Find the value(s) of k so that the lines x - 3/3k + 1 = y + 6/2 =
z + 3 /2k and x + 7/3 = y + 8 /-2k = z + 9/-3 are both perpendicular to the vector u=(10,33,-34)
I will assume you mean:
(x - 3)/(3k + 1) = (y + 6)/2 = (z + 3)/(2k) and (x + 7)/3 = (y + 8)/(-2k) = (z + 9)/-3
---- those brackets are absolutely essential for the question to make sense
so the directions of your given lines are ((3k+1), 2, 2k) and (3,-2k,-3)
and the dot product of each with (10,33,-34) must be zero
((3k+1), 2, 2k) dot (10,33,-34) = 0
30k+30 + 66 - 68k = 0
k = 96/38 = 48/19
AND
(3,-2k,-3) dot (10,33,-34) = 0
30 -66k + 102 = 0
k = 132/66 = 2
Unless I made some kind of arithmetic error, it looks like we have no solution. We had to get the same for k