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I need to find out the relative positions to each other a straight line, first I'm trying to check if they are coplanar but I get an unknown variable.
Can anyone help me on how to solve this part of the resolution?

$r1: x + 3 = \dfrac{2y - 4}4 = \dfrac{z - 1}3 = \lambda$
$r1: x + 3 = \dfrac{y - 2}2 = \dfrac{z - 1}3 = \lambda$

$r1: (x, y, z) = (-3, 2, 1) + \lambda(a, b, c)$
$r2: x = (0, 2, 2) + \lambda(1, 1, -1)$

Checks if are coplanar
$\vec{v_1}\cdot (\vec{v_2} \times \vec{u}) = \begin{vmatrix} a & b & c \\ 1 & 1 & -1 \\ -3 & 0 & 1 \end{vmatrix} = 0?$

Jyrki Lahtonen
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Daniela Morais
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  • If you take a look I made a mistake in the equation r1 ( the initial equation is correct ) and therefore this error occurs in the matrix , there is no incognita variable. If I guessed right, are reversed – Daniela Morais Jul 18 '15 at 04:37
  • @HarshKumar: Do. Not. "Invent" new tags without getting clearance from the expert taxonomists. Discuss any such idea in meta in advance. – Jyrki Lahtonen Jan 06 '17 at 07:13

1 Answers1

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Firstly, to correct some things. The first vector equation is: $$r_1: (x,y,z) = \underbrace{(-3,2,1)}_{P_1} +\lambda \cdot \underbrace{(1,2,3)}_{\mathbf a}, \quad \lambda \in \mathbb R.$$

I suppose that the second vector equation is: $$r_2 : (x,y,z) = \underbrace{(0,2,2)}_{P_2} + \lambda \cdot \underbrace{(a,b,c)}_{\mathbf b}, \quad \lambda \in \mathbb R.$$

The $2$ lines are coplanar $\iff \det(P_1 - P_2,\mathbf a, \mathbf b) =0. $

That is: $$\begin{vmatrix} -3 & 0 & -1 \\ 1 & 2 & 3 \\ a & b & c\end{vmatrix}=0.$$

The above zero determinant implies that all the rows are linearly dependent, thus we have that $(a,b,c)$ can be written as a linear combination of the $2$ other rows, i.e. $$(a,b,c) = \mu \cdot (-3,0,-1) + \nu \cdot (1,2,3),\quad \mu, \nu \in \mathbb R.$$

So, you can consider any $\mu$ and $\nu $ you want (of course it cannot hold $\mu= \nu =0)$, therefore you can define all of $a,b,c$.

thanasissdr
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