http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 02 Jul 2014 14:01:57 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/On the Exam 1 Review sheet, question #13b asks to find an equation of the plane that contains the two lines p(t)=(3+2t,1=t,4+t) and q(t)=(-2+3t,2t,2+t). I found the intersection of these lines to be at (1,2,3). Would the next step be to find the cross product of the direction vectors and use the intersection point as my (a,b,c) in the general equation for a plane? *Comment*: I think you're on the right track!AnonymousWed, 02 Jul 2014 10:32:36 -0500http://calc3.askbot.com/question/47/http://calc3.askbot.com/question/51/number-10-on-the-review-sheet/Hey, guys. Here's where I am with this one. First I solved for a normal vector $$n=\langle p_{1}-p_{2} \rangle \times \langle p_{3}-p_{2}\rangle $$$$= (\langle 1,-2,3 \rangle - \langle -1,2,2 \rangle) \times (\langle 1,1,2 \rangle - \langle -1,2,2 \rangle)$$$$=\langle 2,-4,1 \rangle \times \langle 2,-1,0 \rangle$$$$=\langle 1,2,6 \rangle$$ Then I define a p nought $$p_{0}=p_{1}=\langle 1,-2,3 \rangle$$ So if I let the coefficients of the plane equation be the components of the normal vector I end up here. $$(x-1)+2(y+2)+6(z-3)=0$$$$x+2y+6z=21$$ So I guess the question is did everyone else get the same result? If not how did you do the problem? Is this the best method or is there a simpler way?SpaceManSpiffWed, 02 Jul 2014 14:01:57 -0500http://calc3.askbot.com/question/51/