If \( \vec{a}\times\vec{b}=\vec{a}\times\vec{c} \) and \( \vec{a}\ne\vec{0} \), then which of the following is/are TRUE?
A. \( \vec{b}=\vec{c} \)
B. \( \vec{b}+\vec{c}=\vec{0} \)
C. \( \vec{b}=\vec{c}+\lambda\vec{a} \) for some scalar \( \lambda \)
D. \( \vec{b}+\vec{c}=\lambda\vec{a} \) for some scalar \( \lambda \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A and C only
Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect, and the correct answer should be “C only”, which is not among the options, as explained below.
Since \( \vec a\times\vec b = \vec a\times\vec c \),
\( \vec a\times(\vec b-\vec c) = \vec 0 \)
A cross product is zero only when the two vectors are parallel (or one is zero). As \( \vec a \ne \vec 0 \), \( \vec b-\vec c \) is a multiple of \( \vec a \):
\( \vec b-\vec c = \lambda\vec a \) for some scalar λ
An example that tests each statement: \( \vec a=\hat\imath,\ \vec b=\hat\imath+\hat\jmath,\ \vec c=\hat\jmath \). Then \( \vec a\times\vec b = \vec a\times\vec c = \hat k \).
(A) \( \vec b=\vec c \) — Incorrect. It is only the special case λ = 0. In the example, \( \vec b \ne \vec c \).
(B) \( \vec b+\vec c=\vec 0 \) — Incorrect. In the example, \( \vec b+\vec c = \hat\imath+2\hat\jmath \ne \vec 0 \).
(C) \( \vec b=\vec c+\lambda\vec a \) — Correct. This is exactly the result shown above.
(D) \( \vec b+\vec c=\lambda\vec a \) — Incorrect. In the example, \( \hat\imath+2\hat\jmath \) is not a multiple of \( \hat\imath \).
So only (C) is always true.
Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.