Solving vector equations of planes

Question

Find the line of intersection of two planes denoted by: $r=\overrightarrow{b}+\lambda(\overrightarrow{b}-\overrightarrow{a})+\nu(\overrightarrow{a}+\overrightarrow{c})$ $r=\overrightarrow{c}+\alpha(\overrightarrow{b}-\overrightarrow{c})+\beta(\overrightarrow{a}+\overrightarrow{b})$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non coplanar vectors.

How to go about such parametric problems? Solving a general Cartesian equation of plane is by assuming a either $x$ or $y$ or $z$ as a parameter and so on. Am I to assume these vectors using variables for each $i,j,k$ component?

Answer 1

Yes. Write your vectors like $\vec{b}=b_1 \vec{1}_x + b_2 \vec{1}_y + b_3 \vec{1}_z$ etc, and enforce the equality for every component, which gives you 3 equations in 4 unknowns ($\alpha,\beta,\mu,\nu$). You can eliminate three of the unknowns which gives you a one-dimensional solution space, i.e. a line.

Answer 2

Here is an arguably better approach to the problem:

Let there be a vector $\overrightarrow{r}$ such that

$\overrightarrow{r}=\overrightarrow{b}+\lambda(\overrightarrow{b}-\overrightarrow{a})+\nu(\overrightarrow{a}+\overrightarrow{c}) = \overrightarrow{c}+\alpha(\overrightarrow{b}-\overrightarrow{c})+\beta(\overrightarrow{a}+\overrightarrow{b})$

Since $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non coplanar vectors, condition for $\overrightarrow{r}$ is: (by equating coefficients of every vector)

$1+\lambda = \alpha+\beta$

$-\lambda + \nu=\beta$

$\nu=1-\alpha$

Solving this, you can simplify $\overrightarrow{r}$ to: $\overrightarrow{r}=\overrightarrow{b}+\nu(\overrightarrow{a}+\overrightarrow{c})$