Find the vector equation of the plane through any three points a, b,c

Question

In $\mathbb R^{3}$, how to prove the plane through the points $\mathbf a$, $\mathbf b$, and $\mathbf c$ has the equation $$\mathbf r = (1-\mu-v)\mathbf a+\mu\mathbf b+v\mathbf c$$ I tried to evaluate the formula in this form: $$\mathbf r \cdot(\mathbf a-\mathbf b)\times(\mathbf a-\mathbf c)=\mathbf a\cdot(\mathbf a-\mathbf b)\times(\mathbf a-\mathbf c)$$ which is equivalent to $$[\mathbf r,\mathbf a-\mathbf b,\mathbf a-\mathbf c]=[\mathbf a,\mathbf a-\mathbf b,\mathbf a-\mathbf c]$$ then how should I simplify this into the linear equation above?

Answer 1

Assuming non-collinearity, I would approach it as follows.

$\mathbf {b-a}$ and $\mathbf {c-a}$ span the plane so a linear combination $\mu (\mathbf {b-a}) + \nu (\mathbf {c-a})$ of them reaches all points in the plane. But, you have to 'get to' the plane first and this can be done by adding in $\mathbf a$. The equation is thus $$ \mathbf r = \mathbf a + \mu (\mathbf {b-a}) + \nu (\mathbf {c-a}) $$

Answer 2

If vectors $\mathbf X$, $\mathbf Y$ and $\mathbf Y$ lie in the same plane we can write $$\mathbf X=\mu\mathbf Y+v\mathbf Z$$ for some real $\mu$ and $v$. Now, let $\mathbf r$ be a point that lies in the plane containing points $\mathbf a$, $\mathbf b$ and $\mathbf c$. Then, that means $\mathbf r-\mathbf a$ is a vector that lies in the same plane. But this plane also contains the vectors $\mathbf b-\mathbf a$ and $\mathbf c-\mathbf a$. Hence we can write $$\mathbf r-\mathbf a=\mu(\mathbf b-\mathbf a)+v(\mathbf c-\mathbf a)$$ $$\Rightarrow \mathbf r=(1-\mu-v)\mathbf a+\mu\mathbf b+v\mathbf c$$