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#+title: NST1A Math Example Sheet 1 Solutions
#+filetags: :public:note:project:
* About
This is the solutions to the first example sheet for [[id:c5927205-583c-4106-ae2c-aedd1f74871b][NST1A: Mathematics I (Course B)]].
I have downloaded the file at [[file:/home/atticusk/Downloads/math1A_example_sheet_1.pdf][NST1A Math Example Sheet 1]]
* Solutions
** Vector Addition and Subtraction
*** A1
$\vec{BO}=\vec{BA}+\vec{AO}=-\vec{AB}-\vec{OA}$
*** A2
$\vec{AB} = (4,-1,4)$
$\vec{BC} = (-3,1,3)$
$\vec{AC} = (1,0,7)$
This means that $A$ and $C$ are closest together.
*** A3
(i) $85 km/h north$
(ii) $25\sqrt{29}$ at $\arctan{\frac{2}{5}}$
east of north.
(iii) $5*\sqrt{400*\sqrt{2} + 881}$ at $\arctan(\frac{8\sqrt{2}}{8\sqrt{2} + 25})$ west of north.
*** A4
Let us first prove the forward direction.
Let there exist vectors $\vec{u}$ and $\vec{v}$ which are parallel to the sides of a
parallelogram. The diagonals of the parallelogram are given by $\vec{u} + \vec{v}$ and
$\vec{u} - \vec{v}$.
The midpoint of the first diagonal can be found by $\frac{\vec{u} + \vec{v}}{2}$. The midpoint of the other
diagonal can be found by $\vec{v} + \frac{\vec{u} + \vec{v}}{2} = \frac{\vec{u} + \vec{v}}{2}$. As these two
midpoints are equal, we have proven the forward direciton.
Now let us prove the backwards direction.
Let us assume that the diagonals of a quadrilateral
bisect each other. By side-angle-side congruence,
the 4 triangles that make up the quadrilateral
must be congruence. Because congruent parts
of congruent triangles are congruent, the
inside facing angles must be congurent.
This means that the sides of the quadrilateral
must be parallel, so the quadrilateral is a
parallelogram.
*** A5
(i) $\frac{\vec{b}+\vec{c}}{2}$.
(ii) $\vec{r} = \vec{a} + \lambda(frac{\vec{b}+\vec{c}}{2} - \vec{a})$
(iii)
\[\vec{r} = \vec{c} + \lambda(frac{\vec{a}+\vec{b}}{2} - \vec{c})\]
\[\vec{r} = \vec{b} + \lambda(frac{\vec{a}+\vec{c}}{2} - \vec{b})\]
(iv) The three medians all meet at $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$
*** A6
Let us assume that one vertex of a tetrahedron
is at the origin, and that the other edges
are outlined by vectors
\[\vec{a}, \vec{b}, \vec{c}.\]
The lines connecting
mid-points of opposite edges are given
by the equations
\[\vec{r} = \frac{\vec{a}}{2} + \lambda_{1}(\frac{\vec{b} + \vec{c}}{2}) \]
\[\vec{r} = \frac{\vec{b}}{2} + \lambda_{1}(\frac{\vec{a} + \vec{c}}{2}) \]
\[\vec{r} = \frac{\vec{c}}{2} + \lambda_{1}(\frac{\vec{b} + \vec{a}}{2}) \]
We may check that these lines are concurrent
because all of them meet at the point
\[\vec{r} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}\]
*** A7
The condition that $\alpha\vec{a}, \beta\vec{b}, \gamma\vec{c}$
be co-linear is logically equivalent to the
equation
\[\exists t \in \mathbb{R} \qquad \gamma\vec{c} = \alpha\vec{a} + t(\beta\vec{b} - \alpha\vec{a}).\]
This is logically equivalent to the equation
\[\exists t \in \mathbb{R} \qquad (\alpha - t \alpha) \vec{a} + t\beta \vec{b} - \gamma \vec{c}.\]
We can solve for $\vec{c}$ and equate the
two equations to get
\[-\frac{\alpha(1-t)}{\gamma}\vec{a} - \frac{t\beta}{\gamma}\vec{b} = \vec{c}\]
\[\frac{\lambda}{\nu}\vec{a} + \frac{\mu}{\nu}\vec{b} = \vec{c} \]
\[-\frac{\alpha(1-t)}{\gamma}\vec{a} - \frac{t\beta}{\gamma}\vec{b} =\frac{\lambda}{\nu}\vec{a} + \frac{\mu}{\nu}\vec{b}\]
If we assume that $\vec{a}$ and $\vec{b}$ are
linearly independent, then we can equate
the likewise terms on both sides of the equality,
giving
\[-\frac{\alpha(1-t)}{\gamma} = \frac{\lambda}{\nu} \implies \frac{\lambda}{\alpha} = -\frac{\nu(1-t)}{\gamma}\]
\[-\frac{t\beta}{\gamma} = \frac{\mu}{\nu} \implies \frac{\mu}{\beta} = -\frac{t\nu}{\gamma}\]
Thus, we get
\[\frac{\lambda}{\alpha} + \frac{\mu}{\beta} +
\frac{\nu}{\gamma} = -\frac{\nu(1-t)}{\gamma} - \frac{t \nu}{\gamma} + \frac{\nu}{\gamma} = 0\]
which is what we wanted to show, so we are done.
*** A8
**** i
\[\vec{r} = (1,0,1) + \lambda (0,1,-1)\]
**** ii
\[-x + 1 = y = \frac{-z + 1}{2}\]
** Scalar Product
*** B1
$\vec{a} \cdot \vec{b} = 3-1 = 2$
Conversely,
$\vec{b} \cdot \vec{a} = 3 - 1 = 2$
*** B2
(i) $\cos\theta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}| | \vec{d}|} = \frac{8}{7\sqrt{6}}$
so
$\theta = \cos^{-1}(\frac{4}{3\sqrt{6}}) = 0.995204855182370$
(ii) The vector $\langle 2 , 0 , -1 \rangle$ is perpendicular to $\vec{c}$.
*** B3
**** i
\[|\vec{OA}| = 5\]
\[|\vec{OB}| = \sqrt{14}\]
\[|\vec{AB}| = \sqrt{19}\]
**** ii
\[mAOB = \arccos\frac{|\vec{a}\cdot\vec{b}|}{|\vec{a}| |\vec{b}|} = \arccos\left(\frac{1}{7} \, \sqrt{14}\right)\]
**** iii
\[\vec{OA} = (0,3,4)\]
\[\vec{AB} = (3,-1,-3)\]
\[\theta = \arccos\frac{}{} = \arccos\left(-\frac{3}{19} \, \sqrt{19}\right)\]
*** B4
Let us assume that $A,B,C,D$ have respective
position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.
The givens state that
\[(\vec{d} - \vec{a}) \cdot (\vec{c} - \vec{b}) =
\vec{c}\cdot\vec{d} - \vec{b}\cdot\vec{d} - \vec{a}\cdot\vec{c} + \vec{a}\cdot\vec{b} = 0\]
and
\[(\vec{d} - \vec{b}) \cdot (\vec{c} - \vec{c}) =
\vec{c}\cdot\vec{d} - \vec{c}\cdot\vec{d} - \vec{b}\cdot\vec{c} + \vec{b}\cdot\vec{d} = 0.\]
If we simply subtract the two equations, we get
that
\[\vec{b} \cdot \vec{d} - \vec{a} \cdot \vec{d} -
\vec{c} \cdot \vec{b} + \vec{a} \cdot \vec{c} =
(\vec{d} - \vec{c}) \cdot (\vec{b} - \vec{a}) = 0\]
which is what we wanted to show.
*** B5
This is a plane which passes through the
point $\vec{a}$. It contains vectors which
are orthogonal to $\vec{b}$.
**** i
\[|\vec{c} - \vec{a}| \sin\theta
= |(\vec{c} - \vec{a} ) \times \hat{b}| \]
**** ii
By 1.7.2, we may calculate this as
\[d = |(\vec{c} - \vec{a}) \cdot \frac{\vec{b}}{|\vec{b}|}|\]
*** B6
$|a + b|^2 = (a + b) \cdot (a + b) = |a|^2 + 2|a\cdot b| + |b|^2 \le |a|^2 + 2|a||b|+|b|^2 = (|a|+|b|)^2$
*** B7
**** i
A sphere of radius $k$.
**** ii
A plane
**** iii
A cylinder.
**** iv
A cone
*** B8
We get the equations
\[1 + a = 0\]
and
\[1 + b = 0.\]
Therefore,
\[a = -1 \qquad b = -1.\]
Now we have equations
\[c + d -2 e = 0\]
\[e - 1 = 0\]
\[c + e = 0\]
\[c + e = 0\]
We get that
\[c = -1 , e = 1, d= 3\]
*** B9
See [[id:e4a53d0e-6b4e-43db-8752-e5e17c8b9583][intersection of 2 planes]].
These two plane intersect at
\[3 (-2y -3z) + 2y + z = 0 \implies y +2z = 0\]
\[x + 2y + 3(-3x-2y) = 0 \implies 2x + y = 0\]
So the component form of the line is
\[y = -2z = -2x\]
We can get the vector form of this line as
\[\vec{r} = (1 , -2 , 1)b\]
We can then find the angle it makes with
each axis.
- $\theta_x = \arccos(1/\sqrt{6})$
- $\theta_y = \arccos(-2/\sqrt{6})$
- $\theta_z = \arccos(1/\sqrt{6})$
*** B10
This plane has equation
\[\vec{r} \cdot (5,2,-7) = 9\]
This is equivalent to
\[(\vec{r} - (1,2,0)) \cdot (5,2,-7) = 0\]
This is equivalent to
\[(\vec{r} - (1,2,0)) \cdot (5,2,-7)/\sqrt{78} = 0\]
The first distance is given by
\[((1,-1,3) - (1,2,0)) \cdot (5,2,-7)/\sqrt{78} = -\frac{9}{26}\sqrt{78}\]
The second distance is given by
\[((3,2,3) - (1,2,0)) \cdot (5,2,-7)/\sqrt{78} = -\frac{11}{\sqrt{78}}\]
The points must be on the same side of the plane
because their signed distances have the same sign.
*** B11
Let the tetrahredon be outlined by the
position vectors $\vec{a}, \vec{b}, \vec{c}$.
We want to show that
\[ |\vec{a}^{2}| + |\vec{b}|^{2} + |\vec{c}|^{2}+ |\vec{a}-\vec{b}|^2 + | \vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2 = 4(|\frac{\vec{b}+\vec{c}-\vec{a}}{2} |^2 + |\frac{\vec{a}+\vec{b}-\vec{c}}{2} |^2 + |\frac{\vec{a}+\vec{c}-\vec{b}}{2} |^2) \]
If we expand the left-hand side of the equation,
we get
\[2(|\vec{a}|^{2} + |\vec{b}|^{2} +|\vec{c}|^{2})\]
If we expand the right-hand side of the equation,
we get
\[2(|\vec{b}+\vec{c}-\vec{a} |^2 + |\vec{a}+\vec{b}-\vec{c} |^2 + |\vec{a}+\vec{c}-\vec{b}|^2) = 2(|\vec{a}|^{2} + |\vec{b}|^{2} +|\vec{c}|^{2}) \]
As these two expressions are equal, then we are
done.
*** B12
Since we need to find any distance, I will find
the distance from $\vec{0}$ to the plane
containing positions $\vec{a}, \vec{b}, \vec{c}$
From the notes, we know that the distance from
a point to a plane is given by
\[d = |(\vec{p} - \vec{a}) \cdot \hat{n}|\]
where $\vec{p}$ is a point, $\vec{a}$ is a
point on the plane, and $\hat{n}$ is the unit normal
from the plane.
In our case, let $\vec{p} = \vec{c}$, let
$\vec{a} = \vec{a}$.
To find $\hat{n}$, we will use the
[[id:cedb1489-8203-40db-82b7-48031c23f171][cross product]].
\[\hat{n} = \frac{\vec{a} \times \vec{b}}{l^{2}}\]
So the distance is
\[d = |(\vec{c} -\vec{a}) \cdot \frac{\vec{a} \times \vec{b}}{l^{2}}|\]
This means that
\[d = \frac{[\vec{c}, \vec{a}, \vec{b}]}{l^{2}} - \frac{1}{l}\]
By the volume of a regular tetrahredron,
we may deduce that
\[[\vec{c}, \vec{b}, \vec{a}] = \frac{l^{3}}{2\sqrt{2}}\]
Thus,
\[d = -\frac{1}{4} \, \sqrt{2} l {\left(2 \, \sqrt{2} - 1\right)}\]
*** B13
Two diagonals of a cube are given by the position
vectors
\[\vec{d}_{1} = (1,1,1)\]
and
\[\vec{d}_{2} = (1,1,-1).\]
Using the [[id:51f7af57-31ed-4224-ae23-3e8230d908e7][dot product]], we can find the angle
between them as
\[\theta = \arccos\frac{\vec{d}_{1}\cdot\vec{d}_{2}}{|\vec{d}_{1}||\vec{d}_{2}|} = \arcos\frac{1}{3} \]
** C. Vector Product
*** C1
**** i
$\vec{a} \times \vec{b} = (5,8,-6)$. We also see
that $(5,8,-6) \cdot (2,1,3) = 0$ and
$(5,8,-6) \cdot (6,0,5) = 0$.
**** ii
$\vec{b} \times \vec{a} = (-5, -8, 6) = -(5,8,-6)$
**** iii
\[(\vec{a} +\vec{b} ) \times \vec{c} = (8,1,8) \times (5,3,1) = (-23,32,19) \]
\[\vec{a} \times \vec{c} + \vec{b}\times\vec{c}= (-8,13,1) + (-15,19,18) = (-23,32,19)\]
**** iv
\[(\vec{a} \times \vec{b} ) \times\vec{c}= \left(30,\,-45,\,-35\right)\]
\[\vec{a} \times (\vec{b} \times\vec{c}) =\left(-15,\,-27,\,19\right)
\]
**** v
\[\vec{a} \cdot (\vec{b} \times \vec{c}) =15 \]
\[\vec{b} \cdot (\vec{c} \times \vec{a}) =15 \]
**** vi
\[\vec{a} \cdot \vec{c} \vec{b} \vec{a} \cdot \vec{b} \vec{c} = (-15,-27, 19)\]
\[\vec{a} \times (\vec{b} \times\vec{c}) =\left(-15,\,-27,\,19\right)\]
*** C2
Let
\[\vec{a} = (2,1,1)\]
\[\vec{b} = (3,-1,-5)\]
Then,
\[\theta = \frac{\pi}{2}\]
We can get a vector $\vec{c}$ perpendicular to
both by using the [[id:cedb1489-8203-40db-82b7-48031c23f171][cross product]].
\[\vec{c} = \vec{a} \times \vec{b} = (-4,13,-5)\]
The direction cosines are
\[\cos\theta_{x} = -\frac{2}{105}\sqrt{210}\]
\[\cos\theta_{y} = \frac{13}{210} \sqrt{210}\]
\[\cos\theta_{z} = -\frac{1}{42} \sqrt{210}\]
*** C3
**** i
\begin{align}
&(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})\\
&= \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} \\
&= \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} \\
\end{align}
**** ii
\begin{align}
(\vec{a} \times \vec{b}) \times \vec{c} + (\vec{b} \times \vec{c}) \times \vec{a} + (\vec{c} \times \vec{a}) \times \vec{b} = 0 \\
&= \vec{c}\cdot \vec{a} \vec{b} - \vec{c} \cdot \vec{b} \vec{a} + \vec{a} \cdot \vec{b} \vec{c}
- \vec{a} \cdot \vec{c} \vec{b} + \vec{b} \cdot \vec{c} \vec{a} - \vec{b} \cdot \vec{a}\vec{c} \\
&= (\vec{b} \cdot\vec{c} - \vec{c} \cdot \vec{b}) \vec{a} + (\vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{c})\vec{b} + (\vec{a} \cdot\vec{b} - \vec{b}\cdot\vec{a})\vec{c} \\
&= 0
\end{align}
*** C4
Let us solve the equation
\[(x,y,z) \times (1,1,0) = (1,-1,0)\]
We get
\[z= - 1 \qquad x - y = 0\]
Thus, this locus is a plane satisfying the
equations
\[z = -1 \qquad x = y\]
*** C5
**** i
\[\vec{A} \cdot\vec{a} = \frac{(\vec{b} \times\vec{c}) \cdot \vec{a}}{\vec{a} \cdot (\vec{b} \times \vec{c})} = 1\]
Similarly for $\vec{b}$ and $\vec{c}$.
**** ii
\begin{align}
\vec{A} \cdot \vec{a} \\
&= \frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]} \cdot \vec{a} \\
&= \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]} \\
&= \frac{[\vec{b}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} \\
&= 0
\end{align}
The parallelpiped $[\vec{b}, \vec{b}, \vec{c}]$
is flat, so it has no volume.
The other cases follow similarly.
**** iii
\begin{align}
[\vec{A}, \vec{B}, \vec{C}] \\
&= \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^{3}}[ \vec{b} \times \vec{c} , \vec{c} \times \vec{a} , \vec{a} \times \vec{b}] \\
&= \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^{3}} (\vec{b} \times \vec{c}) \cdot ((\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})) \\
&= \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^{3}} (\vec{b} \times \vec{c}) \cdot [\vec{a}, \vec{b}, \vec{c}] \cdot \vec{a} \\
&= \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^{3}} [\vec{a}, \vec{b}, \vec{c}]^{2} \\
&= \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \\
\end{align}
**** iv
\begin{align}
\frac{\vec{B} \times \vec{C}}{[\vec{A} , \vec{B}, \vec{C}]} \\
&= [\vec{a}, \vec{b}, \vec{c}] (\vec{B} \times \vec{C})
&= \frac{(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})}{[\vec{a}, \vec{b}, \vec{c}]} \\
&= \frac{[\vec{a}, \vec{b}, \vec{c}]\vec{a}}{[\vec{a}, \vec{b}, \vec{c}]} \\
&= \vec{a}
\end{align}
*** C6
To show this, we will show that
\begin{align}
(\frac{\vec{a}}{\alpha} - \frac{\vec{b}}{\beta}) \cdot (\vec{g}) \\
&= (\frac{\vec{a}}{\alpha} - \frac{\vec{b}}{\beta}) \cdot (\alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}) \\
&= \frac{\alpha}{\alpha} \vec{a} \cdot \vec{A} +
\frac{\beta}{\alpha} \vec{a} \cdot \vec{B} +
\frac{\gamma}{\alpha} \vec{a} \cdot \vec{C}
- \frac{\alpha}{\beta} \vec{a} \cdot \vec{A}
- \frac{\beta}{\beta} \vec{b} \cdot \vec{B}
- \frac{\gamma}{\beta} \vec{b} \cdot \vec{C}
\\
&= 1 - 1 \\
&= 0
\end{align}
This means that $\frac{\vec{a}}{\alpha} - \frac{\vec{b}}{\beta}$ is normal to $\vec{g}$.
A similar calculation can be done to show that
$\frac{\vec{c}}{\gamma} - \frac{\vec{b}}{\beta}$ is normal to $\vec{g}$.
As these two vectors define the plane, this
shows that $\vec{g}$ is normal to the plane,
which is what we wanted to show.