http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 02 Jul 2014 15:06:07 -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!Wed, 02 Jul 2014 10:32:36 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=48#post-id-48Anonymous, I approached this problem exactly as you described. I also found the point of intersection to be (1,2,3). I then rewrote the linear equations as follows:$$\vec{p}(t) = \langle3,1,4\rangle + \langle2,-1,1\rangle(t)$$$$\vec{q}(t) = \langle-2,0,2\rangle + \langle3,2,1\rangle(t)$$ From here I took the cross product of the directional vectors since I know they are both on the plane in order to get the normal vector. $$\langle2,-1,1\rangle \times \langle3,2,1\rangle = \langle-3,1,7\rangle$$ plugging what I have into the equation for a plane I get $$\langle-3,1,7\rangle\cdot\langle x-1, y-2, z-3\rangle = 0$$ $$-3(x-1) + (y-2) + 7(z-3) = 0$$ Or $$-3x + y + 7z = 20$$ As usual, I would love comments, critiques, etc. Thanks! **ADDED for Tiffany:** The first thing to do to find point of intersection is to change the parameter of one equation, lets say change $\vec{q}(t)$ to $\vec{q}(s)$, making it $$\vec{q}(s) = \langle-2+3s, 2s, 2+s\rangle$$ Now you set the components of $\vec{p}$ and $\vec{q}$ equal to each other. $$3 + 2t = -2 + 3s$$ $$1 - t = 2s$$ $$4 + t = 2 + s$$ If you solve for one variable then back substitute, you should end up with t = -1 and s = 1. Using these values to plug back into the vector equations gives you $$\vec{p}(-1) = \langle1,2,3\rangle$$$$\vec{q}(1) = \langle1,2,3\rangle$$ Thus the intersection...Does this sound right? Wed, 02 Jul 2014 12:00:37 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=48#post-id-48http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=50#post-id-50Not an answer, but how did y'all find the point of intersection? **Christina** THANKS! Now that you brought it up, I remember him talking about this in class, I just couldn't remember how to do it!Wed, 02 Jul 2014 12:57:20 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=50#post-id-50http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=49#post-id-49Yeah! That's exactly what I did. First I broke $ \vec{p}(t) $ and $ \vec{q}(t) $ into the other form to get their direction vectors: $ \vec{p}(t) = \langle 3,1,4 \rangle + \langle 2,-1,1 \rangle t$ and $ \vec{q}(t) = \langle -2,0,2 \rangle + \langle 3,2,1 \rangle t $ Then I took the cross product of $ \vec{p}(t) \times \vec{q}(t) $: $ \vec{I}\vec{J} \vec{K} $ 2 -1 1 3 2 1 $ \vec{I} (-1-2)- \vec{J}(2-3)+ \vec{K}(4+3)$ $ \langle -3,1,7 \rangle $ Then plugging this for $ \langle a,b,c \rangle $ and the intersection point for $ \langle x_0,y_0,z_0 \rangle $ into the equation: $ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $ $ -3(x-1)+1(y-2)+7(z-3)=0$ And that simplifies to: $ -3x+y+7z= 20$Wed, 02 Jul 2014 12:09:35 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=49#post-id-49http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=57#post-id-57When I was looking for the points of intersection I set the two vectors equal to each other. $$3+2t=-2-3s$$ $$1-t=2s$$ $$4+t=2+s$$ and then I solved for s in the first two equations and got $$(5+2t)/-3=s$$ $$(1-t)/2=s$$ then I put them into the last equation and solved for t and found t=-1. When I put the t that I found back into the equations I found u=1. I put u and t into an equation and it did not work for all of them so I concluded that they do not intersect. Did anyone else have the same results? Wed, 02 Jul 2014 15:06:07 -0500http://calc3.askbot.com/question/47/how-to-find-the-equation-of-a-plane-containing-two-lines/?answer=57#post-id-57