Let bar{a}, bar{b} and bar{c} be non-coplanar | vedclass

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(C) Given position vectors of points $P, Q, R, S$ are:$\vec{p} = -\bar{a} + 4\bar{b} - 3\bar{c}$$\vec{q} = 3\bar{a} + 2\bar{b} - 5\bar{c}$$\vec{r} = -

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$(-1, -1)$

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(C) Given position vectors of points $P, Q, R, S$ are:$\vec{p} = -\bar{a} + 4\bar{b} - 3\bar{c}$$\vec{q} = 3\bar{a} + 2\bar{b} - 5\bar{c}$$\vec{r} = -3\bar{a} + 8\bar{b} - 5\bar{c}$$\vec{s} = -3\bar{a} + 2\bar{b} + \bar{c}$Calculate the displacement vectors:$\overline{PQ} = \vec{q} - \vec{p} = (3\bar{a} + 2\bar{b} - 5\bar{c}) - (-\bar{a} + 4\bar{b} - 3\bar{c}) = 4\bar{a} - 2\bar{b} - 2\bar{c}$$\overline{PR} = \vec{r} - \vec{p} = (-3\bar{a} + 8\bar{b} - 5\bar{c}) - (-\bar{a} + 4\bar{b} - 3\bar{c}) = -2\bar{a} + 4\bar{b} - 2\bar{c}$$\overline{PS} = \vec{s} - \vec{p} = (-3\bar{a} + 2\bar{b} + \bar{c}) - (-\bar{a} + 4\bar{b} - 3\bar{c}) = -2\bar{a} - 2\bar{b} + 4\bar{c}$We need to find $x, y$ such that $\overline{PQ} = x\overline{PR} + y\overline{PS}$:$4\bar{a} - 2\bar{b} - 2\bar{c} = x(-2\bar{a} + 4\bar{b} - 2\bar{c}) + y(-2\bar{a} - 2\bar{b} + 4\bar{c})$$4\bar{a} - 2\bar{b} - 2\bar{c} = (-2x - 2y)\bar{a} + (4x - 2y)\bar{b} + (-2x + 4y)\bar{c}$Comparing coefficients of $\bar{a}, \bar{b}, \bar{c}$:$1) -2x - 2y = 4 \implies x + y = -2$$2) 4x - 2y = -2 \implies 2x - y = -1$Adding $(1)$ and $(2)$:$3x = -3 \implies x = -1$Substituting $x = -1$ into $(1)$:$-1 + y = -2 \implies y = -1$Thus,the ordered pair is $(x, y) = (-1, -1)$.

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