Vectors and the Dot Product

실제 시험에서는 이렇게 출제됩니다

이 개념을 다룬 실제 CSCA 기출문제입니다. 답을 펼치기 전에 먼저 직접 풀어 보세요.

1. CSCA Math December 2025 Past Paper
   Let $O$ be the origin, and let $A$ and $B$ be two points in the plane. If $P$ is the midpoint of $AB$, which expression can represent $\overrightarrow{OP}$?
   • A．$\overrightarrow{OB}-\overrightarrow{OA}$
   • B．$\overrightarrow{OA}+\overrightarrow{OB}$
   • C．$\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})$
   • D．$\overrightarrow{OA}-\overrightarrow{OB}$
   정답 및 풀이 보기

   정답: C

   Answer: C Since $P$ is the midpoint of $AB$, $\overrightarrow{OP}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})$. Choose C.
   CSCA Math December 2025 Past Paper 전체에서 보기
2. CSCA Math June 2026 Past Paper
   Given that plane vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy $|\vec{a}| = 1$, $|\vec{b}| = 1$, $\vec{a} \cdot \vec{b} = -\frac{1}{2}$, and $\vec{c} = \lambda \vec{a} + (1 - \lambda)\vec{b}$ ($\lambda \in \mathbb{R}$), then the minimum value of $|\vec{c}|$ is
   • A．$\frac{1}{2}$
   • B．$\frac{\sqrt{3}}{2}$
   • C．$1$
   • D．$\frac{\sqrt{2}}{2}$
   정답 및 풀이 보기

   정답: A

   $|\vec{c}|^2 = \lambda^2 |\vec{a}|^2 + (1 - \lambda)^2 |\vec{b}|^2 + 2\lambda(1 - \lambda)\vec{a} \cdot \vec{b} = \lambda^2 + (1 - \lambda)^2 - \lambda(1 - \lambda) = 3\lambda^2 - 3\lambda + 1$. This quadratic in $\lambda$ attains its minimum $\frac{1}{4}$ at $\lambda = \frac{1}{2}$, so the minimum of $|\vec{c}|$ is $\frac{1}{2}$. Choose (A).
   CSCA Math June 2026 Past Paper 전체 풀기 — 시간 측정 및 채점