Let \(\vec{a}\) be a non-zero vector, and let \(\vec{b}\) and \(\vec{c}\) be any vectors. Which of the following statements are correct?
(A) If \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\), then \(\vec{b} = \vec{c}\)
(B) If \(\vec{a}\times\vec{b} = \vec{a}\times\vec{c}\), then \(\vec{b} = \vec{c}\)
(C) If \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\) and \(\vec{a}\times\vec{b} = \vec{a}\times\vec{c}\), then \(\vec{b} = \vec{c}\)
(D) If \(\vec{a}\cdot\vec{b} = 0\) and \(\vec{a}\times\vec{b} = \vec{0}\), then \(\vec{b} = \vec{0}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) and (D) only
For the non-zero vector \(\vec{a}\) and any vector \(\vec{v}\):
\(\vec{a}\cdot\vec{v} = 0\) means \(\vec{v}\) is perpendicular to \(\vec{a}\) (or \(\vec{v} = \vec{0}\)).
\(\vec{a}\times\vec{v} = \vec{0}\) means \(\vec{v}\) is parallel to \(\vec{a}\) (or \(\vec{v} = \vec{0}\)).
A non-zero vector cannot be both perpendicular and parallel to \(\vec{a}\).
So each condition alone allows \(\vec{v} \ne \vec{0}\), but both together force \(\vec{v} = \vec{0}\); here \(\vec{v}\) is \(\vec{b}-\vec{c}\) or \(\vec{b}\).
Checking the statements
(A) \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\) gives \(\vec{b} = \vec{c}\) — Incorrect. \(\vec{a} = \hat{\imath}\), \(\vec{b} = \hat{\jmath}\), \(\vec{c} = \hat{k}\): both products are 0 but \(\vec{b} \ne \vec{c}\).
(B) \(\vec{a}\times\vec{b} = \vec{a}\times\vec{c}\) gives \(\vec{b} = \vec{c}\) — Incorrect. \(\vec{a} = \hat{\imath}\), \(\vec{b} = \hat{\jmath}\), \(\vec{c} = \hat{\imath}+\hat{\jmath}\): both products are \(\hat{k}\) but \(\vec{b} \ne \vec{c}\).
(C) Both equalities give \(\vec{b} = \vec{c}\) — Correct. \(\vec{b}-\vec{c}\) is both perpendicular and parallel to \(\vec{a}\), so \(\vec{b}-\vec{c} = \vec{0}\).
(D) \(\vec{a}\cdot\vec{b} = 0\) and \(\vec{a}\times\vec{b} = \vec{0}\) give \(\vec{b} = \vec{0}\) — Correct. \(\vec{b}\) is both perpendicular and parallel to \(\vec{a}\), so \(\vec{b} = \vec{0}\).