Building blocks

## Problem

A plane in 3D can be defined using various combinations of

• points lying in the plane,
• direction vectors in (or parallel to) the plane, and
• normal vectors which are perpendicular to the plane.

For each of the following combinations of data,

• does it define a unique plane?
• does it define a family of possible planes? If so, can you describe the family?

The points $(-1,4,6)$, $(0,7,4)$ and $(3,-1,10)$

The points $(-1,4,6)$, $(0,7,4)$ and $(-3,-2,10)$

The direction vectors $\mathbf{i}+3\mathbf{j}+\mathbf{k}$ and $-2\mathbf{i}-6\mathbf{j}-2\mathbf{k}$

The direction vectors $\mathbf{i}+3\mathbf{j}$,
$\mathbf{i}+\mathbf{j}+2\mathbf{k}$ and
$2\mathbf{i}-\mathbf{j}$

The direction vectors $\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $-2\mathbf{i}-4\mathbf{j}+3\mathbf{k}$ and $3\mathbf{i}+6\mathbf{j}+5\mathbf{k}$

The point $(1,5,3)$ and direction vectors $\mathbf{i}-3\mathbf{j}+2\mathbf{k}$ and $5\mathbf{i}+2\mathbf{j}+2\mathbf{k}$

The points $(1,5,8)$ and $(-2,17,3)$ and direction vector $-3\mathbf{i}+2\mathbf{j}-\mathbf{k}$

The points $(1,-6,2)$ and $(3,12,-4)$ and direction vector $\mathbf{i}+9\mathbf{j}-3\mathbf{k}$

The point $(7,-3,2)$ and normal vector $-2\mathbf{i}+8\mathbf{j}-12\mathbf{k}$

The normal vector $-2\mathbf{i}+\mathbf{j}+4\mathbf{k}$ and points $(-3,3,4)$ and $(1,-1,7)$

The normal vector $-2\mathbf{i}+\mathbf{j}+4\mathbf{k}$ and points $(6,-3,5)$ and $(1,-1,7)$

The direction vector $4\mathbf{i}+8\mathbf{j}-10\mathbf{k}$ and normal vector $3\mathbf{i}+\mathbf{j}+2\mathbf{k}$