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Consider two planes $\vec{r} \cdot \vec{n_{1}}=d_{1}\tag i$ $\vec{r} \cdot \vec{n_2}=d_{2}.\tag {ii}$

We know that their line of intersection's direction vector is pointing in the direction of the cross product of their normals, so we get its equation to be of the form $$\vec{r}=\vec{a}+\lambda(\vec{n_1} \times \vec{n_2}),$$ where $\vec{a}$ is a point on the intersection line.

Is it possible to get the point by using (i) and (ii)?

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  • $\begingroup$ In vector form $x+y+z=1$ is ($x, y, 1-x-y)=(0,0,1)+x(1,0,-1)+y(0,1,-1$. Write your two lanes in vector form (use different parameters for each $x_1, y_1$ and $x_2, y_2$). Equate the two planes and find $x_1$ in terms of $y_1$ and you have your line. $\endgroup$
    – Paul
    Commented Apr 5, 2022 at 13:49
  • $\begingroup$ Nice thanks @Paul $\endgroup$
    – Orion_Pax
    Commented Apr 5, 2022 at 15:36

2 Answers 2

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Yes, you could convert equation (i) to parametric vector form, substitute it into (ii), make one parameter the subject, substitute it back into that parametric vector equation to finally obtain the intersection line's parametric vector equation. Notice that this doesn't actually involve finding $\vec a,$ which can be determined as the next step.

A more direct method (also not requiring finding $\vec a)$ is illustrated by this example: \begin{gather}\vec{r} \cdot \begin{pmatrix} 7\cr2 \cr -3\end{pmatrix}=4 \tag{$\pi_1$}\\\vec{r} \cdot \begin{pmatrix} 2\cr1 \cr 0\end{pmatrix}=5\tag{$\pi_2$} \end{gather}

$\pi_1:\;7x+2y-3z=4\\\pi_2:\;2x+y=5$

Expressing $x$ and $y$ in terms of $z$ $(\pi_1{-}2\pi_2,\,\ldots)$ gives the intersection line of $\pi_1$ and $\pi_2: \\\vec r=\begin{pmatrix} x\cr y \cr z\end{pmatrix}\\=\begin{pmatrix} z-2\cr 9-2z \cr z\end{pmatrix}\\=\begin{pmatrix} -2\cr 9 \cr 0\end{pmatrix}+z\begin{pmatrix} 1 \cr -2 \cr 1\end{pmatrix}\quad\left(z\in\mathbb R\right).$

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Since they ask for intersection, we know that $\vec n_1$ and $\vec n_2$ are not parallel. Then $\vec n_1$, $\vec n_2$, and $\vec n_1\times \vec n_2$ are independent, so they form a basis of $\mathbb R^3$. Then we can write $$\vec a=\alpha\vec n_1+\beta\vec n_2+\gamma\vec n_1\times\vec n_2$$ For simplicity, we assume that $\vec a$ has no component along the cross product direction, so the last term has $\gamma=0$. Now take this form for $\vec a$, plug it into $\vec r$ and then into the first two equations. Using the fact that $|\vec n_1|=|\vec n_2|=1$ and $\vec n_1\cdot(\vec n_1\times\vec n_2)=0$, you get $$\alpha+\beta \vec n_1\cdot\vec n_2=d_1\\\alpha\vec n_1\cdot\vec n_2+\beta=d_2$$ Get $\alpha$ and $\beta$ and plug it into the formula for $\vec a$.

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  • $\begingroup$ We can also let gamma be non zero as such when we finally put all values with unknown lambda we can still add lamda and gamma to get new constant times(n1 * n2) isnt Sir ? $\endgroup$
    – Orion_Pax
    Commented Apr 5, 2022 at 15:34
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    $\begingroup$ Yes you can. But then you need to say how much of that component you have. You only have two equations, so you can only find two unknowns. $\endgroup$
    – Andrei
    Commented Apr 5, 2022 at 15:37
  • $\begingroup$ Understood Sir thanks $\endgroup$
    – Orion_Pax
    Commented Apr 5, 2022 at 17:22

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