Question

CD is trisected at points P and Q. Find the position vectors of points of trisection, if the
position vectors of C and D are c
→
and d
→
respectively

Answer 1

Answer on Question #57582 – Math – Analytic Geometry

Question

CD is trisected at points P and Q. Find the position vectors of points of trisection, if the position vectors of C and D are $c \rightarrow$ and $d \rightarrow$ respectively.

Let O be the origin.

Given:

$$\overline{OC} = \vec{c}$$ $$\overline{OD} = \vec{d}$$ $$CP = PQ = QD$$

Find:

$$\overline{OP} = \vec{p}$$ $$\overline{OQ} = \vec{q}$$

Solution

$$\overline{CD} = \vec{d} - \vec{c}$$

On the one hand,

$$\overline{CP} = \vec{p} - \vec{c}$$ $$\overline{CQ} = \vec{q} - \vec{c}$$

On the other hand,

$$\overline{CP} = \frac{1}{3}(\vec{d} - \vec{c})$$ $$\overline{CQ} = \frac{2}{3}(\vec{d} - \vec{c})$$ $$\vec{p} = \vec{c} + \overline{CP} = \vec{c} + \frac{1}{3}(\vec{d} - \vec{c}) = \frac{2}{3}\vec{c} + \frac{1}{3}\vec{d}$$ $$\vec{q} = \vec{c} + \overline{CQ} = \vec{c} + \frac{2}{3}(\vec{d} - \vec{c}) = \frac{1}{3}\vec{c} + \frac{2}{3}\vec{d}$$

Answer: $\vec{p} = \frac{2}{3}\vec{c} + \frac{1}{3}\vec{d}$, $\vec{q} = \frac{1}{3}\vec{c} + \frac{2}{3}\vec{d}$.

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