# Find the orthogonal projection of the vector $(2, 3, 4)^t \in \mathbb{R^3}$ onto the $xy$ plane

Find the orthogonal projection of the vector $(2, 3, 4)^t \in \mathbb{R^3}$ onto the $xy$ plane where the symmetric bilinear product is given by the matrix

\begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 3 \end{pmatrix}

For finding orthogonal projection of a point on a plane we first find orthonormal basis of that plane, here one matrix is given, how do I use ? Plase help. Thanks.

• Are you having trouble constructing an orthonormal basis? – wilkersmon Feb 26 '18 at 11:26
• Take one simple vector in $xy$, $(1,0,0)$ and find an orthogonal vector like this: $(1,0,0) \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 3 \end{pmatrix} \begin{pmatrix} x\\y\\0 \end{pmatrix} = 0$. An easy computation shows you that $(1,0,0)$ and $(1,-1,0)$ is such a pair. Now normalize them and you got yourself an orthonormal basis. – wilkersmon Feb 26 '18 at 11:30
• @wilkersmon, how do you get $(1, -1, 0)$ – 1256 Feb 26 '18 at 13:09

We don't need an orthonormal basis of the $(x,y)$-plane, but we need a vector $n=(n_1,n_2,n_3)$ which is orthogonal to this plane. To this end we compute $$n\cdot e_1=\sum_{i=1}^3 n_i g_{i1}=n_1+n_2,\quad n\cdot e_2=\sum_{i=1}^3 n_i g_{i2}=n_1+2n_2+n_3\ .$$ Since we want $n\cdot e_1=n\cdot e_2=0$ it becomes clear that, e.g., $n=(1,-1,1)$ does the job. We now have to determine $t\in{\mathbb R}$ such that $$(2,3,4)+t(1,-1,1)=(p,q,0)\ .$$ From $4+t=0$ we obtain $t=-4$, so that the projected vector becomes $(-2,7,0)$.