Linear Algebra/Vectors in Space/Solutions

Solutions

This exercise is recommended for all readers.

Problem 1

Find the canonical name for each vector.

1. the vector from $(2,1)$ to $(4,2)$ in $\mathbb {R} ^{2}$
2. the vector from $(3,3)$ to $(2,5)$ in $\mathbb {R} ^{2}$
3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\mathbb {R} ^{3}$
4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\mathbb {R} ^{3}$

Answer

1. ${\begin{pmatrix}2\\1\end{pmatrix}}$
2. ${\begin{pmatrix}-1\\2\end{pmatrix}}$
3. ${\begin{pmatrix}4\\0\\-3\end{pmatrix}}$
4. ${\begin{pmatrix}0\\0\\0\end{pmatrix}}$

This exercise is recommended for all readers.

Problem 2

Decide if the two vectors are equal.

1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$
2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$

Answer

1. No, their canonical positions are different.

   ${\begin{pmatrix}1\\-1\end{pmatrix}}\qquad {\begin{pmatrix}0\\3\end{pmatrix}}$

2. Yes, their canonical positions are the same.

   ${\begin{pmatrix}1\\-1\\3\end{pmatrix}}$

This exercise is recommended for all readers.

Problem 3

Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$?

Answer

That line is this set.

$\{{\begin{pmatrix}-2\\1\\1\\0\end{pmatrix}}+{\begin{pmatrix}7\\9\\-2\\4\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}$

Note that this system

${\begin{array}{*{2}{rc}r}-2&+&7t&=&1\\1&+&9t&=&0\\1&-&2t&=&2\\0&+&4t&=&1\end{array}}$

has no solution. Thus the given point is not in the line.

This exercise is recommended for all readers.

Problem 4

1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$.
2. Is the origin in that plane?

Answer

1. Note that

   ${\begin{pmatrix}2\\2\\2\\0\end{pmatrix}}-{\begin{pmatrix}1\\1\\5\\-1\end{pmatrix}}={\begin{pmatrix}1\\1\\-3\\1\end{pmatrix}}\qquad {\begin{pmatrix}3\\1\\0\\4\end{pmatrix}}-{\begin{pmatrix}1\\1\\5\\-1\end{pmatrix}}={\begin{pmatrix}2\\0\\-5\\5\end{pmatrix}}$

   and so the plane is this set.

   $\{{\begin{pmatrix}1\\1\\5\\-1\end{pmatrix}}+{\begin{pmatrix}1\\1\\-3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\0\\-5\\5\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}$

2. No; this system

   ${\begin{array}{*{3}{rc}r}1&+&1t&+&2s&=&0\\1&+&1t&&&=&0\\5&-&3t&-&5s&=&0\\-1&+&1t&+&5s&=&0\end{array}}$

   has no solution.

Problem 5

Describe the plane that contains this point and line.

${\begin{pmatrix}2\\0\\3\end{pmatrix}}\qquad \{{\begin{pmatrix}-1\\0\\-4\end{pmatrix}}+{\begin{pmatrix}1\\1\\2\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}$

Answer

The vector

${\begin{pmatrix}2\\0\\3\end{pmatrix}}$

is not in the line. Because

${\begin{pmatrix}2\\0\\3\end{pmatrix}}-{\begin{pmatrix}-1\\0\\-4\end{pmatrix}}={\begin{pmatrix}3\\0\\7\end{pmatrix}}$

that plane can be described in this way.

$\{{\begin{pmatrix}-1\\0\\-4\end{pmatrix}}+m{\begin{pmatrix}1\\1\\2\end{pmatrix}}+n{\begin{pmatrix}3\\0\\7\end{pmatrix}}\,{\big |}\,m,n\in \mathbb {R} \}$

This exercise is recommended for all readers.

Problem 6

Intersect these planes.

$\{{\begin{pmatrix}1\\1\\1\end{pmatrix}}t+{\begin{pmatrix}0\\1\\3\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}\qquad \{{\begin{pmatrix}1\\1\\0\end{pmatrix}}+{\begin{pmatrix}0\\3\\0\end{pmatrix}}k+{\begin{pmatrix}2\\0\\4\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}$

Answer

The points of coincidence are solutions of this system.

${\begin{array}{*{2}{rc}r}t&&&=&1+2m\\t&+&s&=&1+3k\\t&+&3s&=&4m\end{array}}$

Gauss' method

$\left({\begin{array}{*{4}{c}|c}1&0&0&-2&1\\1&1&-3&0&1\\1&3&0&-4&0\end{array}}\right)\;{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-\rho _{1}+\rho _{2}}}\;\left({\begin{array}{*{4}{c}|c}1&0&0&-2&1\\0&1&-3&2&0\\0&3&0&-2&-1\end{array}}\right)\;{\xrightarrow[{}]{-3\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&0&0&-2&1\\0&1&-3&2&0\\0&0&9&-8&-1\end{array}}\right)$

gives $k=-(1/9)+(8/9)m$, so $s=-(1/3)+(2/3)m$ and $t=1+2m$. The intersection is this.

$\{{\begin{pmatrix}1\\1\\0\end{pmatrix}}+{\begin{pmatrix}0\\3\\0\end{pmatrix}}(-{\frac {1}{9}}+{\frac {8}{9}}m)+{\begin{pmatrix}2\\0\\4\end{pmatrix}}m\,{\big |}\,m\in \mathbb {R} \}=\{{\begin{pmatrix}1\\2/3\\0\end{pmatrix}}+{\begin{pmatrix}2\\8/3\\4\end{pmatrix}}m\,{\big |}\,m\in \mathbb {R} \}$

This exercise is recommended for all readers.

Problem 7

Intersect each pair, if possible.

1. $\{{\begin{pmatrix}1\\1\\2\end{pmatrix}}+t{\begin{pmatrix}0\\1\\1\end{pmatrix}}\,{\big |}\,t\in \mathbb {R} \}$, $\{{\begin{pmatrix}1\\3\\-2\end{pmatrix}}+s{\begin{pmatrix}0\\1\\2\end{pmatrix}}\,{\big |}\,s\in \mathbb {R} \}$
2. $\{{\begin{pmatrix}2\\0\\1\end{pmatrix}}+t{\begin{pmatrix}1\\1\\-1\end{pmatrix}}\,{\big |}\,t\in \mathbb {R} \}$, $\{s{\begin{pmatrix}0\\1\\2\end{pmatrix}}+w{\begin{pmatrix}0\\4\\1\end{pmatrix}}\,{\big |}\,s,w\in \mathbb {R} \}$

Answer

1. The system

   ${\begin{array}{*{1}{rc}r}1&=&1\\1+t&=&3+s\\2+t&=&-2+2s\end{array}}$

   gives $s=6$ and $t=8$, so this is the solution set.

   $\{{\begin{pmatrix}1\\9\\10\end{pmatrix}}\}$

2. This system

   ${\begin{array}{*{1}{rc}r}2+t&=&0\\t&=&s+4w\\1-t&=&2s+w\end{array}}$

   gives $t=-2$, $w=-1$, and $s=2$ so their intersection is this point.

   ${\begin{pmatrix}0\\-2\\3\end{pmatrix}}$

Problem 8

When a plane does not pass through the origin, performing operations on vectors whose bodies lie in it is more complicated than when the plane passes through the origin. Consider the picture in this subsection of the plane

$\{{\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\1\\0\end{pmatrix}}y+{\begin{pmatrix}-0.5\\0\\1\end{pmatrix}}z\,{\big |}\,y,z\in \mathbb {R} \}$

and the three vectors it shows, with endpoints $(2,0,0)$, $(1.5,1,0)$, and $(1.5,0,1)$.

1. Redraw the picture, including the vector in the plane that is twice as long as the one with endpoint $(1.5,1,0)$. The endpoint of your vector is not $(3,2,0)$; what is it?
2. Redraw the picture, including the parallelogram in the plane that shows the sum of the vectors ending at $(1.5,0,1)$ and $(1.5,1,0)$. The endpoint of the sum, on the diagonal, is not $(3,1,1)$; what is it?

Answer

1. The vector shown

   

   is not the result of doubling

   ${\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\1\\0\end{pmatrix}}\cdot 1$

   instead it is

   ${\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\1\\0\end{pmatrix}}\cdot 2={\begin{pmatrix}1\\2\\0\end{pmatrix}}$

   which has a parameter twice as large.

2. The vector

   

   is not the result of adding

   $({\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\1\\0\end{pmatrix}}\cdot 1)+({\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\0\\1\end{pmatrix}}\cdot 1)$

   instead it is

   ${\begin{pmatrix}2\\0\\0\end{pmatrix}}+{\begin{pmatrix}-0.5\\1\\0\end{pmatrix}}\cdot 1+{\begin{pmatrix}-0.5\\0\\1\end{pmatrix}}\cdot 1={\begin{pmatrix}1\\1\\1\end{pmatrix}}$

   which adds the parameters.

Problem 9

Show that the line segments ${\overline {(a_{1},a_{2})(b_{1},b_{2})}}$ and ${\overline {(c_{1},c_{2})(d_{1},d_{2})}}$ have the same lengths and slopes if $b_{1}-a_{1}=d_{1}-c_{1}$ and $b_{2}-a_{2}=d_{2}-c_{2}$. Is that only if?

Answer

The "if" half is straightforward. If $b_{1}-a_{1}=d_{1}-c_{1}$ and $b_{2}-a_{2}=d_{2}-c_{2}$ then

${\sqrt {(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}}}={\sqrt {(d_{1}-c_{1})^{2}+(d_{2}-c_{2})^{2}}}$

so they have the same lengths, and the slopes are just as easy:

${\frac {b_{2}-a_{2}}{b_{1}-a_{1}}}={\frac {d_{2}-c_{2}}{d_{1}-a_{1}}}$

(if the denominators are $0$ they both have undefined slopes).

For "only if", assume that the two segments have the same length and slope (the case of undefined slopes is easy; we will do the case where both segments have a slope $m$). Also assume, without loss of generality, that $a_{1}<b_{1}$ and that $c_{1}<d_{1}$. The first segment is ${\overline {(a_{1},a_{2})(b_{1},b_{2})}}=\{(x,y)\,{\big |}\,y=mx+n_{1},\;x\in [a_{1}..b_{1}]\}$ (for some intercept $n_{1}$) and the second segment is ${\overline {(c_{1},c_{2})(d_{1},d_{2})}}=\{(x,y)\,{\big |}\,y=mx+n_{2},\;x\in [c_{1}..d_{1}]\}$ (for some $n_{2}$). Then the lengths of those segments are

${\sqrt {(b_{1}-a_{1})^{2}+((mb_{1}+n_{1})-(ma_{1}+n_{1}))^{2}}}={\sqrt {(1+m^{2})(b_{1}-a_{1})^{2}}}$

and, similarly, ${\sqrt {(1+m^{2})(d_{1}-c_{1})^{2}}}$. Therefore, $|b_{1}-a_{1}|=|d_{1}-c_{1}|$. Thus, as we assumed that $a_{1}<b_{1}$ and $c_{1}<d_{1}$, we have that $b_{1}-a_{1}=d_{1}-c_{1}$.

The other equality is similar.

Problem 10

How should $\mathbb {R} ^{0}$ be defined?

Answer

We shall later define it to be a set with one element— an "origin".

This exercise is recommended for all readers.

Problem 11

A person traveling eastward at a rate of $3$ miles per hour finds that the wind appears to blow directly from the north. On doubling his speed it appears to come from the north east. What was the wind's velocity? (Klamkin 1957)

Answer

Consider the person traveling at 3 miles per hour, the same person traveling at 6 miles per hour, the true wind, the apparent wind when the person is traveling at 3 miles per hour and the apparent wind when he is traveling at 6 miles per hour, respectively, as the vectors ${\vec {p}}_{1}=\left(3,0\right),{\vec {p}}_{2}=\left(6,0\right),{\vec {w}}=\left(x,y\right),{\vec {a}}_{1}$ and ${\vec {a}}_{2}$ in a 2-dimensional space where east and north are in the positive directions of the x and y axises.

From the previous consideration and from the fact that the apparent wind is the vector sum of the velocity of the true wind minus the velocity of the person motion, we have ${\begin{cases}{\vec {a}}_{1}=\left(x,y\right)-\left(3,0\right)\\{\vec {a}}_{2}=\left(x,y\right)-\left(6,0\right)\end{cases}}$.

We know from the problem that ${\vec {a}}_{1}$ is orthogonal to ${\vec {p}}_{1}$. That means ${\vec {a}}_{1}\cdot {\vec {p}}_{1}=0\implies \left(\left(x,y\right)-\left(3,0\right)\right)\cdot \left(3,0\right)=0\implies \left(x-3\right)3+\left(y-0\right)0=0\implies 3x-9=0\implies x=3$.

The problem states also that the line whose direction vector is ${\vec {a}}_{2}$ forms a 45° angle with the line whose direction vector is ${\vec {p}}_{2}$. It means ${\vec {a}}_{2}\cdot {\vec {p}}_{2}={\frac {\sqrt {2}}{2}}\left|{\vec {a}}_{2}\right|\left|{\vec {p}}_{2}\right|\implies \left(\left(x,y\right)-\left(6,0\right)\right)\cdot \left(6,0\right)={\frac {\sqrt {2}}{2}}\left|\left(x,y\right)-\left(6,0\right)\right|\left|\left(6,0\right)\right|\implies \left(x-6\right)6+\left(y-0\right)0={\frac {\sqrt {2}}{2}}\left({\sqrt {\left(x-6\right)^{2}+\left(y-0\right)^{2}}}\right)6\implies x-6={\frac {\sqrt {2}}{2}}{\sqrt {x^{2}-12x+36+y^{2}}}$.

We know from above that $x=3$, so:$x-6={\frac {\sqrt {2}}{2}}{\sqrt {x^{2}-12x+36+y^{2}}}\implies 3-6={\frac {\sqrt {2}}{2}}{\sqrt {9-36+36+y^{2}}}\implies -3={\frac {\sqrt {2}}{2}}{\sqrt {9+y^{2}}}\implies 9={\frac {9+y^{2}}{2}}\implies 9=y^{2}\implies y=\pm 3$

The equation ${\vec {a}}_{1}=\left(x,y\right)-\left(3,0\right)$ and the fact that the apparent wind ${\vec {a}}_{1}$ comes from north implies $y<0$, so $y=-3$

The velocity of the true wind ${\vec {w}}$ is $\left(3,-3\right)$, whose magnitude is ${\sqrt {3^{2}+\left(-3\right)^{2}}}={\sqrt {18}}=3{\sqrt {2}}$.

This is how the answer was given in the cited source.

The vector triangle is as follows, so ${\vec {w}}=3{\sqrt {2}}$ from the north west.

This exercise is recommended for all readers.

Problem 12

Euclid describes a plane as "a surface which lies evenly with the straight lines on itself". Commentators (e.g., Heron) have interpreted this to mean "(A plane surface is) such that, if a straight line pass through two points on it, the line coincides wholly with it at every spot, all ways". (Translations from Heath 1956, pp. 171-172.) Do planes, as described in this section, have that property? Does this description adequately define planes?

Answer

Euclid no doubt is picturing a plane inside of $\mathbb {R} ^{3}$. Observe, however, that both $\mathbb {R} ^{1}$ and $\mathbb {R} ^{3}$ also satisfy that definition.

References

• Klamkin, M. S. (proposer) (1957), "Trickie T-27", Mathematics Magazine, 30 (3): 173 {{citation}}: Unknown parameter |month= ignored (help).
• Heath, T. (1956), Euclid's Elements, vol. 1, Dover.