Find vector defined by two points on two separate lines

[daster]

The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point PQ.


Can anyone help?

 

[The Bob]

The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point PQ.


Can anyone help?

I get the feeling that you need to subtract the vectors to get one of the answers but I am really not sure at all. I never was any good at vectors.

Sorry

The Bob (2004 ©)

 

[daster]

I hate them too.

 

[daster]

Anyone...?

 

[HallsofIvy]

The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k.

Okay, any point on that line, in particular P, must have coordinates x= 1+ 2t, y= 1+ t, z= 2- t for some number t.

The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k.

And any point on this line, in particular Q, must have coordinates x= 1+ s, y= 1+ s, z= 4- 2s for some number s.

The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point PQ.

The vector from P to Q is given by ((1+2t)-(1+s))i+ ((1+t)-(1+s))j+ ((2-t)-(4-2s))k= (2t-s)i+ (t-s)j+ (-2+2s-t)k.

Since that is to be perpendicular to the first line, we must have the dot product of vectors, (2t-s)(2)+ (t-s)(1)+ (-2+2s-t)(-1)= 4t-2s+t-s+2-2s+t= 6t-5s+ 2= 0.
Since that is to be perpendicular to the second line, we must have the dot product of vectors, (2t-s)(1)+ (t-s)(1)+ (-2+2s-t)(-2)= 2t-s+t-s+4-4s+2t= 5t- 6s+ 4= 0.

Solve those two equations for s and t. Then you can find the coordinates of the points P and Q. Once you know those you can find the equation of the line from P to Q and the midpoint of the line segment.

 

[daster]

Thank you!