ABCD is a parallelogram having vertices A, B, C, D with position vectors \( -\hat{\imath}+\displaystyle\frac{1}{2}\hat{\jmath}+4\hat{k} \), \( \hat{\imath}+\displaystyle\frac{1}{2}\hat{\jmath}+4\hat{k} \), \( \hat{\imath}-\displaystyle\frac{1}{2}\hat{\jmath}+4\hat{k} \), \( -\hat{\imath}-\displaystyle\frac{1}{2}\hat{\jmath}+4\hat{k} \) respectively, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \overrightarrow{AB} \) | (I) | \( 2\hat{\imath}-\hat{\jmath} \) |
| (B) | \( \overrightarrow{AD} \) | (II) | \( 2\hat{\imath} \) |
| (C) | \( \overrightarrow{AB}\times\overrightarrow{AD} \) | (III) | \( -\hat{\jmath} \) |
| (D) | \( \overrightarrow{AC} \) | (IV) | \( -2\hat{k} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
Each vector is (position vector of the head) − (position vector of the tail).
(A) \( \overrightarrow{AB} \)
This matches (II).
\( \displaystyle \overrightarrow{AB}=(1-(-1))\hat{\imath}+\left(\frac{1}{2}-\frac{1}{2}\right)\hat{\jmath}+(4-4)\hat{k} \)
\( \displaystyle =2\hat{\imath} \)
(B) \( \overrightarrow{AD} \)
This matches (III).
\( \displaystyle \overrightarrow{AD}=(-1-(-1))\hat{\imath}+\left(-\frac{1}{2}-\frac{1}{2}\right)\hat{\jmath}+(4-4)\hat{k} \)
\( \displaystyle =-\hat{\jmath} \)
(C) \( \overrightarrow{AB}\times\overrightarrow{AD} \)
This matches (IV). Using \( \hat{\imath}\times\hat{\jmath}=\hat{k} \):
\( \displaystyle \overrightarrow{AB}\times\overrightarrow{AD}=2\hat{\imath}\times(-\hat{\jmath}) \)
\( \displaystyle =-2\hat{k} \)
(D) \( \overrightarrow{AC} \)
This matches (I).
\( \displaystyle \overrightarrow{AC}=(1-(-1))\hat{\imath}+\left(-\frac{1}{2}-\frac{1}{2}\right)\hat{\jmath}+(4-4)\hat{k} \)
\( \displaystyle =2\hat{\imath}-\hat{\jmath} \)
So the matching is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).