With respect to the origin O, the points A, B, C, D have position vectors given by $$ \overrightarrow {OA} = \mathbf i + 3\mathbf j + \mathbf k $$
$$\overrightarrow {OB} = \mathbf 2i + \mathbf j - \mathbf k$$
$$\overrightarrow {OC} = \mathbf 2i + 4\mathbf j + \mathbf k$$
$$\overrightarrow {OD} = \mathbf 3i + \mathbf j + 2\mathbf k$$
$(i) $Find the equation of the plane containing A, B and C, giving your answer in the form $ax + by + cz = d$
How do I tackle these type of questions?
The plane containing $A,B,C$
Point $D$ is irrelevant.
First find two vectors in the plane:
$\vec {AB} = 1\mathbf i - 2\mathbf j - 2\mathbf k\\ \vec {AC} = 1\mathbf i + 1\mathbf j$
I prefer the $(1,-2,-2), (1,1,0)$ notation to the $\mathbf i,\mathbf j,\mathbf k$ notation.
Now we need to find a vector that is perpendicular to those two (normal to the plane). The cross product is the most "mechanical" and reliable way to do this, but is not necessarily the easiest.
$(2,-2,3)$
$2x -2y + 3z = d$ now plug any of the points for $A,B,$ or $C$ to solve for $d.$
$2x -2y + 3z = -1$