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JEE Main 2000 Mathematics question on Vector Algebra

If the vectors a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} from the sides BCBC, CACA and ABAB respectively of a △ABC\triangle ABC, then
  1. A.
    a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=0\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0
  2. B.
    a⃗×b⃗=b⃗×c⃗=c⃗×a⃗\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}
  3. C.
    a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a}
  4. D.
    a⃗×b⃗+b⃗×c⃗+c⃗×a⃗=0⃗\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0}
Show answer and solution

Correct option: B

Solution

Since a, b, c are the sides BC, CA, AB respectively, they form a closed triangle, so a + b + c = 0.

Take the cross product of both sides with a: a × (a + b + c) = 0, giving a × b + a × c = 0, so a × b = c × a. Similarly crossing with b gives a × b = b × c. Thus a × b = b × c = c × a.

This works because each cross product magnitude equals twice the area of the triangle, and their directions are all the same (perpendicular to the plane of the triangle), making all three equal as vectors.

[2000]

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