Consider a plane in $\mathbb{R}^{3}$ which intersects the origin and is orthogonal to $$ \mathbf{v}=\left(\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right) .$$
(a) Find a $3 \times 3$ matrix $P_{\mathbf{v}}$ which projects vectors onto this plane. $\tag{5 marks }$
(b) Given the matrix $P_{\mathbf{v}}$ from part (a), describe the geometric significance of the (nonlinear) function $f(\mathbf{w})=\sqrt{\mathbf{w}^{T} P_{\mathbf{v}} \mathbf{w}}$. $\tag{5 marks }$
My attempt : The plane which is orthogonal to the vector $x =\left ( \begin{matrix} 1 \\ -2 \\ 1 \\ \end{matrix} \right )$ is $x-2y+z = 0.$
Now the points on this plane belongs to this set - $\{ c_1\left ( \begin{matrix} 2 \\ 1 \\ 0 \\ \end{matrix} \right ) + c_2\left ( \begin{matrix} 0 \\ 1 \\ 2 \\ \end{matrix} \right ) : c_1 , c_2 \in \mathbb R\}$ . $\left ( \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right )$ also belongs to this set.
So we can say the matrix which projects all the vectors on the plane $x-2y+z = 0$ is $\left ( \begin{matrix} 0 & 2 &1\\ 1 & 1 & 1\\ 2 & 0 & 1\\ \end{matrix} \right ) $
Because $\left ( \begin{matrix} 1 & -2 & 1 \end{matrix} \right ) \times \left ( \begin{matrix} 0 & 2 &1\\ 1 & 1 & 1\\ 2 & 0 & 1\\ \end{matrix} \right ) \times \left ( \begin{matrix} x \\ y \\ z \\ \end{matrix} \right )= 0$ for all $ x, y , z \in \mathbb R$
Can anyone please check if my solution is correct and give some hints for the second part ?
If you have a basis $\beta$ for a $2-$dimensional subspace $W$ conatined in $\mathbb{R}^3$, you should put these two vectors in the columns of a matrix $A$ and then compute $$P=A(A^TA)^{-1}A^T$$ It turns out that this matrix is the matrix of the transformation that projects vectors in $\mathbb{R}^3$ onto $W$.
The column space of the matrix you provided, namely $$\left ( \begin{matrix} 0 & 2 &1\\ 1 & 1 & 1\\ 2 & 0 & 1\\ \end{matrix} \right )$$ is in fact the plane $\{x-2y+z=0\}\cap\mathbb{R}^3$ but it doesn't act on $\mathbb{R}^3$ by projecting vectors onto this plane. Can you see why? For a hint on part (b), I suggest you draw a picture and use the fact that $$\vec{u} \cdot \vec{v}=||\vec{u}||\times ||\vec{v}||\times\cos(\theta)$$ where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.