straight passes by point

**Apprentice123** · June 28th 2008, 10:23 AM

Finding the equation of the straight (l1), which passes in point $P(2,1,-1)$ and perpendicular to the straight (l2): $(x,y,z)=(2,0,0)+t(3,1,-1)$

Answer:
$(x,y,z)=(2,1,-1)+h(2,-3,3)$

**earboth** · June 28th 2008, 11:18 AM

Originally Posted by Apprentice123

Finding the equation of the straight (l1), which passes in point $P(2,1,-1)$ and perpendicular to the straight (l2): $(x,y,z)=(2,0,0)+t(3,1,-1)$

Answer:
$(x,y,z)=(2,1,-1)+h(2,-3,3)$

There are 2 completely different ways to do this problem:

1. By calculus: Let Q denote a point on $l_2 : Q \in l_2$
Then calculate the distance $|\overline{PQ}|$ which will give you an equation of a function with respect to t. Calculate the minimum distance using the first derivative.

$(d(t))^2=2-4t+11t^2$

If d(t) has an extreme value then (d(t))^2 has an extreme value too. Therefore it is sufficient if you examine (d(t))^2.
Solve $((d(t))^2)' = 0$ for t.
Calculate the coordinates of Q.
Calculate the equation of the line PQ.

2. By analytic geometry: Calculate the equation of an auxiliar plane a perpendicular to $l_2$ and containing P.
Then calculate the intersection of $l_2$ and a which is the point Q.
Calculate the equation of the line PQ.

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