Let $\\boldsymbol{v} = \\begin{pmatrix} a \\\\ b \\end{pmatrix}$ where $a,b \\in \\mathbb{R}$.
\nWrite down an expression for the norm $||\\boldsymbol{v} ||$ in terms of $a$ and $b$. (If you wish to express something like $x^{\\frac{1}{2}}$ then write x^(1/2) .)
$||\\boldsymbol{v} || =$ [[0]] .
Now consider the vectors $ \\begin{pmatrix} c \\\\ d \\end{pmatrix}$ and $\\begin{pmatrix} f \\\\ g \\end{pmatrix}$,where $c,d,f,g \\in \\mathbb{R}$. Express the dot product of these two vectors in terms of $c,d,f,$ and $g$. (If you wish to write something like $cg$ then please write c*g and not cg, as the system may not correcly interpret the latter.)
$\\begin{pmatrix} c \\\\ d \\end{pmatrix} \\cdot \\begin{pmatrix} f \\\\ g \\end{pmatrix} =$ [[1]] .
Now find an expression for $||\\boldsymbol{v} ||$ in terms of a suitable dot product.
\n$||\\boldsymbol{v} || = \\Big($ [[2]] $\\cdot$ [[3]] $\\Big)^{\\frac{1}{2}}$.
\n", "unitTests": [], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "gaps": [{"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "sqrt(a^2 + b^2)", "expectedVariableNames": [], "checkVariableNames": false, "marks": 1, "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": false, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "c*f + d*g", "expectedVariableNames": [], "checkVariableNames": false, "marks": 1, "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": false, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "v", "expectedVariableNames": [], "checkVariableNames": false, "marks": "1", "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": false, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "v", "expectedVariableNames": [], "checkVariableNames": false, "marks": "1", "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": false, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}], "marks": 0, "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "prompt": "In parts b to d, we will use the vector paramtric equation of a line in $\\mathbb{R}^2$ to obtain the cartesian description of the line.
\nConsider the line in $\\mathbb{R}^2$ described by the folowing vector parametric equation.
\n$\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} + \\begin{pmatrix} \\var{{f}} \\\\ \\var{{h}} \\end{pmatrix} t$ where $t$ takes values in $\\mathbb{R}$.
\nThis equation tells us that $\\begin{pmatrix} \\var{{f}} \\\\ \\var{{h}} \\end{pmatrix}$ is a direction vector for our line, and $\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix}$ is a position vector for our line.
\nFirst, let us express $t$ in terms of $x$, and then express $t$ in terms of $y$. (For the former use the first row of the equation above, and for the latter use the second row).
\n$t =$ [[0]] ,
\n$t =$ [[1]] .
\nPlease press the \"Show feedback\" button once you have completed this part of the question.
", "unitTests": [], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {"mark": {"order": "after", "script": "this.markingFeedback = [];\n//This clears the feedback that would normally be given automatically by the built in script. We just want to show our own feedback.\n\n//For the code below: \"this.credit tells us the proportion of the marks that the student got for this part of the question. For example, if the student scored 1 mark out of three, then \"this.credit\" would equal to 1/3.\nif (this.credit < 1){\n this.markingComment('Not quite. Try again and have a look at this feedback section once each gap has been answered correctly as there will be some useful feedback.');\n}else if (this.credit == 1){\n this.markingComment('Good! We can now equate the two equations to get an expression involving only $x$ and $y$. This is a valid cartesian equation, but we will manipulate it further to get it in a form which has more intuitive value.');\n}"}}, "gaps": [{"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "(x-{d})/{f}", "expectedVariableNames": [], "checkVariableNames": false, "marks": 1, "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": true, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "(y-{g})/{h}", "expectedVariableNames": [], "checkVariableNames": false, "marks": 1, "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": true, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}], "marks": 0, "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "prompt": "Now, let us equate the two equations above (as described in the feedback section of the part above). We then obtain the following equation:
\n$y=$ [[0]] .
\nWe can express this euqation using dot products as follows:
\n[[1]]
\nPlease press the \"Show feedback\" button once you have completed this part of the question.
", "unitTests": [], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {"mark": {"order": "after", "script": "this.markingFeedback = [];\n//This clears the feedback that would normally be given automatically by the built in script. We just want to show our own feedback.\n\n//For the code below: \"this.credit tells us the proportion of the marks that the student got for this part of the question. For example, if the student scored 1 mark out of three, then \"this.credit\" would equal to 1/3.\nif (this.credit < 1){\n this.markingComment('Not quite. Try again and have a look at this feedback section once each gap has been answered correctly as there will be some useful feedback.');\n}else if (this.credit == 1){\n this.markingComment('Good! In the next part of the question, we will find an intuitive explanation of this equation. Notice that $(\\\\var{{h}} , \\\\simplify{-{f}})^{\\\\text{T}}$ is a normal vector to our line, and $(\\\\var{{d}} , \\\\var{{g}})^{\\\\text{T}}$ is a position vector of our line. You may wish to investigate whether the use of any normal vector and any position vector, in the equation above, would still represent the same line.');\n}"}}, "gaps": [{"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "x*{h}/{f} - {d}*{h}/{f} + {g}", "expectedVariableNames": [], "checkVariableNames": false, "marks": 1, "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": true, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0, "matrix": ["1", 0, 0], "distractors": ["", "", ""], "choices": ["$\\begin{pmatrix} x \\\\ y \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{-{f}} \\end{pmatrix}=\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{-{f}} \\end{pmatrix}$
", "$\\begin{pmatrix} x \\\\ y \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{f}} \\\\ \\simplify{{h}} \\end{pmatrix}=\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{f}} \\\\ \\simplify{{h}} \\end{pmatrix}$
", "$\\begin{pmatrix} x \\\\ y \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{{f}} \\end{pmatrix}=\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} \\cdot\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{{f}} \\end{pmatrix}$
"], "displayColumns": "1", "showFeedbackIcon": true, "showCorrectAnswer": true, "minMarks": 0, "scripts": {}, "displayType": "radiogroup", "shuffleChoices": true, "unitTests": [], "maxMarks": 0, "customMarkingAlgorithm": "", "type": "1_n_2"}], "marks": 0, "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "prompt": "So far, we have used valid mathematical reaosning to obtain the equation above, but we have not given any intuitive explanation of this equation. We will do this now.
\nFirst notice that $\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{-{f}} \\end{pmatrix}$ is orthogonal to the direction vector of the the line (see part b); that is, it is a normal vector of our line. Also, $\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix}$ is a position vector of the line (it is the position corresponding to $t=0$ in the parametric equation).
\nGiven that $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ is also a point on our line (a general point), we can see that $\\Bigg( \\begin{pmatrix} x \\\\ y \\end{pmatrix} - \\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} \\Bigg)$ is a [[0]] of our line. Hence it must be [[1]] to the normal vector $\\begin{pmatrix} \\var{{h}} \\\\ \\simplify{-{f}} \\end{pmatrix}$. Hence we have that
\n$\\Bigg( \\begin{pmatrix} x \\\\ y \\end{pmatrix} -\\begin{pmatrix} \\var{{d}} \\\\ \\var{{g}} \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} \\var{{h}} \\\\ \\simplify{-{f}} \\end{pmatrix} =$ [[2]].
\nWe can easily see that this is equivalent to our equation from part c, but this time we did not really have to do any calculations; we just used our geometric intuition to obtain the equation. (This would not constitute a rigorous proof though. Whereas the calculations in parts b and c would.)
\nPlease press the \"Show feedback\" button once you have completed this part of the question.
", "unitTests": [], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {"mark": {"order": "after", "script": "this.markingFeedback = [];\n//This clears the feedback that would normally be given automatically by the built in script. We just want to show our own feedback.\n\n//For the code below: \"this.credit tells us the proportion of the marks that the student got for this part of the question. For example, if the student scored 1 mark out of three, then \"this.credit\" would equal to 1/3.\nif (this.credit < 1){\n this.markingComment('Not quite. Try again and have a look at this feedback section once each gap has been answered correctly as there will be some useful feedback.');\n}else if (this.credit == 1){\n this.markingComment('Good! Notice that right from the start our direction vector $(\\\\var{{f}} , \\\\var{{h}})^{\\\\text{T}}$ did not contain any zeroes in its entries. You may wish to investigate whether we can apply the same explanation as in parts b to d, if our direction vector contains exactly one zero entry. Can we still get a cartesian equation involving a normal vector and a position vector, as we did above?');\n}"}}, "gaps": [{"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0, "matrix": [0, 0, "1"], "distractors": ["", "", ""], "choices": ["position vector
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"], "displayColumns": 0, "showFeedbackIcon": true, "showCorrectAnswer": false, "minMarks": 0, "scripts": {}, "displayType": "dropdownlist", "shuffleChoices": true, "unitTests": [], "maxMarks": 0, "customMarkingAlgorithm": "", "type": "1_n_2"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0, "matrix": ["0.5", 0, 0], "distractors": ["", "", ""], "choices": ["orthogonal
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"], "displayColumns": 0, "showFeedbackIcon": true, "showCorrectAnswer": false, "minMarks": 0, "scripts": {}, "displayType": "dropdownlist", "shuffleChoices": true, "unitTests": [], "maxMarks": 0, "customMarkingAlgorithm": "", "type": "1_n_2"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showCorrectAnswer": false, "mustBeReduced": false, "unitTests": [], "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "correctAnswerFraction": false, "mustBeReducedPC": 0, "minValue": "0", "allowFractions": false, "correctAnswerStyle": "plain", "scripts": {}, "maxValue": "0", "marks": "0.5", "customMarkingAlgorithm": "", "type": "numberentry"}], "marks": 0, "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "prompt": "For the remainder of the question, we will look at lines, planes, and plane intersections in $\\mathbb{R}^3$.
\nConsider the line $l_1$ in $\\mathbb{R}^3$ described by the following parametric equation:
\n$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} \\var{{j}} \\\\ \\var{{k}} \\\\ \\var{{l}} \\end{pmatrix} + \\begin{pmatrix} \\var{{m}} \\\\ \\var{{n}} \\\\ \\var{{o}} \\end{pmatrix} t$ where $t$ takes values in $\\mathbb{R}$.
\nNow consider the vertical plane $\\Pi_1$ (vertical means it is parallel to the $z$-axis) which contains the line $l_1$. We want to obtain a cartesian description for this plane. Recall that the cartesian description of a plane can be expressed as $(\\boldsymbol{x} - \\boldsymbol{p} ) \\cdot \\boldsymbol{n} = 0$ where $\\boldsymbol{x} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ is a general point on the plane, $\\boldsymbol{p}$ is a fixed point on the plane, and $\\boldsymbol{n}$ is a normal to the plane. Hence, we need only find a position vector $\\boldsymbol{p}$ and a normal vector $\\boldsymbol{n}$ of the plane $\\Pi_1$.
\nConsidering that the plane $\\Pi_1$ contains the line $l_1$, we can see that any position vector of the line is also a position vector of the plane. Looking at the parametric equation of the line, one particular position vector can easily be obtained without even carrying out any calculations:
$\\boldsymbol{p} =$ [[0]] .
Now, with regards to the normal vector of the plane, we recall that a normal vector can be obtained by taking the cross product of two non-parallel directional vectors of the plane. Considering that the plane $\\Pi_1$ contains the line $l_1$, we can see that any directional vector of the line is also a directional vector of the plane. Looking at the parametric equation of the line, one particular directional vector can easily be obtained without even carrying out any calculations:
$\\boldsymbol{d_1} = $ [[1]] .
Furthermore, we are told that $\\Pi_1$ is vertical, i.e. parallel to the z-axis. Hence, the most obvious other directional vector is $\\boldsymbol{d_2} = $ [[2]] .
\n(Note that we should check that $\\boldsymbol{d_1}$ and $\\boldsymbol{d_2}$ are non-parallel; if they are parallel then the following cross product will will give us the zero vector which is a not helpful for us here).
\nWe now take the cross product of these two direcional vectors to obtain a normal vector to the plane $\\Pi_1$:
$\\boldsymbol{n} = \\boldsymbol{d_1} \\times \\boldsymbol{d_2} =$ [[3]] .
We now expand the cartesian equation $(\\boldsymbol{x} - \\boldsymbol{p} ) \\cdot \\boldsymbol{n} = 0$ (with our values of $\\boldsymbol{p}$ and $\\boldsymbol{n}$ substituted in) to get the following cartesian equation for the plane $\\Pi_1$:
$y=$ [[4]]
Now consider the horixontal plane $\\Pi_2$ described by $z=0$. We wish to find the line of intersection of $\\Pi_1$ and $\\Pi_2$ in vector parametric form. We will call this line $l_2$.
\nNow, since the plane $\\Pi_1$ contains the line $l_2$, we can see that any point $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ on $l_2$ must satisfy the following equation: $y=$ [[0]] (Hint: consider what you found in part e).
\nAlso, since the plane $\\Pi_2$ contains the line $l_2$, we can see that any point $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ on $l_2$ must satisfy the following equation: $z=$ [[1]] .
\nSo, we have two equations that $l_2$ must satisfy. Recall that a one-dimensional object (e.g. a line such a s $l_2$) in three-dimensional space must have exactly two unique equations in its cartesian description. Hence, we can see that the two equations above, together, form the cartesian description for $l_2$. However, it is the vector parametric equation that we want, but we can use the cartesian equation to obtain it.
\nFirst, we can express the two equations
$y=$ [[0]] and
$z=$ [[1]] ,
of the cartesian description of $l_2$, in the matrix form $A \\boldsymbol{x} = \\boldsymbol{b}$, where
$\\boldsymbol{x} = \\begin{pmatrix} x \\\\ y \\\\z \\end{pmatrix}$, $b= \\begin{pmatrix} \\simplify{- {n}*{j}/{m} + {m}*{k}/{m}} \\\\ 0 \\end{pmatrix}$ and $A = $ [[2]] .
Finding the reduced row echelon form of the augmented matrix $(A|\\boldsymbol{b})$ will help us find the parametric equation of $l_2$.
$\\textbf{RRE}(A|\\boldsymbol{b}) =$ [[3]] .
We can see that $x$ and $z$ are the leading variables, and the variable $y$ will play the role of the free variable. That is, we let $y = t$ and find $x$ and $z$ in terms of $t$. This will give us the following parametric equation for our line $l_2$:
\n$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = $ [[4]] $+$ [[5]] $t$.
\n", "unitTests": [], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "gaps": [{"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "{n}*x/{m} - {n}*{j}/{m} + {m}*{k}/{m}", "expectedVariableNames": [], "checkVariableNames": false, "marks": "0.5", "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": true, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "checkingType": "absdiff", "answer": "0", "expectedVariableNames": [], "checkVariableNames": false, "marks": "0.5", "showFeedbackIcon": true, "checkingAccuracy": 0.001, "showCorrectAnswer": true, "vsetRange": [0, 1], "scripts": {}, "vsetRangePoints": 5, "showPreview": true, "failureRate": 1, "unitTests": [], "customMarkingAlgorithm": "", "type": "jme"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFractions": false, "numColumns": "3", "unitTests": [], "markPerCell": false, "showFeedbackIcon": true, "allowResize": false, "showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "numRows": "2", "correctAnswer": "matrix([ -{n}/{m} , 1 , 0 ] , [ 0 , 0 , 1 ])", "tolerance": 0, "marks": "1", "customMarkingAlgorithm": "", "type": "matrix"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFractions": true, "numColumns": "4", "unitTests": [], "markPerCell": false, "showFeedbackIcon": true, "allowResize": false, "showCorrectAnswer": true, "allowFractions": true, "scripts": {}, "numRows": "2", "correctAnswer": "matrix([ 1 , -{m}/{n} , 0 , {j} - {m}*{k}/{n} ] , [ 0 , 0 , 1 , 0])", "tolerance": 0, "marks": 1, "customMarkingAlgorithm": "", "type": "matrix"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFractions": true, "numColumns": "1", "unitTests": [], "markPerCell": false, "showFeedbackIcon": true, "allowResize": false, "showCorrectAnswer": true, "allowFractions": true, "scripts": {}, "numRows": "3", "correctAnswer": "matrix([ {j} - {m}*{k}/{n} ] , [ 0 ] , [0])", "tolerance": 0, "marks": 1, "customMarkingAlgorithm": "", "type": "matrix"}, {"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFractions": true, "numColumns": "1", "unitTests": [], "markPerCell": false, "showFeedbackIcon": true, "allowResize": false, "showCorrectAnswer": true, "allowFractions": true, "scripts": {}, "numRows": "3", "correctAnswer": "matrix([ {m}/{n} ] , [ 1 ] , [0])", "tolerance": 0, "marks": 1, "customMarkingAlgorithm": "", "type": "matrix"}], "marks": 0, "customMarkingAlgorithm": "", "type": "gapfill"}], "statement": "Week 7 (lectures 13 and 14): In this question, you will look at the parametric and cartesian descriptions of lines and planes in $\\mathbb{R}^2$ and $\\mathbb{R}^3$.
\nIf you have not provided an answer to every input gap of a question or part of the question, and you try to submit your answers to the question or part, then you will see the message \"Can not submit answer - check for errors\". In reality your answer has been submitted, but the system is just concerned that you have not submitted an answer to every input gap. For this reason, please ensure that you provide an answer to every input gap in the question or part before submitting. Even if you are unsure of the answer, write down what you think is most likely to be correct; you can always change your answer or retry the question.
\nAs with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.
Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.
Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.
This is the question for week 7 of the MA100 course at the LSE. It looks at material from chapters 13 and 14.
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"}}, "ungrouped_variables": [], "extensions": [], "preamble": {"js": "", "css": ""}, "type": "question", "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}]}], "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}