aurelegena

2022-09-07

Is this true or false? If false, why is it false? An equation of the form $a{x}_{1}+b{x}_{2}+c{x}_{3}=d$ can be visualized as a line in ${\mathbb{R}}^{3}$. I think it's false because an equation in this form it should be a plane in ${\mathbb{R}}^{3}$ not a line. Is that right?

ralharn

You can see that the points $\left(\frac{d}{a},0,0\right),\left(0,\frac{d}{b},0\right),\left(0,0,\frac{d}{c}\right)$ all satisfy the equation. The first two points create a line that z=0. However, the third point doesn't lie on the line, as $\frac{d}{c}\ne 0$
($a,b,c,d$ might be equal to 0, but you can find other points that do not lie on the same line as well.)

Diana Lamb

One of the coefficients must be nonzero. We can let the other two variables be anything, in which case we can solve for this third variable, so there are two "degrees of freedom" (a plane) instead of one (a line). So your argument is correct. More geometrically:
Indeed, normalize the equation so it is of the form $\mathbf{n}\cdot \mathbf{x}=r$, where n is a unit normal vector. (This means dividing by $\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$ essentially.) This may be rewritten as $\mathbf{n}\cdot \left(\mathbf{x}-\mathbf{p}\right)=0$ where $\mathbf{p}=r\mathbf{n}$. Thus, we can say the displacement vector x−p is any vector orthogonal to n, in which case the solution set $\left\{\mathbf{x}\right\}$ is the plane orthogonal to $\mathbf{n}$ and r units it that direction (possibly backwards if r<0). Thus, it's an affine plane.

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