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For nonzero $x$, let $\varphi$ be its angle to the unit vector $q_1$. The scalar projection on the oriented $q_1$ axis is $\Vert x\Vert \cos \varphi$. Since $\Vert q_1\Vert =1$, the dot product identity gives $q_1^{T}x=\Vert x\Vert \cos \varphi$. Multiplying that signed scalar by $q_1$ places the foot at $p=(q_1^{T}x)q_1$.

The residual is perpendicular to the axis, because $q_1^T(x-p)=q_1^{T}x-(q_1^{T}x)(q_1^T{q}_1)=0$. An obtuse angle gives a negative scalar and a foot on the negative side of the line; $x=0$ has projection zero without needing an angle.

The first panel is a constructed geometric illustration. The second illustrates the directional stretch used later in Question 6.

Rotate the specified line onto the horizontal axis, stretch horizontally by two while leaving the orthogonal component unchanged, then rotate back. With $\theta =\pi /3$ and counterclockwise rotation matrix $R(\theta )$, the requested product isThe rightmost rotation acts first. In particular, $R(\theta )=\left(\begin{array}{cc}1/2& -\sqrt{3}/2\\ \sqrt{3}/2& 1/2\end{array}\right)$ and $R(-\theta )=R(\theta {)}^T$. For the unit direction $u=(1/2,\sqrt{3}/2{)}^T$, the same transformation is $T=I+u{u}^T$. It sends $u$ to $2u$ and fixes every orthogonal vector, verifying that the product is a directional stretch rather than a shear mapping or scaling in every direction. The constructed second panel in the geometric figure in Question 5 illustrates this separation.