Hey, the question is as follows:
A line goes through the point (1,1,0) and has got the direction vector v=(1,2,2). What's the shortest distance from the point (1,3,1) to the line?
The suggested solution is as follows: [MATH]P=(1,1,0), Q=(1,3,1), u=Q-P=(0,2,1).[/MATH]Shortest distance is given by [MATH]\mid u-(\frac{u.v}{v.v})v \mid =1[/MATH]. And as this did not make any sense to me I looked around and I found an alternate method that gives the same answer from Wikipedia.
I'm guessing that we do a-p to first create a line that intersects point p and then we find any given point on the line a-p and the line, we are already given the direction vector and we have the point a-p intersects, my understanding is that we should now find a normal to both lines which [MATH](a-p)\times n[/MATH] gives us but shouldn't we use orthogonal projection to find the shortest distance or is that exactly what we have done but we cancelled an n because they're scalars due to the absolute values?
A line goes through the point (1,1,0) and has got the direction vector v=(1,2,2). What's the shortest distance from the point (1,3,1) to the line?
The suggested solution is as follows: [MATH]P=(1,1,0), Q=(1,3,1), u=Q-P=(0,2,1).[/MATH]Shortest distance is given by [MATH]\mid u-(\frac{u.v}{v.v})v \mid =1[/MATH]. And as this did not make any sense to me I looked around and I found an alternate method that gives the same answer from Wikipedia.
I'm guessing that we do a-p to first create a line that intersects point p and then we find any given point on the line a-p and the line, we are already given the direction vector and we have the point a-p intersects, my understanding is that we should now find a normal to both lines which [MATH](a-p)\times n[/MATH] gives us but shouldn't we use orthogonal projection to find the shortest distance or is that exactly what we have done but we cancelled an n because they're scalars due to the absolute values?