If \( \theta \) is angle between \( \vec{a} \) & \( \vec{b} \), then for vectors \( \vec{a},\vec{b} \) and \( \vec{c} \),
Match the LIST-I with LIST-II
| LIST-I | LIST-II |
| A. \( \vec{a}\times\vec{b}=\vec{a}\times\vec{c},\ \vec{a}\ne 0,\ \vec{b}\ne\vec{c} \) | I. \( \theta=\displaystyle\frac{\pi}{4} \) |
| B. \( \vec{a}\times\vec{b}=0 \) | II. \( \vec{a}\parallel(\vec{b}-\vec{c}) \) |
| C. \( |\vec{a}.\vec{b}|=|\vec{a}\times\vec{b}| \) | III. \( \vec{b}=\vec{c} \) |
| D. \( \vec{a}.\vec{b}=\vec{a}.\vec{c} \) and \( \vec{a}\times\vec{b}=\vec{a}\times\vec{c}, \) \( \vec{a}\ne 0 \) | IV. \( \vec{a}=0,\vec{b}=0 \) or \( \vec{a}\parallel\vec{b} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-IV, C-I, D-III
(A) \( \vec{a}\times\vec{b}=\vec{a}\times\vec{c},\ \vec{a}\ne 0,\ \vec{b}\ne\vec{c} \)
This matches II. \( \vec{a}\parallel(\vec{b}-\vec{c}) \). Here \( \vec{a}\times(\vec{b}-\vec{c})=\vec{0} \) with \( \vec{a} \) and \( \vec{b}-\vec{c} \) both non-zero, so they are parallel.
(B) \( \vec{a}\times\vec{b}=0 \)
This matches IV. \( \vec{a}=0,\vec{b}=0 \) or \( \vec{a}\parallel\vec{b} \). A cross product is zero when one of the vectors is zero or the two are parallel.
(C) \( |\vec{a}.\vec{b}|=|\vec{a}\times\vec{b}| \)
This matches I. \( \theta=\frac{\pi}{4} \). For non-zero vectors,
\( \displaystyle |\vec{a}||\vec{b}|\,|\cos\theta|=|\vec{a}||\vec{b}|\sin\theta \)
\( \displaystyle |\tan\theta|=1 \)
so \( \theta=\frac{\pi}{4} \) (taking the acute angle).
(D) \( \vec{a}.\vec{b}=\vec{a}.\vec{c} \) and \( \vec{a}\times\vec{b}=\vec{a}\times\vec{c}, \) \( \vec{a}\ne 0 \)
This matches III. \( \vec{b}=\vec{c} \). \( \vec{b}-\vec{c} \) is both perpendicular and parallel to the non-zero vector \( \vec{a} \), which is possible only if \( \vec{b}-\vec{c}=\vec{0} \).
So the matching is A-II, B-IV, C-I, D-III.