// Numbas version: exam_results_page_options {"name": "Cartesian equation of a plane", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "steps": [{"showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "type": "information", "extendBaseMarkingAlgorithm": true, "prompt": "

We can write a vector equation of the plane in the form:

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$\\boldsymbol{r}=\\boldsymbol{r_1}+\\lambda (\\boldsymbol{r_2}-\\boldsymbol{r_1}) + \\mu (\\boldsymbol{r_3}-\\boldsymbol{r_1})$

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Note that three points determine a plane and

\n
\n
1. $\\lambda=0,\\;\\;\\mu=0$ gives $\\boldsymbol{r}=\\boldsymbol{r_1}$
2. \n
3. $\\lambda=1,\\;\\;\\mu=0$ gives $\\boldsymbol{r}=\\boldsymbol{r_2}$
4. \n
5. $\\lambda=0,\\;\\;\\mu=1$ gives $\\boldsymbol{r}=\\boldsymbol{r_3}$
6. \n
\n

Note that if we let

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\$\\boldsymbol{n}=(\\boldsymbol{r_2}-\\boldsymbol{r_1})\\times (\\boldsymbol{r_3}-\\boldsymbol{r_1})\$

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then $\\boldsymbol{n}\\cdot (\\boldsymbol{r_2}-\\boldsymbol{r_1})=0$ and $\\boldsymbol{n}\\cdot (\\boldsymbol{r_3}-\\boldsymbol{r_1})=0$.

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If $\\boldsymbol{r} = (x,\\;y,\\;z)$ are the Cartesian coordinates of a point on the line, it follows that

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\$\\boldsymbol{r}\\cdot \\boldsymbol{n}=(x,\\;y,\\;z)\\cdot \\boldsymbol{n}=\\boldsymbol{r_1}\\cdot \\boldsymbol{n} \$

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Find the Cartesian equation of this plane, in the form $ax+by+cz = d$, with $a$, $b$ and $c$ integers, not decimals.

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Equation of the plane: [[0]] $=$ [[1]]

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You can get help by clicking on Show steps.

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Input numbers as integers and not decimals

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Normal to the plane

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A plane goes through the points:

\n

\\begin{align}
\\boldsymbol{r_1} &= \\var{r_1}, & \\boldsymbol{r_2} &= \\var{r_2}, & \\boldsymbol{r_3} &= \\var{r_3}
\\end{align}

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A plane goes through three given non-collinear points in 3-space. Find the Cartesian equation of the plane in the form $ax+by+cz=d$.

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There is a problem in that this equation can be multiplied by a constant and be correct. Perhaps d can be given as this makes a,b and c unique and the method of the question will give the correct solution.

We can write a vector equation of the plane in the form:

\n

$\\boldsymbol{r}=\\boldsymbol{r_1}+\\lambda (\\boldsymbol{r_2}-\\boldsymbol{r_1}) + \\mu (\\boldsymbol{r_3}-\\boldsymbol{r_1})$

\n

Note that three points determine a plane and

\n
\n
1. $\\lambda=0,\\;\\;\\mu=0$ gives $\\boldsymbol{r}=\\boldsymbol{r_1}$
2. \n
3. $\\lambda=1,\\;\\;\\mu=0$ gives $\\boldsymbol{r}=\\boldsymbol{r_2}$
4. \n
5. $\\lambda=0,\\;\\;\\mu=1$ gives $\\boldsymbol{r}=\\boldsymbol{r_3}$
6. \n
\n

Note that if we let

\n

\$\\boldsymbol{n}=(\\boldsymbol{r_2}-\\boldsymbol{r_1})\\times (\\boldsymbol{r_3}-\\boldsymbol{r_1})\$

\n

then $\\boldsymbol{n}\\cdot (\\boldsymbol{r_2}-\\boldsymbol{r_1})=0$ and $\\boldsymbol{n}\\cdot (\\boldsymbol{r_3}-\\boldsymbol{r_1})=0$.

\n

Hence $\\boldsymbol{r}\\cdot \\boldsymbol{n}=\\boldsymbol{r_1}\\cdot \\boldsymbol{n}$.

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If $\\boldsymbol{r} = (x,\\;y,\\;z)$ are the Cartesian coordinates of a point on the line, it follows that

\n

\$\\boldsymbol{r}\\cdot \\boldsymbol{n}=(x,\\;y,\\;z)\\cdot \\boldsymbol{n}=\\boldsymbol{r_1}\\cdot \\boldsymbol{n} \$

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If $\\boldsymbol{r}=(x,\\;y,\\;z)$ are the Cartesian coordinates of a point on the line, it follows that the equation of the plane is given by $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\cdot \\boldsymbol{n} = \\boldsymbol{r_1} \\cdot \\boldsymbol{n}$.

\n

We have:

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\$\\boldsymbol{n}=(\\boldsymbol{r_2}-\\boldsymbol{r_1})\\times (\\boldsymbol{r_3}-\\boldsymbol{r_1}) = \\var{r_2-r_1} \\times \\var{r_3-r_1} = \\var{n} \$

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Hence, $\\boldsymbol{r_1} \\cdot \\boldsymbol{n} = \\var{con}$.

\n

So the Cartesian equation of the plane is

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\$\\simplify[all,!noLeadingMinus]{{coeffx}x + {coeffy}y + {coeffz}z = {con}} \$

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