# Thread: Finding the equation of a plane

1. ## Finding the equation of a plane

Could somebody please explain me how I should solve the following?

Determine the vector equation of the plane P(a) containing the line AB and perpendicular to the vector v(a).

Given

A(1,0,1) B(0,-1,1) and v(a) = i - j + (a)k

2. Let $\displaystyle \vec u = (1,1,0) \parallel \vec {AB}$

Then $\displaystyle \vec w = \vec u\times \vec v=(a,-a,-2)$ will be perpendicular to both $\displaystyle \vec u$ and $\displaystyle \vec v$.

$\displaystyle \left(\begin{array}{ccc}x\\y\\z\end{array}\right)= \left(\begin{array}{ccc}1\\0\\1\end{array}\right)+ s\left(\begin{array}{ccc}1\\1\\0\end{array}\right) +t\left(\begin{array}{ccc}a\\-a\\-2\end{array}\right)$

3. A plane perpendicular to vector <A, B, C> and containing point $\displaystyle (x_0,y_0, z_0)$ has equation $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$

Here (1, 0, 1) is a point in the plane and <1, -1, a> is perpendicular to the plane so (x-1)- y+ a(z-1)= 0. We can write that as x- y+ az= a+1. We can solve for x: x= y- az+ a+ 1 and, using y and z as parameters write
$\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}a+1 \\ 0 \\ 0\end{bmatrix}+ \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}y+ \begin{bmatrix}-a \\ 0 \\ 1\end{bmatrix}z$.