"NST1A Math Example Sheet 1 Solutions"

Written By Atticus Kuhn
Tags: "public", "note", "project"
:PROPERTIES: :ID: e2b62eb7-2e66-44dc-90ed-69fee167bbb4 :mtime: 20231022073623 20231022061902 20231021043959 20231020071447 20231018060318 20231018042021 20231017072157 20231016102811 20231016042430 20231014100647 20231007024229 20231007002711 20231006011212 :ctime: 20231006011134 :END: #+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.

See Also

NST1A: Mathematics I (Course B)intersection of 2 planesvector productVector Scalar Productvector product

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