# Homework Help: A question about vectors

1. Apr 10, 2007

### oahsen

1. The problem statement, all variables and given/known data

1) Find the equation of the line L which passes through the point P0(1, 0, 2) and
intersects the line L1 :(x − 1)/2=(y + 1)/3, z = 1 orthogonally.

3. The attempt at a solution

ı thought the line L1 as a parametrisation like x=2t+1;y=3t-1;z=1; then I took a vector parallel to L as ai+bj+ck. Then since the parallel vector of L1 (which is 2i+3j) orthogonal to ai+bj+ck then the dot product of them will 0. This lead 2a+3b=0; Therefore, I can think L as ; x=1+3bt/2 and y=bt. However, my question is how about c? how can ı relate it to a and b? Or is my approach to question is wrong?

2. Apr 10, 2007

### HallsofIvy

That's perfectly valid but there is no way to relate c to a and b. First, notice that t is not "unique". Multiplying t by any number just changes the length of the parallel vector giving different parametric equations for the same line.

The line x= 1+ 3bt/2, y= bt, z= ct+ 1 is orthogonal to L1 for all b and c but does not necessarily intersect it. In order to intersect it, for some t, you must have (x-1)/2= 3bt/4= (y+1)/2= (bt+1)/3, z= ct+1= 1. From the last, either c= 0 or t= 0. If you put t= 0 into 3bt/4= (bt+1)/3 you get 1/3= 0 which is not true. Therefore, c= 0 and z= 1 for all t. Now, since t is arbitrary, assume your parmeterization is such that your new line intersects L1 when t= 1. Solve 3b/4= (b+1)/3 for b.

3. Apr 10, 2007

### oahsen

it should z=ct+2 isn't it? Because the vector passes the point (1,0,2). And if it is so then ct+2=1 and ct=-1; then we could not fnd exact values to c or t . then how could we continue? ;