Linear Algebra: Parallel, Perpendicular, or Neither?

AI Thread Summary
The line l, defined by point P=(1,-1,1) and direction vector d=[2,3,-1], is parallel to the normal vector n of the plane given by the equation 2x+3y-z=1. Since the direction vector and normal vector are scalar multiples of each other, they are parallel. Consequently, the line l is perpendicular to the plane. This relationship is confirmed through the use of the cross product, which yields zero, indicating parallelism. Therefore, the line and the plane are indeed perpendicular.
sdoyle
Messages
18
Reaction score
0

Homework Statement



The line l passes through the point P=(1,-1,1) and has direction vector d=[2,3,-1]. Determine whether l and P are parallel, perpendicular, or neither to 2x+3y-z=1.

Homework Equations


n.p=n.x, cross product, dot product


The Attempt at a Solution


Would you just relate the direction vector, d, to the normal vector, n[2,3,-1] ?
 
Physics news on Phys.org
Yes, you would just relate the normal vector to the direction vector. How do they relate?
 
they would be parallel because they are scalar multiples of one another (1 is the multiple) also the cross product would be zero indicating parallel vectors.
 
That means that the NORMAL to the plane is parallel to the line. So what's the relation between the line and the plane?
 
that they are perpendicular?
 
Yes, they are.
 
How would you prove that mathematically?
 
What's to prove? The normal vector of the plane is parallel to the line. You proved that by showing they are scalar multiples or using the cross product. You're done.
 

Similar threads

Back
Top