- Note: Both Question and Solution are provided, I need clarification (bottom page).
- Question:
Based on the provided solution available online, could you please explain why do we use the unitary vectors j and k? Why not the unitary vector i(1, 0, 0) for finding the vector equation of the plane? For example, x = P + ti ?
The reason why I think like that is because if we assume unitary vector i to be the normal vector (n) to the plane, any vector equation of a plane parallel to the normal vector n will result in a plane perpendicular to a surface (it can be an axis or another plane). Am I thinking right?
Best Answer
The unit vector $\mathbf{i}=\langle1,0,0\rangle$ is normal (perpendicular, orthogonal) to the plane. Using it we can set up a so-called standard equation of the plane: if $P_0(x_0,y_0,z_0)$ is a point in the plane and $\mathbf{N}=\langle A,B,C\rangle$ is a vector normal to the plane, then its equation is $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0,$$ which can be further simplified to $Ax+By+Cz+D=0$. In this example, we have an initial point $P_0(1,1,3)$ and a normal vector $\mathbf{N}=\mathbf{i}=\langle1,0,0\rangle$, so we would get $$1(x-1)+0(y-1)+0(z-3)=0 \quad \text{or} \quad x=1.$$ This would be a correct equation of the desired plane… except for it's not in the required vector form.
So that's your mistake: to use a single vector to set up an equation of a plane (in $\mathbb{R}^3$), you need a normal vector, and you'd end up with a standard equation, and it could be fine — but that's not what you're asked to set up in this question.
Also, what you're suggesting is wrong for yet another reason: in $\mathbb{R}^3$, a vector equation of the form $\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}$ describes a line, not a plane, because (informally speaking) such an equation describes all points obtained by moving from $\mathbf{r}_0$ in the direction of the vector $\mathbf{v}$ only, i.e. along only a line. To create a plane, you have to allow moving in two different directions, so a vector equation of a plane has to be of the form $$\mathbf{r}(t)=\mathbf{r}_0+t_1\mathbf{v}_1+t_2\mathbf{v}_2,$$ where $\mathbf{v}_1$ and $\mathbf{v}_2$ are two vectors lying in the plane (not normal to it).
So that's what the solution says: we need to find two vectors lying in the plane, and from the provided description we see that $\mathbf{i}=\langle1,0,0\rangle$ and $\mathbf{j}=\langle0,1,0\rangle$ are in it, so they would do the job. Using the given point and these vectors we can set up the desired vector equation of the plane.