# Thread: Algebra - Vectors

1. ## Algebra - Vectors

a) By looking at their intersection and their direction vectors, show that the following lines are skew:
(x,y,z) = (2, -1, 6) + t(1, 2, 0)
and
(x, y, z) = (1, 4, 4) + u(2, -3, 1).

b) Find the closest point to the origin on the line (x,y,z) = (2, -1, 6) + t(1, 2, 0) and the shortest distance from the origin to that line.

2. For b), first rewrite the line as $\displaystyle \displaystyle (x, y, z) = (2 + t, -1 + 2t, 6)$.

The distance from the origin to the line is $\displaystyle \displaystyle D = \sqrt{(2 + t - 0)^2 + (-1 + 2t - 0)^2 + (6-0)^2}$

$\displaystyle \displaystyle = \sqrt{(2 + t)^2 + (-1 + 2t)^2 + 6^2}$

$\displaystyle \displaystyle = \sqrt{4 + 4t + t^2 + 1 - 4t + 4t^2 + 36}$

$\displaystyle \displaystyle = \sqrt{5t^2 + 41}$.

The minimum distance will the value of $\displaystyle \displaystyle t$ for which the derivative is $\displaystyle \displaystyle 0$ and where the second derivative is positive.

3. What ideas have you had for a)?