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Checking whether a given set is a plane or not. Depends on whether two vectors are parallel or not. Then checking whether the plane goes through the origin. This is not always obvious from the presentation.
\nNot randomised because it's the same as in our workbook.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Determine which of the following are planes.
\n
For each plane, also say whether it goes through the origin or not.
CAREFUL: while we said that if the plane (or line) goes through the origin, we don’t use the fixed first vector (the one that does not have a parameter in front of it), the following planes may not be written in the most efficient way, so you have to check explicitely whether they go through the origin \\(\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}\\) or not.
", "advice": "Given \\(\\left\\{ a+\\lambda u + \\mu v \\middle| \\lambda,\\mu \\in \\mathbb{R}\\right\\}\\), we can tell whether this is a plane or not by checking whether the vectors \\(u\\) and \\(v\\) are parallel: if they are parallel, then it is just a line, if they are not parallel, then it is a plane.
\nTo check whether the plane goes through the origin or not, we have to work out whether we can pick any values for \\(\\lambda\\) and \\(\\mu\\) that make the whole expression \\(a+\\lambda u + \\mu v =0\\). For most cases, this can be done by making use of the \\(0\\) entries. We will learn how to solve it in general in Chapter 2.
\na) Because of the \\(0\\) entries, it is clear that \\(\\var{u1}\\) and \\(\\var{v1}\\) are not parallel. So this is a plane. Setting \\(\\lambda=\\mu=0\\) we see that we get \\(\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}\\), so this plane goes through the origin.
\nb) Because of the \\(0\\) entries, it is clear that \\(\\var{u2}\\) and \\(\\var{v2}\\) are not parallel. So this is a plane. If we try to get \\(\\var{a2}+\\lambda \\var{u2}+\\mu \\var{v2}=\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}\\), we would need \\(\\lambda = -1\\) to sort out the top row, and \\(\\mu=-\\frac{1}{2}\\) to sort out the bottom row, but this leaves the middle row non-zero. So this plane does not go through the origin.
\nc) Here \\(\\var{u3}\\) and \\(\\var{v3}\\) are parallel: \\(\\var{v3}=-2\\var{u3}\\). So this is not a plane.
\nd) The third entry ensures that \\(\\var{u4}\\) and \\(\\var{v4}\\) are not parallel, so this is a plane. But we can easily see that \\(\\lambda=0\\) and \\(\\mu=-1\\) gives us the zero vector, so this plane does go through the origin.
\ne) Again the third entry ensures that \\(\\var{u5}\\) and \\(\\var[fractionNumbers]{v5}\\) are not parallel, so this is a plane. This time it is perhaps a bit more tricky to find if it goes through the origin. But if you spot that \\(\\var{u5}+\\var[fractionNumbers]{v5}=\\var[fractionNumbers]{a5}\\), then it is clear that \\(\\lambda=\\mu=-1\\) gives the zero vector. So this plane also goes through the origin.
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