Appeared in: 2026 (2 sets)
For any two vectors $\vec{a}$ and $\vec{b}$, which of the following statements is always true ?
(A) $\vec{a} \cdot \vec{b} \le |\vec{a}|\,|\vec{b}|$
(B) $|\vec{a} + \vec{b}| \ge |\vec{a}| + |\vec{b}|$
(C) $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$
(D) $|\vec{a} \times \vec{b}| \ge |\vec{a}|\,|\vec{b}|$
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Official Evaluation Steps
Step 1 (Formula / Governing Law): (1/2 Mark)
The Cauchy-Schwarz inequality for vectors states that for any two vectors $\vec{a}$ and $\vec{b}$ in Euclidean space, the absolute value of their dot product is less than or equal to the product of their magnitudes:
$$|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|$$Since $\vec{a} \cdot \vec{b} \le |\vec{a} \cdot \vec{b}|$, it follows that $\vec{a} \cdot \vec{b} \le |\vec{a}| |\vec{b}|$.
Step 2 (Step-by-Step Working): (1/4 Mark)
- Option (A): $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Since $-1 \le \cos \theta \le 1$, then $\vec{a} \cdot \vec{b} \le |\vec{a}| |\vec{b}|$. This is always true.
- Option (B): By the Triangle Inequality, $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. Thus, the statement $|\vec{a} + \vec{b}| \ge |\vec{a}| + |\vec{b}|$ is generally false.
- Option (C): This is only true if the vectors are collinear and in the same direction, not for any two vectors.
- Option (D): Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$ and $|\sin \theta| \le 1$, it follows that $|\vec{a} \times \vec{b}| \le |\vec{a}| |\vec{b}|$. Thus, the statement is false.
Step 3 (Final Answer): (1/4 Mark)
The correct statement is (A).
Final Answer: (A) $\vec{a} \cdot \vec{b} \le |\vec{a}|\,|\vec{b}|$