Finding the point of intersection of straight through the points A (3,-5,2) and B (11,-3,6) with the straight (r): (x,y,z) = (5,-3,2) + t (4,-2,4).

Answer:

R (7,-4,4)

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- Jun 22nd 2008, 07:15 PMApprentice123Point of intersection
Finding the point of intersection of straight through the points A (3,-5,2) and B (11,-3,6) with the straight (r): (x,y,z) = (5,-3,2) + t (4,-2,4).

Answer:

R (7,-4,4) - Jun 22nd 2008, 07:56 PMo_O
$\displaystyle \vec{s} = \vec{s}_{0} + \lambda \vec{v}$

First find the direction vector of your line, i.e. $\displaystyle \vec{v} = \vec{AB} = (8, 2, 4)$.

Now, $\displaystyle s_{0}$ can simply the vector from the origin to the point A (or B, doesn't matter as long as it's on the line). So, our line is: $\displaystyle \vec{s} = (3, -5, 2) + \lambda(8,2,4)$

Now to see where lines**s**and**r**intersect, we set the them equal to each other, i.e. $\displaystyle (3, -5, 2) + \lambda(8,2,4) = (5,-3,2) + t (4,-2,4)$. - Jun 23rd 2008, 03:06 PMApprentice123
ok, tanks. I managed to solve