Orthogonal projection onto a subspace

Orthonormal basis of $F$ by Gram–Schmidt. Take $a_1=(1,1,0)$ and $a_2=(0,1,1)$.
- $u_1=\dfrac{a_1}{\|a_1\|}=\dfrac{(1,1,0)}{\sqrt2}$.
- $\langle a_2,u_1\rangle=\dfrac{1}{\sqrt2}$, so
$$a_2-\langle a_2,u_1\rangle u_1=(0,1,1)-\frac12(1,1,0)=\left(-\frac12,\frac12,1\right),$$
whose norm is $\sqrt{\tfrac14+\tfrac14+1}=\sqrt{\tfrac32}$. Hence
$$u_2=\frac{\left(-1/2,\,1/2,\,1\right)}{\sqrt{3/2}}=\frac{(-1,1,2)}{\sqrt6}.$$

Projections.
$$\langle v,u_1\rangle=\frac{1+2+0}{\sqrt2}=\frac{3}{\sqrt2},\qquad \langle v,u_2\rangle=\frac{-1+2+6}{\sqrt6}=\frac{7}{\sqrt6}.$$

Projection.
$$p_F(v)=\frac{3}{\sqrt2}\,u_1+\frac{7}{\sqrt6}\,u_2 =\frac32(1,1,0)+\frac76(-1,1,2).$$
Componentwise:
$$p_F(v)=\left(\frac32-\frac76,\ \frac32+\frac76,\ \frac{14}{6}\right) =\left(\frac13,\ \frac83,\ \frac73\right).$$