The position vectors of four points P, Q, R, | vedclass

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(A) Given position vectors are:$\vec{p} = 2a + 4c$$\vec{q} = 5a + 3\sqrt{3}b + 4c$$\vec{r} = -2\sqrt{3}b + c$$\vec{s} = 2a + c$Calculate $\overrightar

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$\overrightarrow{PQ}$ is parallel to $\overrightarrow{RS}$

Vedclass · Answer

(A) Given position vectors are:$\vec{p} = 2a + 4c$$\vec{q} = 5a + 3\sqrt{3}b + 4c$$\vec{r} = -2\sqrt{3}b + c$$\vec{s} = 2a + c$Calculate $\overrightarrow{PQ} = \vec{q} - \vec{p} = (5a + 3\sqrt{3}b + 4c) - (2a + 4c) = 3a + 3\sqrt{3}b = 3(a + \sqrt{3}b)$.Calculate $\overrightarrow{RS} = \vec{s} - \vec{r} = (2a + c) - (-2\sqrt{3}b + c) = 2a + 2\sqrt{3}b = 2(a + \sqrt{3}b)$.Since $\overrightarrow{PQ} = \frac{3}{2} \overrightarrow{RS}$,the vectors are scalar multiples of each other.Therefore,$\overrightarrow{PQ}$ is parallel to $\overrightarrow{RS}$.

The position vectors of four points $P, Q, R, S$ are $2a + 4c$,$5a + 3\sqrt{3}b + 4c$,$-2\sqrt{3}b + c$,and $2a + c$ respectively. Then:

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