Calculus/3d vectors

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Expert: - 2/9/2009


Question
the line l_1 passes through the point A,whose position vector is i-j-5k, & is parallel to the vector i-j-4k. The line l_2 passes through the point B,whose position vector is 2i-9j-14k, & is parallel to the vector 2i+5j+6k.The point P on l_1 & the point Q on l_2 are such that PQ is perpendicular to both l_1 & l_2.a) find the length of PQ.b) find a vector perpendicular to the plane ¥ which contains PQ & l_2. I dont understand this topic. Can you explain? I'm stuck. Much appreciated

Answer
Line L1 is (1,-1,5) + (1,-1,-4)t.
Line L2 is (2,-9,-14) + (2,5,6)t.
The two points P and Q are P = (1+t,-1-t,5-4t) and
Q = (2-2t,-9+5t,-14+6t).

The direction vectors for L1 and PQ have a cross product that is 0
since they are perpendicular.

The direction vectors for L2 and QP have a cross product that is 0
since they are perpendicular.

Once P and Q have been found, construct E=P-Q.
The length of E is √(Ex² + Ey² + Ez²).

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