Tutorial note 11

Problem 1.  You can see the picture here.

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Problem 2. The length of $\vec v$ is $\|\vec v\|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$. Also, a vector $\vec u /\!\!/ \vec v$ if and only if \[

\vec u = k\vec v,\quad \exists k\in \R,

\] i.e., $\{k\vec v:k\in \R\}$ is the collection of all vectors parallel to $\vec v$.

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Problem 3. Recall that \[

\vec u\cdot \vec v=\|\vec u\| \|\vec v\| \cos \theta,

\] for some $\theta\in [0,180^\circ]$ the angle between vectors $\vec u$ and $\vec v$. Hence if we use this formula to $\vec u$ and $\vec v$ given in this problem, we have \[

\cos \theta =\frac{\vec u\cdot \vec v}{\|\vec u\|\cdot \|\vec v\|}=\frac{1\cdot 4+2\cdot 5 +3\cdot 6}{\sqrt{14}\cdot \sqrt{77}}=\frac{33}{7\sqrt{22}},

\] so approximately, $\theta = 12.93^\circ$.

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Problem 4. The vector projection is: \[

 \brac{\vec u \cdot \frac{\vec v}{\|\vec v\|}} \frac{\vec v}{\|\vec v\|}=\brac{\inner{9,1,1}\cdot \frac{\inner{7,4,7}}{\sqrt{114}}} \frac{\inner{7,4,7}}{\sqrt{114}}=\frac{37}{57} \inner{7,4,7}=\inner{\frac{37\cdot 7}{57},\frac{37\cdot 4}{57},\frac{37\cdot 7}{57}}.

\]

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Problem 5. $P,Q,R$ are collinear if and only if \[

(0,0,0), \quad Q-P=(1,2,3), \quad R-P=(3,2,3)

\] are colinear. Obviously it is impossible since $(3,2,3)\neq k (1,2,3,)$ for every $k\in \R$.

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Problem 6. $\vec{BP}=\vec P -\vec B$. The length $\ell$ of $P$ projected onto the line $AB$ is \[

\ell =\frac{(\vec P-\vec B)\cdot (\vec A-\vec B)}{\|\vec A-\vec B\|}=\left|\frac{(0,-3,0)\cdot (-3,-6,0)}{\sqrt{9+36}}\right|=\frac{18}{\sqrt{45}}=\frac{6\sqrt{5}}{5}.

\]

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Problem 7.\begin{align*}

\vec u\times \vec v& =\begin{vmatrix}
\vec i&\vec j&\vec k\\
2&2&-4\\
4&0&-3
\end{vmatrix}\\
&=\vec i\begin{vmatrix} 2&-4\\0&-3\end{vmatrix} -\vec j \begin{vmatrix}2&-4\\4&-3\end{vmatrix}+\vec k\begin{vmatrix}f2&2\\4&0 \end{vmatrix}\\
&=-6\vec i - 10 \vec j -8 \vec k.

\end{align*}

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Problem 8. The vectors $\vec u\times \vec v$ and $-\vec u\times \vec v$ will do.

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Problem 9. The area is \begin{align*}

&\|\big(\inner{1,1,0}-\inner{0,1,0}\big)\times \big(\inner{2,13,0}-\inner{1,1,0}\big)\|\\
&=\|\inner{1,0,0}\times \inner{1,12,0}\|\\
&=\left\|\begin{vmatrix}
\vec i&\vec j &\vec k\\
1&0&0\\1&12&0
\end{vmatrix}\right\|\\
&=\left\|\vec i \begin{vmatrix}0&0\\12&0\end{vmatrix} - \vec j \begin{vmatrix}1&0\\1&0\end{vmatrix}+\vec k \begin{vmatrix}1&0\\1&12\end{vmatrix}\right\|\\
&=\|12\vec k\| = \|\inner{0,0,12}\|=12.
\end{align*}

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Problem 10. \begin{align*}

\text{Volume}&=\abs{\|\vec{PQ}\times \vec{PR}\| \brac{\vec{PS}\cdot \frac{\vec{PQ}\times \vec{PR}}{\|\vec{PQ}\times \vec{PR}\|}}}\\
&=|\vec{PS}\cdot (\vec{PQ}\times \vec{PR})|\\
&=|(6,-2,2)\cdot \big( (2,3,3)\times (-1,-1,-1)\big)|\\
&=\begin{vmatrix}
6&2&-1\\-2&3&-1\\
2&3&-1
\end{vmatrix} \\
&=6\begin{vmatrix}
3&-1\\
3&-1
\end{vmatrix} -2\begin{vmatrix}
-2&-1\\2&-1
\end{vmatrix} - \begin{vmatrix}
-2&3\\2&3
\end{vmatrix}\\
&=4

\end{align*}