How to find intersect of two parametric equations in 3D space?

Question

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.

Answer 1

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

• $\lambda = \frac12$
• $a=-1$

Answer 2

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.