Given two vectors: $\vec{u} = \langle -2,1,2\rangle$ and $\vec{v}=\langle 4,0,1 \rangle$
(a)
Find the dot product $\vec{u} \cdot \vec{v}$
Solution:
$$
\vec{u} \cdot \vec{v} = -2\cdot 4 + 1\cdot 0 +2\cdot 1=-6
$$
(b)
Find the cross product $\vec{u} \times \vec{v}$
Solution:
$$
\vec{u}\times \vec{v} = \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-2 & 1 & 2 \\
4 & 0 & 1 \end{array} \right|
=
\vec{i} \cdot \left| \begin{array}{cc}
1 & 2 \\
0 & 1 \end{array} \right|
-\vec{j}\cdot
\left| \begin{array}{cc}
-2 & 2 \\
4 & 1 \end{array} \right|
+\vec{k}\cdot
\left| \begin{array}{cc}
-2 & 1 \\
4 & 0 \end{array} \right| = 1\cdot \vec{i}+10 \cdot \vec{j}-4\cdot \vec{k}=\langle 1,10,-4 \rangle
$$
(c)
Find the projection of $\vec{u}$ onto $\vec{v}$
Solution:
$$
\operatorname{Proj}_{\vec{v}}(\vec{u})=\frac{\vec{u}\cdot \vec{v}}{|\vec{v}|^{2}}\vec{v}=\frac{-6}{4^{2}+0^{2}+1^{2}}\langle 4,0,1\rangle=
\frac{-6}{17}\langle 4,0,1\rangle
$$
(d)
Find the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$
Solution:
Lets recall that absolute value of the cross product of two vectors $\vec{a}, \vec{b}$ just by definition is
$|\vec{a}\times\vec{b}|=|a|\cdot |b|\sin \alpha$, which in turns is the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$.
$$
\operatorname{Area} = |\vec{u}\times \vec{v}| = |\langle 1,10,-4 \rangle| = \sqrt{117}
$$
(e)
Find the equation of the stright line which is parallel $\vec{u}-\vec{v}$ and passes throiugh the point $P=(1,1,3)$
Solution:
$$
\vec{r}(t) = \vec{OP}+t(\vec{u}-\vec{v}) = \langle 1,1,3\rangle + t \left( \langle -2,1,2\rangle-\langle 4,0,1 \rangle\right)
=\langle 1,1,3\rangle + t \langle -6,1,1\rangle
$$
(a)
Find the equation of the plane that contains the triangle: $\Delta ABC,$
where $A=(2,1,0)$, $B=(-2,1,3)$, $C=(1,0,2)$ are 3 points in space
Solution:
See the solution of the
problem 1
from the quiz 2 and show that $3x+5y+4z=11$ is the equation of the desired plane
(b)
Find the distance between the point $P=(3,2,-4)$ and the plane in (a).
Solution:
Take any point in the plane, say $P_{0}=A=(2,1,0)$ then we know that
$$
\operatorname{distance} = \frac{| \vec{P_{0}P}\cdot \vec{n}|}{|\vec{n}|}
$$
where $n$ is the normal vector of the plane $\vec{n} =\langle 3,5,4\rangle$. So
$$
\operatorname{distance} = \frac{| \vec{P_{0}P}\cdot \vec{n}|}{|\vec{n}|} =
\frac{|\langle 1,1,-4\rangle \cdot \langle 3,5,4 \rangle| }{|\langle 3,5,4 \rangle|} = \frac{8}{\sqrt{50}}
$$
Sketch and identify the surface: $x^{2}+y^{2}-z^{2}+1=0$
Solution:
For each fixed $z$ such that $|z|\geq 1 $ we have $ x^{2}+y^{2}=1-z^{2}$, so the set of points $(x,y)$ are circles.
So this is Hyperboloid of two sheets.
(a)
Let $A=(-1,2,0)$ and $B=(2,1,-2)$ (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:
See the solution of the
problem 2 from quiz 2.
(b)
How much does it take for the bird to fly from $A$ to $B$?
Solution:
$$
|AB| = \sqrt{14}.
$$
So
$$
\operatorname{time} = \frac{\operatorname{distance}}{\operatorname{speed}} = \frac{\sqrt{14}}{2} \operatorname{sec}
$$
Given a curve $\vec{r}(t) = \sin t \cdot \vec{i} + (t^{2}-\cos t)\cdot \vec{j} + (t-1) \cdot \vec{k}$. Find the parametric equation of the line which is tangent to the curve at
point $t=\frac{\pi}{2}$
Solution:
This line must pass through the point $\vec{r}\left(\frac{\pi}{2}\right)$ and must be parallel to the vector $\vec{r}'\left(\frac{\pi}{2}\right)$.
So the equation of the line will be
\begin{align*}
&\vec{r}\left(\frac{\pi}{2}\right) + t\cdot \vec{r}'\left(\frac{\pi}{2}\right)=
\left\langle\sin\left(\frac{\pi}{2}\right), \left(\frac{\pi}{2}\right)^{2}-\cos \left(\frac{\pi}{2}\right), \frac{\pi}{2}-1 \right\rangle +
t\cdot \left\langle\cos\left(\frac{\pi}{2}\right), 2\cdot \frac{\pi}{2}+\sin \left(\frac{\pi}{2}\right), 1 \right\rangle=\\
&\left\langle 1, \frac{\pi^{2}}{4}, \frac{\pi-2}{2} \right\rangle+
t\cdot \left\langle 0, \pi+1, 1 \right\rangle
\end{align*}
Let $\vec{r}(t)$ be a 3D curve. Show that if $\vec{r}'(t)\cdot \vec{r}(t)=0$ for all $t \in \mathbb{R}$ then $|\vec{r}(t)$ is equal to a constant
for all $t$.
Solution:
$$
(|\vec{r}(t)|^{2})' = (\vec{r}(t)\cdot \vec{r}(t))' = \vec{r}'(t)\cdot \vec{r}(t)+\vec{r}(t)\cdot \vec{r}'(t)=2\vec{r}'(t)\cdot \vec{r}(t)=0
$$
So $(|\vec{r}(t)|^{2})'=0$ which means that $|\vec{r}(t)|^{2}$ is constant and hence $|\vec{r}(t)|$ is constant.