A, B and C are the points (1, −1, 2), (3, 0, 4) and (1, −1, 6) respectively.
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|\overrightarrow{AC}|\) | (I) | 3 |
| (B) | \(\overrightarrow{AB}\cdot\overrightarrow{AC}\) | (II) | \(2\sqrt{5}\) |
| (C) | \(|\overrightarrow{BC}|\) | (III) | 8 |
| (D) | Area of triangle ABC (in square units) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
The side vectors are found as position vector of the end minus that of the start:
\(\overrightarrow{AB} = 2\hat{\imath}+\hat{\jmath}+2\hat{k}, \quad \overrightarrow{AC} = 4\hat{k}\)
\(\overrightarrow{BC} = -2\hat{\imath}-\hat{\jmath}+2\hat{k}\)
(A) \(|\overrightarrow{AC}|\)
This matches (IV) 4.
\(|4\hat{k}| = 4\)
(B) \(\overrightarrow{AB}\cdot\overrightarrow{AC}\)
This matches (III) 8.
\((2)(0)+(1)(0)+(2)(4) = 8\)
(C) \(|\overrightarrow{BC}|\)
This matches (I) 3.
\(\sqrt{4+1+4} = 3\)
(D) Area of triangle ABC
This matches (II) \(2\sqrt{5}\).
\( \displaystyle \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 0 & 4 \end{vmatrix} = 4\hat{\imath}-8\hat{\jmath} \)
\(|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{16+64} = 4\sqrt{5}\)
The area of the triangle is half of this:
\( \displaystyle \frac{1}{2}\times 4\sqrt{5} = 2\sqrt{5} \)