Calculate Sum of Vector Components in Non Coplanar Basis

Given that $\vec{p}$, $\vec{q}$, and $\vec{r}$ are three non-coplanar vectors in $\mathbb{R}^3$, they form a basis for $\mathbb{R}^3$.

The components of vector $\vec{s}$ along $\vec{p}$, $\vec{q}$, and $\vec{r}$ are $4$, $3$, and $5$, respectively. Therefore, $\vec{s}$ can be uniquely expressed as: $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$

Similarly, given that $x$, $y$, and $z$ are the components of $\vec{s}$ along $(-\vec{p} + \vec{q} + \vec{r})$, $(\vec{p} - \vec{q} + \vec{r})$, and $(-\vec{p} - \vec{q} + \vec{r})$ respectively, we can express $\vec{s}$ as: $\vec{s} = x(-\vec{p} + \vec{q} + \vec{r}) + y(\vec{p} - \vec{q} + \vec{r}) + z(-\vec{p} - \vec{q} + \vec{r})$

Rearranging the terms by grouping $\vec{p}$, $\vec{q}$, and $\vec{r}$: $\vec{s} = (-x + y - z)\vec{p} + (x - y - z)\vec{q} + (x + y + z)\vec{r}$

Since $\vec{p}$, $\vec{q}$, and $\vec{r}$ are non-coplanar, the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$ in both representations must be equal. Equating the respective coefficients, we get the following system of linear equations:

1. $-x + y - z = 4$
2. $x - y - z = 3$
3. $x + y + z = 5$

To solve for $x$, $y$, and $z$:

• Add Equation (2) and Equation (3): $(x - y - z) + (x + y + z) = 3 + 5$ $2x = 8 \implies x = 4$

• Add Equation (1) and Equation (3): $(-x + y - z) + (x + y + z) = 4 + 5$ $2y = 9 \implies y = \frac{9}{2}$

• Substitute $x = 4$ and $y = \frac{9}{2}$ into Equation (3): $4 + \frac{9}{2} + z = 5$ $z = 5 - 4 - \frac{9}{2} = 1 - \frac{9}{2} = -\frac{7}{2}$

Now, we calculate the value of $2x + y + z$: $2x + y + z = 2(4) + \frac{9}{2} + \left(-\frac{7}{2}\right)$ $2x + y + z = 8 + \frac{9 - 7}{2} = 8 + 1 = 9$

Thus, the required value of $2x + y + z$ is $9$.