Projection of a vector on a plane in $\Bbb{R}^4$

Question

I have to find the projection of $v=(1, −1, 1, 1)$ on $S=\langle(0, 1, 1, 1) , (1, 1, 1, 0)\rangle$. This is what i'd do in 3 dimensions:

• Find a base of $S$
• Find a vector perpendicular to the plane generated by $S$ using cross product
• Project $v$ on this last vector to get $w$
• The projection of $v$ on $S$ is: $v - w$

The problem is that cross product, at least the one i know, only works in 3 dimensions, so i can't find a normal vector for the plane.

Answer 1

You can apply Gram-Schmidt to obtain an orthonormal basis of $S$. What I got was $\{e_1,e_2\}$, with$$e_1=\frac1{\sqrt3}(0,1,1,1)\text{ and }e_2=\frac1{\sqrt{15}}(3,1,1,-2).$$Then, the projection of $v$ is$$\langle v,e_1\rangle e_1+\langle v,e_2\rangle e_2=\frac15(1,2,2,1).$$

Answer 2

For general $n$-dimensional Euclidean spaces the inner product of two $n\times1$ vectors $u=(u_1~u_2~\cdots~u_n)^t,v=(v_1~v_2~\cdots~v_n)^t$, is defined as $$<u,v>=u^tv=\sum_{i=1}^nu_iv_i$$