Area Of Triangle
The position vectors of three points are A(1,1,1), B(2,3,5) and C(3,5,1); find the area of triangle ABC formed by these points.
Select the correct option:
Solution
45
The area of a triangle with two edge vectors emanating from a common vertex equals half the magnitude of their cross product, since the cross product gives the parallelogram area. This is the standard JEE Advanced bridge between vector products and planar areas. Take AB = B - A = (1, 2, 4) and AC = C - A = (2, 4, 0). Their cross product is AB x AC = i[(2)(0) - (4)(4)] - j[(1)(0) - (4)(2)] + k[(1)(4) - (2)(2)] = i(0 - 16) - j(0 - 8) + k(4 - 4) = -16i + 8j + 0k. The magnitude is sqrt((-16)^2 + 8^2 + 0^2) = sqrt(256 + 64) = sqrt(320). The area is half of this: (1/2)sqrt(320) = sqrt(320/4) = sqrt(80) = 4 sqrt(5). Option 2 sqrt(61) omits the one-half factor and uses a wrong cross product; option sqrt(122) doubles inside the root; option sqrt(30) uses wrong edge vectors. Plausibility check: the cross product's k-component being zero shows AB and AC have equal third-coordinate spread, a geometric consistency that does not invalidate the nonzero area; numerically 4 sqrt(5) is about 8.94.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- mathematics
- Chapter
- vector algebra
- Topic
- area of triangle
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
45
The area of a triangle with two edge vectors emanating from a common vertex equals half the magnitude of their cross product, since the cross product gives the parallelogram area. This is the standard JEE Advanced bridge between vector products and planar areas. Take AB = B - A = (1, 2, 4) and AC = C - A = (2, 4, 0). Their cross product is AB x AC = i[(2)(0) - (4)(4)] - j[(1)(0) - (4)(2)] + k[(1)(4) - (2)(2)] = i(0 - 16) - j(0 - 8) + k(4 - 4) = -16i + 8j + 0k. The magnitude is sqrt((-16)^2 + 8^2 + 0^2) = sqrt(256 + 64) = sqrt(320). The area is half of this: (1/2)sqrt(320) = sqrt(320/4) = sqrt(80) = 4 sqrt(5). Option 2 sqrt(61) omits the one-half factor and uses a wrong cross product; option sqrt(122) doubles inside the root; option sqrt(30) uses wrong edge vectors. Plausibility check: the cross product's k-component being zero shows AB and AC have equal third-coordinate spread, a geometric consistency that does not invalidate the nonzero area; numerically 4 sqrt(5) is about 8.94.
This medium difficulty mathematics question is from the chapter vector algebra, covering the topic of area of triangle. It appeared in the 2025 exam.
Looking for more practice? Explore all mathematics questions or browse vector algebra questions on RankGuru.