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Considering the plane $V$ given by: $$ V : x + 2y + 3z = 1 $$

How would one determine vectors a and b that are perpendicular and parallel (respectively) to the plane V such that $a+ b = (-1, 0, 1).$

I know that the normal vector is given by $(1,2,3)$ but I do not know how to approach determining the parallel vector. Any help or hints would be appreciated.

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A perpendicular vector has the form $$ \mathbf{a}=\begin{pmatrix}t\\2t\\3t \end{pmatrix} $$ a parallel vector is a vector in the plane $x+2y+3z=0$ so it has the form: $$ \mathbf{b}=\begin{pmatrix}-2u-3v\\u\\v \end{pmatrix} $$

and you want: $$\begin{pmatrix}t\\2t\\3t \end{pmatrix}+ \begin{pmatrix}-2u-3v\\u\\v \end{pmatrix}= \begin{pmatrix}-1\\0\\1 \end{pmatrix} $$

so you have a system of three equations in three unknown.

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Let $$\underline{a}=\lambda\left(\begin{matrix}1\\2\\3\end{matrix}\right)$$ And let $$\underline{b}=\left(\begin{matrix}p\\q\\r\end{matrix}\right)$$ So that$$\underline{a}+\underline{b}=\left(\begin{matrix}-1\\0\\1\end{matrix}\right)$$ $$\implies\left(\begin{matrix}p\\q\\r\end{matrix}\right)=\left(\begin{matrix}-1-\lambda\\-2\lambda\\1-3\lambda\end{matrix}\right)$$

We require $$\underline{b}\cdot\left(\begin{matrix}1\\2\\3\end{matrix}\right)=0$$

$$\implies-1-\lambda+2(-2\lambda)+3(1-3\lambda)=0\implies\lambda=\frac 17$$

Hence $$\underline{a}=\left(\begin{matrix}\frac 17\\ \frac 27\\ \frac 37\end{matrix}\right)$$ and $$\underline{b}=\left(\begin{matrix}-\frac 87\\-\frac 27\\ \frac 47\end{matrix}\right)$$

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