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

Answer:

$\displaystyle (x,y,z)=(2,1,-1)+h(2,-3,3)$

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- Jun 28th 2008, 10:23 AMApprentice123straight passes by point
Finding the equation of the straight (l1), which passes in point $\displaystyle P(2,1,-1)$ and perpendicular to the straight (l2): $\displaystyle (x,y,z)=(2,0,0)+t(3,1,-1)$

Answer:

$\displaystyle (x,y,z)=(2,1,-1)+h(2,-3,3)$ - Jun 28th 2008, 11:18 AMearboth
There are 2 completely different ways to do this problem:

1. By calculus: Let Q denote a point on $\displaystyle l_2 : Q \in l_2$

Then calculate the distance $\displaystyle |\overline{PQ}|$ which will give you an equation of a function with respect to t. Calculate the minimum distance using the first derivative.

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

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 $\displaystyle ((d(t))^2)' = 0$ 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 $\displaystyle l_2$ and containing P.

Then calculate the intersection of $\displaystyle l_2$ and a which is the point Q.

Calculate the equation of the line PQ.