Let \(\vec{a}\) and \(\vec{b}\) be non-zero vectors. Which of the following statements are correct?
(A) \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\) if and only if \(\vec{a}\) is perpendicular to \(\vec{b}\)
(B) \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = 0\) if and only if \(|\vec{a}| = |\vec{b}|\)
(C) If \(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\), then \(\vec{a}\cdot\vec{b} = 0\)
(D) If \(\vec{a}\times\vec{b} = \vec{0}\), then \(\vec{a}\cdot\vec{b} = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Expanding by the distributive law:
\(|\vec{a}+\vec{b}|^2-|\vec{a}-\vec{b}|^2 = 4\,\vec{a}\cdot\vec{b}\)
\((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2-|\vec{b}|^2\)
\(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta, \quad \vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\)
So \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\) exactly when \(\vec{a}\cdot\vec{b} = 0\), the product in (B) is \(|\vec{a}|^2-|\vec{b}|^2\), \(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\) means \(\sin\theta = 1\), and \(\vec{a}\times\vec{b} = \vec{0}\) means \(\sin\theta = 0\).
Checking the statements
(A) Equal magnitudes exactly when perpendicular — Correct. The difference of the squares is \(4\,\vec{a}\cdot\vec{b}\), which is 0 exactly when \(\vec{a}\) is perpendicular to \(\vec{b}\).
(B) Product 0 exactly when \(|\vec{a}| = |\vec{b}|\) — Correct. The product is \(|\vec{a}|^2-|\vec{b}|^2\).
(C) \(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\) gives \(\vec{a}\cdot\vec{b} = 0\) — Correct. \(\sin\theta = 1\) gives \(\theta = \displaystyle\frac{\pi}{2}\), so \(\cos\theta = 0\).
(D) \(\vec{a}\times\vec{b} = \vec{0}\) gives \(\vec{a}\cdot\vec{b} = 0\) — Incorrect. \(\sin\theta = 0\) gives \(\theta = 0\) or \(\pi\), so \(\vec{a}\cdot\vec{b} = \pm|\vec{a}|\,|\vec{b}| \ne 0\).