Area Of Triangle

Question

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.

RankGuru · Accepted Answer

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.

Area Of Triangle

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