Find the equation of the plane that contains the triangle: $\Delta ABC$, where $ A=(1,1,0) $ , $ B=(-2,1,1) $ and $ C=(0,-1,2) $
Solution
The normal vector of the plane must be perpendicular to the triangle $\Delta ABC$,
in other words instead of normal vector $\vec{n}$ of the plane we can choose
$\vec{n} = \vec{AB}\times \vec{AC}$. Once we find this vector $\vec{n}=\langle n_{1},n_{2},n_{3}\rangle$ then we can easly restore the plane equation
$n_{1}(x-x_{0})+n_{2}(y-y_{0})+n_{3}(z-z_{0})=0$ where $(x_{0},y_{0},z_{0})$ is any point which belongs to the plane, for example the point $A$ will work.
$$
\vec{AB} = \vec{AO}+\vec{OB}=\vec{OB}-\vec{OA} =\langle-2,1,1 \rangle - \langle 1,1,0 \rangle = \langle -3,0,1\rangle
$$
$$
\vec{AC} = \vec{AO}+\vec{OC}=\vec{OC}-\vec{OA} =\langle 0,-1,2 \rangle - \langle 1,1,0 \rangle = \langle -1,-2,2\rangle
$$
$$
\vec{n} = \vec{AB}\times \vec{AC} = \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-3 & 0 & 1 \\
-1 & -2 & 2 \end{array} \right|
=
\vec{i} \cdot \left| \begin{array}{cc}
0 & 1 \\
-2 & 2 \end{array} \right|
-\vec{j}\cdot
\left| \begin{array}{cc}
-3 & 1 \\
-1 & 2 \end{array} \right|
+\vec{k}\cdot
\left| \begin{array}{cc}
-3 & 0 \\
-1 & -2 \end{array} \right| = 2\cdot \vec{i}+5 \cdot \vec{j}+6\cdot \vec{k}=\langle 2,5,6 \rangle
$$
Thus, the plane equation will be
$$
2(x-1)+5(y-1)+6z=0
$$
Let $A=(-1,2,0)$ and $B=(3,1,-2)$ (in unit ft).
If a bird travels from point $A$ to point $B$ along a stright line with speed $2$ ft/sec,
find the parametric equation of the line with the parameter as the time $t$ in sec..
Solution:
If the particle moves by parametric equation $\vec{r} (t) = \vec{x}_{0} + t \cdot \vec{v}$ then the speed of this particle is
$|\vec{v}|$.
The line which passes throigh the points $A$ and $B$ and goes from $A$ to $B$ has the following parametric representation
$$
\vec{r} (t) = \vec{OA} + t\cdot \vec{AB} = \langle -1,2,0 \rangle + t\cdot \langle 4,-1,-2 \rangle
$$
Unfortunately the speed in this case is $|\vec{AB}| = \sqrt{4^{2}+(-1)^{2}+(-2)^{2}} = \sqrt{21}$ which is not $2$ as it is required in the problem.
So let's multiply the vector $\vec{AB}$ by some positive number $k$ so that the speed of this new parametric equation will become $2$.
We should not multiply by negative number becasue we want to save direction. So let's consider
$$
\vec{r} (t) = \vec{OA} + t\cdot \vec{AB} k
$$
Then the speed is $|\vec{AB} k| = k |\vec{AB}| = k\sqrt{21}$ on the other hand it must be $2$. So $k\sqrt{21} =2$ hence $k=\frac{2}{\sqrt{21}}$. Thus our parametric equation is
$$
\vec{r} (t) = \vec{OA} + t\cdot \vec{AB} \frac{2}{\sqrt{21}} = \langle -1,2,0 \rangle + t\cdot \langle 4,-1,-2 \rangle \frac{2}{\sqrt{21}}
$$