A and B are the points (1, −2, 3) and (4, 1, −3) respectively.
Match List-I with List-II
| List-I | List-II (position vector) | ||
| (A) | Mid-point of AB | (I) | \(3\hat{\imath}-\hat{k}\) |
| (B) | Point dividing AB internally in the ratio 1 : 2 | (II) | \(7\hat{\imath}+4\hat{\jmath}-9\hat{k}\) |
| (C) | Point dividing AB internally in the ratio 2 : 1 | (III) | \(2\hat{\imath}-\hat{\jmath}+\hat{k}\) |
| (D) | Point dividing AB externally in the ratio 2 : 1 | (IV) | \(\displaystyle\frac{5}{2}\hat{\imath}-\frac{1}{2}\hat{\jmath}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Let \(\vec{a} = \hat{\imath}-2\hat{\jmath}+3\hat{k}\) and \(\vec{b} = 4\hat{\imath}+\hat{\jmath}-3\hat{k}\). By the section formula, internal division in the ratio m : n gives \(\frac{m\vec{b}+n\vec{a}}{m+n}\) and external division gives \(\frac{m\vec{b}-n\vec{a}}{m-n}\).
(A) Mid-point of AB
This matches (IV) \(\frac{5}{2}\hat{\imath}-\frac{1}{2}\hat{\jmath}\).
\( \displaystyle \frac{\vec{a}+\vec{b}}{2} = \frac{5\hat{\imath}-\hat{\jmath}+0\hat{k}}{2} \)
(B) Point dividing AB internally in the ratio 1 : 2
This matches (III) \(2\hat{\imath}-\hat{\jmath}+\hat{k}\).
\( \displaystyle \frac{\vec{b}+2\vec{a}}{3} = \frac{6\hat{\imath}-3\hat{\jmath}+3\hat{k}}{3} = 2\hat{\imath}-\hat{\jmath}+\hat{k} \)
(C) Point dividing AB internally in the ratio 2 : 1
This matches (I) \(3\hat{\imath}-\hat{k}\).
\( \displaystyle \frac{2\vec{b}+\vec{a}}{3} = \frac{9\hat{\imath}+0\hat{\jmath}-3\hat{k}}{3} = 3\hat{\imath}-\hat{k} \)
(D) Point dividing AB externally in the ratio 2 : 1
This matches (II) \(7\hat{\imath}+4\hat{\jmath}-9\hat{k}\).
\( \displaystyle \frac{2\vec{b}-\vec{a}}{2-1} = (8-1)\hat{\imath}+(2+2)\hat{\jmath}+(-6-3)\hat{k} \)
\( = 7\hat{\imath}+4\hat{\jmath}-9\hat{k}\)