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.
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\nCAREFUL: 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 plane may not be written in the most efficient way, so you have to check explicitely whether it 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, or spotting patterns. We will learn how to solve it in general in Chapter 2.
\nClearly \\(\\var{u}\\) and \\(\\var{v}\\) are parallel: \\(\\var{v}=\\var{lambda}\\var{u}\\). So this is not a plane.
\nBecause of the \\(0\\) entries, it is clear that \\(\\var{u}\\) and \\(\\var{v}\\) are not parallel. So this is a plane. If we try to get \\(\\var{a}+\\lambda \\var{u}+\\mu \\var{v}=\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}\\), we would need \\(\\lambda = \\simplify[fractionNumbers]{{-a[2]/u[2]}}\\) to sort out the bottom row, and \\(\\mu=\\simplify[fractionNumbers]{{-a[0]/v[0]}}\\) to sort out the top row, but this leaves the middle row non-zero: \\(\\simplify[fractionNumbers]{{a}+{-a[2]/u[2]} {u}+{-a[0]/v[0]} {v}}= \\simplify[fractionNumbers]{{a-a[2]/u[2]*u-a[0]/v[0]*v}}\\). So this plane does not go through the origin.
\nThe third entry ensures that \\(\\var{u}\\) and \\(\\var{v}\\) are not parallel, so this is a plane. But since \\(\\var{a}=\\var{lambda}\\var{v}\\), we see that \\(\\lambda=0\\) and \\(\\mu=\\var{-lambda}\\) gives us the zero vector, so this plane does go through the origin.
\nThe first entry ensures that \\(\\var{u}\\) and \\(\\var{v}\\) are not parallel, so this is a plane. In this case it is perhaps a bit more tricky to find if it goes through the origin. But if you spot that \\(\\var{u}+\\var{v}=\\var{a}\\), then it is clear that \\(\\lambda=\\mu=-1\\) gives the zero vector. So this plane goes through the origin.
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