Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be non-zero vectors such that \(\vec{a}+\vec{b}+\vec{c} = \vec{0}\), where \(\vec{a}\) and \(\vec{b}\) are not parallel. Which of the following statements are correct?
(A) \(\vec{a}\times\vec{b} = \vec{b}\times\vec{c}\)
(B) \(\vec{a}\times\vec{b} = \vec{c}\times\vec{a}\)
(C) \(\vec{a}\times\vec{b}+\vec{b}\times\vec{c}+\vec{c}\times\vec{a} = \vec{0}\)
(D) \((\vec{a}+\vec{b})\cdot\vec{c} = |\vec{c}|^2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
From \(\vec{a}+\vec{b}+\vec{c} = \vec{0}\), we have \(\vec{c} = -\vec{a}-\vec{b}\). Using \(\vec{v}\times\vec{v} = \vec{0}\) and \(\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}\):
\(\vec{b}\times\vec{c} = \vec{b}\times(-\vec{a}-\vec{b}) = -\vec{b}\times\vec{a} = \vec{a}\times\vec{b}\)
\(\vec{c}\times\vec{a} = (-\vec{a}-\vec{b})\times\vec{a} = -\vec{b}\times\vec{a} = \vec{a}\times\vec{b}\)
Since \(\vec{a}\) and \(\vec{b}\) are not parallel, \(\vec{a}\times\vec{b} \ne \vec{0}\). Also \(\vec{a}+\vec{b} = -\vec{c}\).
So \(\vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a} \ne \vec{0}\), and \((\vec{a}+\vec{b})\cdot\vec{c} = -|\vec{c}|^2\).
Checking the statements
(A) \(\vec{a}\times\vec{b} = \vec{b}\times\vec{c}\) — Correct. \(\vec{b}\times(-\vec{a}-\vec{b}) = -\vec{b}\times\vec{a} = \vec{a}\times\vec{b}\).
(B) \(\vec{a}\times\vec{b} = \vec{c}\times\vec{a}\) — Correct. \((-\vec{a}-\vec{b})\times\vec{a} = -\vec{b}\times\vec{a} = \vec{a}\times\vec{b}\).
(C) The sum of the three cross products is \(\vec{0}\) — Incorrect. The sum is \(3(\vec{a}\times\vec{b})\), which is not \(\vec{0}\) for non-parallel \(\vec{a}\), \(\vec{b}\).
(D) \((\vec{a}+\vec{b})\cdot\vec{c} = |\vec{c}|^2\) — Incorrect. \((\vec{a}+\vec{b})\cdot\vec{c} = (-\vec{c})\cdot\vec{c} = -|\vec{c}|^2\).