Subjects vector algebra

Vector Midpoint 0B68A1

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1. **Stating the problem:** In the figure, the tip of vector $\mathbf{c}$ and the tail of vector $\mathbf{d}$ are both the midpoint of segment $QR$. We need to express $\mathbf{c}$ and $\mathbf{d}$ in terms of vectors $\mathbf{a}$ and $\mathbf{b}$. 2. **Understanding the setup:** Points $Q$, $P$, and $R$ are on a line with $P$ between $Q$ and $R$. Vectors $\mathbf{a}$ and $\mathbf{b}$ represent $\overrightarrow{QP}$ and $\overrightarrow{PR}$ respectively. 3. **Key fact:** The midpoint $M$ of $QR$ satisfies: $$\overrightarrow{QM} = \frac{1}{2} \overrightarrow{QR}$$ Since $\overrightarrow{QR} = \overrightarrow{QP} + \overrightarrow{PR} = \mathbf{a} + \mathbf{b}$, we have: $$\overrightarrow{QM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$ 4. **Expressing $\mathbf{c}$:** Vector $\mathbf{c}$ goes from $Q$ to $P$, so: $$\mathbf{c} = \overrightarrow{QP} = \mathbf{a}$$ But the problem states the tip of $\mathbf{c}$ is the midpoint $M$, so $\mathbf{c}$ must be: $$\mathbf{c} = \overrightarrow{QM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$ 5. **Expressing $\mathbf{d}$:** Vector $\mathbf{d}$ starts at $M$ and ends at $R$, so: $$\mathbf{d} = \overrightarrow{MR} = \overrightarrow{QR} - \overrightarrow{QM} = (\mathbf{a} + \mathbf{b}) - \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$ 6. **Final expressions:** $$\boxed{\mathbf{c} = \frac{1}{2}(\mathbf{a} + \mathbf{b}), \quad \mathbf{d} = \frac{1}{2}(\mathbf{a} + \mathbf{b})}$$ Both vectors $\mathbf{c}$ and $\mathbf{d}$ equal half the sum of $\mathbf{a}$ and $\mathbf{b}$ because they connect $Q$ to the midpoint and the midpoint to $R$ respectively.