How do you find an equation for the plane that contains the line with parametric equations #l=(8 - 7t, -5 - 2t , 5 - t)# and is parallel to the line with parametric equations #x=3 + t, y=-7 + 9t, z=8 - 6t#?

Question

5171 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-30T10:56:43

#Pi->21 x - 43 y - 61 z-78=0#

Given two lines #L_1,L_2# determine a plane #Pi# such that

#L_1 in Pi# and #L_2 !in Pi#

Here

#L_1 -> p = p_1 + lambda vec v_1 #
#L_2 -> p = p_2 + lambda vec v_2 #

with

#p_1 = (8,-5,5), vec v_1 = (-7,-2,-1)#
#p_2 = (3,-7,8), vec v_2 = (1,9,-6)#

The plane equation is

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

#L_1 in Pi# ( make #lambda_2=0# ) and is parallel to #L_2# ( making #lambda_2=0#)

The nonparametric plane equation can be easily obtained remembering that for the plane

#<< p-p_1,vec n >> = 0#

here #p = (x,y,z)# and #vec n = vec v_1 xx vec v_2 = (21,-43,-61)# so

#Pi->21 x - 43 y - 61 z-78=0#