Problem source

The vectors ${\bf a}$ and ${\bf b}$ lie in the plane $\Pi\,$.  Given that $\vert {\bf a} \vert= 1$  and ${\bf a}.{\bf b} = 3,$ find, in terms of ${\bf a}$ and ${\bf b}\,$, a vector ${\bf p}$ parallel to ${\bf a}$  and a vector ${\bf q}$  perpendicular to ${\bf a}\,$, both lying in the plane $\Pi\,$, such that $${\bf p}+{\bf q}={\bf a}+{\bf b}\;.$$
The vector ${\bf c}$  is not parallel to the plane $\Pi$ and is such that ${\bf a}.{\bf c} = -2$ and ${\bf b}.{\bf c} = 2\,$. Given  that $\vert {\bf b} \vert = 5\,$, find, in terms of ${\bf a}, {\bf b}$  and ${\bf c},$ vectors ${\bf P}$, ${\bf Q}$ and ${\bf R}$  such that ${\bf P}$ and ${\bf Q}$ are parallel to ${\bf p}$ and ${\bf q},$  respectively, ${\bf R}$ is perpendicular to the plane $\Pi$  and
$${\bf P} + {\bf Q} + {\bf R} =  {\bf a}+{\bf b}+{\bf c}\;.$$

Solution source

Suppose ${\bf p} = \lambda {\bf a}$ and  ${\bf p} + {\bf q} = {\bf a} + {\bf b}$ then

\begin{align*}
{\bf a} \cdot : && {\bf a} \cdot {\bf p} +  {\bf a} \cdot {\bf p} &=  {\bf a} \cdot {\bf a} +  {\bf a} \cdot {\bf b} \\
&&  \lambda + 0 &= 1 + 3 = 4 \\
\Rightarrow && \mathbf{p} &= 4 \mathbf{a} \\
&& \mathbf{q} &= \mathbf{b} - 3\mathbf{a} \\
\\
&& \mathbf{P} &= 4p\mathbf{a} \\
&& \mathbf{Q} &= q\mathbf{b} - 3q\mathbf{a} \\
\\
\mathbf{a} \cdot : && \mathbf{a} \cdot \mathbf{P} + \mathbf{a} \cdot \mathbf{Q} + \mathbf{a} \cdot \mathbf{R} &= \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \\
&& 4p &= 1+3-2 \\
\Rightarrow && p &= \tfrac12 \\
\\

&& {\bf P} + {\bf Q} + {\bf R} &=  {\bf a}+{\bf b}+{\bf c} \\
\mathbf{b} \cdot : && \mathbf{b} \cdot \mathbf{P} + \mathbf{b} \cdot \mathbf{Q} + \mathbf{b} \cdot \mathbf{R} &= \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} \\
&& 12p + 25q - 9q &= 3+25+2 \\
&& 6+16q &= 30 \\
\Rightarrow && q &=  \tfrac{3}{2}\\
&&  \\
&& \mathbf{P} &= 2\mathbf{a} \\
&& \mathbf{Q} &= \tfrac32 \mathbf{b} - \tfrac92 \mathbf{a} \\
&& \mathbf{R} &= \tfrac72\mathbf{a} -\tfrac12 \mathbf{b} + \mathbf{c}
\end{align*}