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Here is a question I need to practise but I am struggling with the line S and how to find the intersect. Can you help?

Consider the following line equations in the 3D space:

$$r:\begin{cases} x=3-\lambda \\ y=1+2\lambda \\ z=-4+3\lambda \\ \end{cases} s:\begin{cases} x+2y+3z=a \\ 3x-2y+z=-a \\ \end{cases}$$

a) Find the value of 'a' knowing that both lines intersect at one point. Find the coordinated of the intersection point.

b) For the previous value of 'a', find the plane equation \pi that contains both lines.

c) Represent the plane, the intersection point and both lines using Matlab.

Owen Morris
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    Plugg the left x y z in right expressions. – hamam_Abdallah Mar 04 '18 at 21:02
  • Please remember that you can choose an aswer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 09 '18 at 23:03

2 Answers2

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HINT

  • plug $x,y,z$ from $r$ in $s$ and find $\lambda$ and $a$
  • find normal vector $\vec n$ of $\pi$ by cross product of $\vec v\in r$ and $\vec w \in s$

notably we obtain

  • $12\lambda-7=a$
  • $-4\lambda+3=-a$

thus

  • $\lambda = \frac12$
  • $a=-1$
user
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  • Could you please expand on this?

    I have plugged in $x,y,x$ from $r$ and found:

    $6\lambda-4=a$

    $4\lambda-3=-a$

    Do I solve as simulatenous equations for a and -a?

     – Owen Morris Mar 04 '18 at 22:04
  • You need to find $\lambda$ and $a$, I check the result. – user Mar 04 '18 at 22:06
  • @OwenMorris check again your calculation you should obtain the equation and result I've just added if I'm not wrong. – user Mar 04 '18 at 22:11
  • You're not wrong. My bad, I've seen my mistake in summing up. I now have $\lambda=1/2$ and $a=-1$.

    How do I go about finding the vectors?

    Thanks for all your help.

     – Owen Morris Mar 04 '18 at 22:25
  • @OwenMorris As explained in the hint, one way is ti find two vectors and calculate the cross product, in this way you find a vector normal to both line and thus orthogonal to the plane containing the two lines. Are you aware about cross product and about plane equation? – user Mar 04 '18 at 22:33
  • I don't know how to apply the cross product and plane equation. – Owen Morris Mar 04 '18 at 23:13
  • @OwenMorris for cross product take a look here https://math.stackexchange.com/questions/2590734/finding-the-3rd-axis-in-3d-space/2590738#2590738 – user Mar 04 '18 at 23:18
  • @OwenMorris and for plane equation https://math.stackexchange.com/questions/2658300/computing-the-normal-of-a-plane-given-3-points/2658302#2658302 – user Mar 04 '18 at 23:19
  • Thanks for your help on this. I know from the previous work we did that the intercept coordinates are 2.5,2,-2.5.

    I've read through cross product and plane equation, I don't understand how to apply this

     – Owen Morris Mar 13 '18 at 23:25
  • @OwenMorris Now that you know the equation os the two lines ypu can need to find $\vec v\in r$ and $\vec w \in s$. HINT is $P_1$ and $P_2$ are 2 points $\in r$ then $\vec v = P_2-P_1$. – user Mar 13 '18 at 23:40
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Consider $\lambda$ and $a$ to be also variables, together with $x,y,z$.
You have a linear, non homogeneous, system of five equations in five variables, to be solved by any method you prefer.

G Cab
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