Suppose that $\vec{p}, \vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $R^{3}$. Let the com

Question

Suppose that $\vec{p}, \vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $R^{3}$. Let the components of a vector $\vec{s}$ along $\vec{p}, \vec{q}$ and $\vec{r}$ be 4,3 and 5 respectively. If the components of this vector $\vec{s}$ along $(-\vec{p}+\vec{q}+\vec{r}),(\vec{p}-\vec{q}+\vec{r})$ and $(-\vec{p}-\vec{q}+\vec{r})$ are $x, y$ and $z$ respectively, then the value of $2 x+y+z$ is $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(9) : $\vec{s}=4 \vec{p}+3 \vec{q}+5 \vec{r}$ Let $-\vec{p}+\vec{q}+\vec{r}=\vec{\alpha}, \vec{p}-\vec{q}+\vec{r}=\vec{\beta}$ and\ $-\vec{p}-\vec{q}+\vec{r}=\vec{\gamma}$\ Then solving for $\vec{p}, \vec{q}$ and $\vec{r}$, we have\ $\vec{p}=\frac{\vec{\beta}-\vec{\gamma}}{2}, \vec{q}=\frac{\vec{\alpha}-\vec{\gamma}}{2}, \vec{r}=\frac{\vec{\alpha}+\vec{\beta}}{2}$\ Rewrite,\ $\vec{s}=4\left(\frac{\vec{\beta}-\vec{\gamma}}{2}ight)+3\left(\frac{\vec{\alpha}-\vec{\gamma}}{2}ight)+5\left(\frac{\vec{\alpha}+\vec{\beta}}{2}ight)$\ $=4 \vec{\alpha}+\frac{9}{2} \vec{\beta}-\frac{7}{2} \vec{\gamma} \quad 	herefore \quad x=4, y=\frac{9}{2}, z=\frac{-7}{2}$\ Thus, $2 x+y+z=9$\