equal to

Line (x-2)/(-2) = (y+1)/(-3) = (z-5)/1 intersects the plane 3x+4y+z = 3 in the point

Find the value of (( x -y )^3 + ( y - z )^3 + ( z - x )^3 )/ (9 ( x - y )( y - z ) ( z - x )) 1 . 0 2 . 1/9 3 . 1/3 4. 1

2x + 3y-5z = 7, x + y + z = 6,3x-4y + 2z = 1, then x =

The distance between the line (x-2)/(1)=(y-3)/(1)=(z-4)/(5) and the plane 2x+3y-z+5=0 is

If x_(1) = 3y_(1) + 2y_(2) -y_(3), " " y_(1)=z_(1) - z_(2) + z_(3) x_(2) = -y_(1) + 4y_(2) + 5y_(3), y_(2)= z_(2) + 3z_(3) x_( 3)= y_(1) -y_(2) + 3y_(3)," " y_(3) = 2z_(1) + z_(2) express x_(1), x_(2), x_(3) in terms of z_(1) ,z_(2),z_(3) .