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What is the difference between a scalar and a linear equation of a plane? In my textbook it says that a scalar equation is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ and a linear equation is $ax+by+cz=d$ How do they differ in terms of what they describe? If a line on the plane dot the normal vector $= 0$ why do we have $d$ for the linear equation?

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The two equations describe precisely the same sets. Suppose that a non-zero vector $n = (a, b, c)$ and a point $p_{1} = (x_{1}, y_{1}, z_{1})$ are given, and let $p = (x, y, z)$ denote an arbitrary point of $\mathbf{R}^{3}$.

Expanding the "scalar" equation gives \begin{align*} 0 &= a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) \\ &= ax - ax_{1} + by - by_{1} + cz - cz_{1} \\ &= ax + by + cz - \underbrace{(ax_{1} + by_{1} + cz_{1})}_{\text{Call this $d$}}, \end{align*} or $ax + by + cz = d$, what you call the "linear equation".

There is no unique way to go "backward" from the linear equation to a scalar equation: If $(x_{1}, y_{1}, z_{1})$ lies on the plane with "linear" equation $ax + by + cz = d$, namely if $ax_{1} + by_{1} + cz_{1} = d$, then $$ a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0. $$

Each can be written as a vector equation: $$ 0 = n \cdot (p - p_{1})\quad\text{("scalar" form)};\qquad n \cdot p = n \cdot p_{1}\quad \text{("linear" form)}. $$

The "scalar" equation is natural to write down when $n = (a, b, c)$ and a point $p_{1} = (x_{1}, y_{1}, z_{1})$ are known. The "linear" equation is sometimes a bit easier to use in computations with specific planes.

Neither equation is more correct than the other; they're both descriptions of the plane in space through the point $p_{1}$ and with normal vector $n$.

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  • $\begingroup$ thank you! that was really helpful:) $\endgroup$ Apr 11 '15 at 21:15
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The equation $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ means that a vector $\vec x -\vec x_1$ is orthogonal to the vector $\vec v =(a,b,c)^T$, so it represents the plane orthogonal to $\vec v$ and passing thorough $\vec x_1$. It is the same as $ax+by+cz=d=ax_1+by_1+cz_1$.

The first form ( ''scalar") shows immediately a point between wich passes the plane, the second form (''linear'') shows the intercepts of the plane with the coordinates axis $ (d/a,d/b,d/c)$.

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