Thread: Equations of perpeddicual vectors in three-space

the question goes as such: write the parametric equations of the line that goes through point (6, -2, 1) and is perpendicular to both

f: [x,y,z]=[1, 4, -2] +t[3,-1,1] and
u: [x,y,z]=[9,5,-3]+s[1,-3,7]

I think that if a vector is perpendicular to two vectors with different directions then it must be perpendicular at the point where vectors f and u intersect. I think that in this situation I would use the cross product between the two vector f x u to produce the direction vector of the unknown third vector and go from there.

Originally Posted by darksoulzero

the question goes as such: write the parametric equations of the line that goes through point (6, -2, 1) and is perpendicular to both

f: [x,y,z]=[1, 4, -2] +t[3,-1,1] and
u: [x,y,z]=[9,5,-3]+s[1,-3,7]

I think that if a vector is perpendicular to two vectors with different directions then it must be perpendicular at the point where vectors f and u intersect. I think that in this situation I would use the cross product between the two vectors f x u to produce the direction vector of the unknown third vector and go from there.

sounds like a plan

Originally Posted by darksoulzero

the question goes as such: write the parametric equations of the line that goes through point (6, -2, 1) and is perpendicular to both
f: [x,y,z]=[1, 4, -2] +t[3,-1,1] and
u: [x,y,z]=[9,5,-3]+s[1,-3,7]

You really need to review this question. Check all the the numbers.
There is a unique line that is perpendicular to these skew lines.
However the point (6,-2,1) is not on that line.

Originally Posted by Plato

You really need to review this question. Check all the the numbers.
There is a unique line that is perpendicular to these skew lines.
However the point (6,-2,1) is not on that line.

I thought it was weird also, but the numbers are correct. I think there may have been a problem in the printing in the textbook.