Question #90c2e

Question

1398 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-29T17:33:11

$3$

Given

$L_1->p=p_1+lambda_1 vec v_1$ and
$L_2->p=p_2+lambda_2 vec v_2$

with

$p_1 = (1,-3,0)$
$vec v_1 = (4,1,-1)$
$p_2 = (1,1,1)$
$vec v_2 = (2,0,-1)$

The sought plane $Pi$ contains $L_1$ and does not contain $L_2$ , being parallel to it, so their distance is constant.

Then

$Pi-> p = p_1 + lambda_1 vec v_1 + lambda_2 vec v_2$

And $p_0$ the $p_2$ projection onto $Pi$ is the solution for

${( p_0 = p_1 + lambda_1 vec v_1 + lambda_2 vec v_2),( << p_2-p_0, vec v_1 >> = 0),( << p_2-p_0, vec v_2 >> = 0):}$

Giving $lambda_1 = 8/3, lambda_2 = -5$

so $p_0=(5/3, -1/3, 7/3)$ and the distance $d = norm(p_2-p_0) = 2$