Question #90c2e

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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#