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Find the projection of \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) onto the plane \(3x - y + 4z = 0\).

\(\text{Let $\vec{P}= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$}\)

1.

Normal vector \(\mathbf{\vec{n}}\)

\(\begin{array}{|rcll|} \hline \vec{n} &=& \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \quad | \quad \text{plane: }{\color{red}3}x {\color{red}-1}y + {\color{red}4}z = 0 \\ \hline \end{array}\) 

2. \(\mathbf{\vec{P}_{Proj.}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{P}_{Proj.}} &=& \mathbf{\vec{P}+\lambda\vec{n}} \\\\ \vec{P}_{Proj.} &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \\\\ x_{Proj.} &=& 1+3\lambda \\ y_{Proj.} &=& 2-\lambda \\ z_{Proj.} &=& 3+4\lambda \\ \hline \end{array} \)

3. \(\mathbf{\lambda}\)

\(\begin{array}{|rcll|} \hline 3x_{Proj.} - y_{Proj.} + 4z_{Proj.} &=& 0 \\ 3(1+3\lambda) - (2-\lambda) + 4(3+4\lambda) &=& 0 \\ 3+9\lambda-2+\lambda+12+16\lambda &=& 0 \\ 26\lambda &=& -13 \\\\ \lambda &=& -\dfrac{13}{26} \\\\ \mathbf{\lambda} &=& -\mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

4.
\(\mathbf{\vec{P}_{Proj.}}\)

\(\begin{array}{|rcll|} \hline \vec{P}_{Proj.} &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \quad | \quad \lambda = -\dfrac{1}{2} \\\\ &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} -\dfrac{1}{2}\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \\\\ &=& \begin{pmatrix} 1-\dfrac{3}{2} \\\\ 2+\dfrac{1}{2} \\\\ 3-\dfrac{4}{2} \end{pmatrix} \\\\ \mathbf{\vec{P}_{Proj.}}&=& \begin{pmatrix} \mathbf{-\dfrac{1}{2}} \\\\ \mathbf{\dfrac{5}{2}} \\\\ \mathbf{1} \end{pmatrix} \\ \hline \end{array}\)