Q.Find the vector equation of the plane passing through the points $A(1, 0, 1)$, $B(1, -1, 1)$ and $C(4, -3, 2)$.

Find two vectors in the plane using the three points, take their cross product to get the normal, then form $\bar r\cdot\bar n = \bar a\cdot\bar n$.

Given $A(1,0,1)$, $B(1,-1,1)$, $C(4,-3,2)$.

$\overrightarrow{AB} = B-A = (0,-1,0), \qquad \overrightarrow{AC} = C-A = (3,-3,1)$

The normal to the plane is:

$\bar n = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & -1 & 0 \\ 3 & -3 & 1 \end{vmatrix}$

$= \hat i\big[(-1)(1)-(0)(-3)\big] - \hat j\big[(0)(1)-(0)(3)\big] + \hat k\big[(0)(-3)-(-1)(3)\big]$

$= \hat i(-1) - \hat j(0) + \hat k(3) = -\hat i+3\hat k$

Using point $A$ (position vector $\bar a = \hat i+\hat k$): …