The Vector Equation of the Plane Passing through Point A(a-bar) and Parallel to b-bar and c-bar

Theorem 6.10. The plane through the point $A(\vec a)$ that is parallel to two given non-zero, non-parallel vectors $\vec b$ and $\vec c$ has the equation $\left(\vec r - \vec a\right)\cdot\left(\vec b\times\vec c\right) = 0$, equivalently $\vec r\cdot(\vec b\times\vec c) = \vec a\cdot(\vec b\times\vec c)$.

Why this is true: because $\vec b$ and $\vec c$ both lie parallel to the plane (rather than in some other orientation), their cross product $\vec b\times\vec c$ is perpendicular to both of them and therefore perpendicular to the whole plane — it is a normal vector, even though $\vec b$ and $\vec c$ individually are not. For a general point $P(\vec r)$ of the plane, $\overrightarrow{AP}$ lies in the plane, so $\overrightarrow{AP}$ must be perpendicular to $\vec b\times\vec c$: $\overrightarrow{AP}\cdot(\vec b\times\vec c) = 0$. Substituting $\overrightarrow{AP} = \vec r - \vec a$ gives the stated equation (Fig. 6.9 shows $\vec b$ and $\vec c$ as two arrows lying flat in the plane from $A$, and $\vec b \times \vec c$ standing up perpendicular to the plane).

A useful alternative comes from noticing that $\overrightarrow{AP}$, $\vec b$ and $\vec c$ are all parallel to the same plane, so they are coplanar; hence $\overrightarrow{AP}$ can always be written as a linear combination $\overrightarrow{AP} = \lambda\vec b + \mu\vec c$ for some scalars $\lambda, \mu$. This gives $\vec r - \vec a = \lambda\vec b + \mu\vec c$, i.e. $\vec r = \vec a + \lambda\vec b + \mu\vec c$ — the vector equation of a plane in parametric form, with $\lambda$ and $\mu$ free real parameters that sweep out every point of the plane.

Ex.(4): the plane through $A(-1, 2, -5)$ parallel to $\vec b = 4\hat i - \hat j + 3\hat k$ and $\vec c = \hat i + \hat j - \hat k$. Here $\vec a = -\hat i + 2\hat j - 5\hat k$, and $\vec b \times \vec c = \begin{vmatrix}\hat i & \hat j & \hat k \\ 4 & -1 & 3 \\ 1 & 1 & -1\end{vmatrix} = -2\hat i + 7\hat j + 5\hat k$. Then $\vec a\cdot(\vec b\times\vec c) = (-1)(-2) + (2)(7) + (-5)(5) = 2 + 14 - 25 = -9$, so the plane's equation is $\vec r\cdot(-2\hat i + 7\hat j + 5\hat k) = -9$. …