Vectors 1 Question 31
32. Suppose that $\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ are three non-coplanar vectors in $R^{3}$. Let the components of a vector $\mathbf{s}$ along $\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ be 4,3 and 5 , respectively. If the components of this vector $\mathbf{s}$ along $(-\mathbf{p}+\mathbf{q}+\mathbf{r})$, $(\mathbf{p}-\mathbf{q}+\mathbf{r})$ and $(-\mathbf{p}-\mathbf{q}+\mathbf{r})$ are $x, y$ and $z$ respectively, then the value of $2 x+y+z$ is
(2015 Adv.)
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Solution:
- Here, $\mathbf{s}=4 \mathbf{p}+3 \mathbf{q}+5 \mathbf{r}$
and $\mathbf{s}=(-\mathbf{p}+\mathbf{q}+\mathbf{r}) x+(\mathbf{p}-\mathbf{q}+\mathbf{r}) y+(-\mathbf{p}-\mathbf{q}+\mathbf{r}) z$
$\therefore 4 \mathbf{p}+3 \mathbf{q}+5 \mathbf{r}=\mathbf{p}(-x+y-z)+\mathbf{q}(x-y-z)+\mathbf{r}(x+y+z)$
On comparing both sides, we get
$-x+y-z=4, x-y-z=3$ and $x+y+z=5$
On solving above equations, we get
$$ \begin{aligned} x & =4, y=\frac{9}{2}, z=\frac{-7}{2} \\ \therefore \quad 2 x+y+z & =8+\frac{9}{2}-\frac{7}{2}=9 \end{aligned} $$
