Find the equation of the plane through a given point, with given normal vector

Question

I need to find the equation of the plane through the point $(−1,.5, 3)$ with normal vector $ + 4 + $. I know that the equation will look something like this: $1 (x + 1) + 4 (y - .5) + 1 (z - 3) = 0$.

Answer 1

Hint: Inner product + use the fact that your point is a point of the plane.

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Edit:

Since you know a normal vector of your plane, you know that every point on the plane you want is orthogonal to this normal vector. In other words, if I call $(\mathcal{P})$ the plane you are looking for and $\mathbf{n} = \mathbf{i} + 4\mathbf{j} + \mathbf{k}$, $$ \forall \mathbf{p} \in (\mathcal{P}),\ \langle \mathbf{p}, \mathbf{n} \rangle = 0$$ where $\langle \cdot,\cdot\rangle$ denotes the inner product.

Furthermore you now that the point $A = (-1, 0.5, 3)$ is a point of $(\mathcal{P})$. Thus if I assume that $\mathbf{p} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$. I get two equations.

$$\begin{align} x + 4y + z + d = 0 \\ -1 + 4*0.5 + 3 + d = 0 \end{align}$$ I introduced $d$ because since you're talking about points I suppose you're in affine space and thus seek an affine plane.

From there you can find, $d = -4$.

Final answer,

$$( \mathcal{P} ): x + 4y + z - 4 = 0$$