The lines l1 and l2 have equations r= 6i -3j + s(3i - 4j - 2k ) and r= 2i -j - 4k + t(i-3j-k )
The position vector of a point P on line L1 is 3i+j+2k. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Find the position vector of Q.
This is the part I'm having trouble with :
Find, in the form r = a + kb + uc, an equation of the plane which passes through P and is perpendicular to L1.
Okay, so I need two linearly independent vectors lying in the plane. One is PQ, as it is perpendicular to l1, which means that it lies in the plane. Another can be the cross product of the direction vector of l1, and any other vector. The resultant vector will be perpendicular to l1, and will also lie in the plane. But this will give me a different answer depending on the second vector I use.. Why is this not correct?