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Thread: Calculus 3 help! Iintersection point of two lines in 3D and finding coplanarity

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    Calculus 3 help! Iintersection point of two lines in 3D and finding coplanarity

    Here is the question as written on a practice sheet:

    Consider the line l1 : ~r1(t) =< −2, 1, 1 > +t < −2, −3, 1 >and line l2 : ~r2(t) =< 1, 2, −4 > +t < 1, −2, −4 >.

    (a) Show that these two lines intersect at a point. Find the intersection point.
    (b) Show that these two lines are coplanar by finding the plane which contains these two lines


    I understood part a. but id appreciate if someone can verify their intersection point is (0,4,0) ?

    Part b. is where I'm lost. I understand as far as having to do cross product of these two vectors and then plugging in that result into the standard formula for a plane. However, I know that you also need the rectangular points of those lines to include into that plane equation as well. My main question is:

    Where and how the heck do i figure out the rectangular points of these vectors?
    Thanks!
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    Re: Calculus 3 help! Iintersection point of two lines in 3D and finding coplanarity

    "Verifying" a solution is much easier than finding the solution: r1= <-2- 2t, 1- 3t. 1+ t>= < 0, 4, 0> if and only if -2- 2t= 0, 1- 3t= 4, 1+ t= 0. The first gives 2t= -2 so t= -1. The two other equations are 1- 3(-1)= 1+ 3= 4 and 1- 1= 0 both true: (0, 4, 0) is on the first line. For this point to be on the second line, we must have 1+ t= 0, 2- 2t= 4, and -4- 4t= 0. From the first, t= -1. Then the other two equations are 2- 2(-t)= 2+ 2= 4 and -4- 4(-1)= -4+ 4= 0. Those are true so the point is on the second line. The point is on both lines so the point is the point of intersection.

    (Here, the parameter in both equations happened to be t= -1. The parameters in the two sets of equations does not have to be the same.)

    I don't know what you mean by "rectangular points" for a vector nor do I know what vectors you are referring to. You are give two lines, not vectors. Perhaps this is what you mean: given two vectors, there cross product is perpendicular to the plane and the equation of the plane is A(x- a)+ B(y- b)+ C(z- c)= 0 where <A, B, C> is a vector perpendicular to the plane and (a, b, c) is a point in the plane. Here, you know that (0, 4, 0) is in the plane. The vector multiplying the parameter "t" is a vector in the direction of the plane: <-2, -3, 1> for the first line, <1, -2, -4> for the second line.

    Personally, I wouldn't do it that way. Knowing that any plane, containing the point (0, 4, 0) can be written in the form Ax+ B(y- 4)+ Cz= 0, and that we can always divide through by one of those coefficients, say, A to get x+ (B/A)(x- 4)+ (C/A)= x+ B'(y- 4)+ C'z= 0 so we only need two equations to determine B' and C'. We can get two equations by knowing two more points on the plane. Taking t= 0 in the parametric equations, (-2, 1, 1) and (1, 2, -4) are two points, one on each line so two more points in the plane. Setting x= -2, y= 1, z= 1 in the equation of the plane gives -2+ B'+ C'= 0. Setting x= 1, y= 2, z= -4 gives 1+ 2B'- 4C'= 0. Solve B'+ C'= 2 and -2B+ 4C'= 1 for B' an C'.
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    Re: Calculus 3 help! Iintersection point of two lines in 3D and finding coplanarity

    Quote Originally Posted by eddiemoon101 View Post
    Where and how the heck do i figure out the rectangular points of these vectors?
    Suppose that $\vec{N}=<n_1,n_2,n_3>$ is a non-zero vector and $P: (a,b,c)~\&~R: (x,y,z)$
    If the question were: Write the equation of a plane which contains point $P$ and has normal $\vec{N}$
    $ \begin{align*}\vec{N}\cdot(\vec{R}-\vec{P})&=n_1(x-a)+n_2(y-b)+n_3(z-c)=0 \\n_1x+n_2y+n_3z &=n_1a+n_2b+n_3c \end{align*}$
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