Let a, b, c be unit vectors such that a + b + c = 0. Which one of the following is correct?

a × b, b × c, c × a are mutually perpendicular

Let a, b, c be unit vectors such that a + b + c = 0. Which… — JEE Main 2007 Mathematics PYQ

Let

\vec{a}

\vec{b}

\vec{c}

be unit vectors such that

\vec{a} + \vec{b} + \vec{c} = \vec{0}

. Which one of the following is correct?

A.
$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} = \vec{0}$
B.
$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq \vec{0}$
C.
$\vec{b} \times \vec{b} = \vec{b} \times \vec{c} = \vec{a} \times \vec{c} = \vec{0}$
D.
$\vec{a} \times \vec{b}$ , $\vec{b} \times \vec{c}$ , $\vec{c} \times \vec{a}$ are mutually perpendicular

Show answer and solution

Correct option: B

Solution

From a + b + c = 0, dot each vector with itself and others: |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0, giving a·b + b·c + c·a = -3/2. Since all are unit vectors, each dot product equals -1/2, meaning the angle between any two vectors is 120°.

Now, a × b = |a||b|sin(120°) n̂ = (√3/2)n̂ ≠ 0, so option 1 is out. From a + b + c = 0, cross both sides with a: a×b + a×c = 0, so a×b = c×a. Cross with b: a×b + b×c = 0, so a×b = b×c... wait, b×c = a×b. Thus a×b = b×c = c×a, and since the angle between any two unit vectors is 120° (not 0° or 180°), these cross products are all nonzero. This confirms option 2.

[AIEEE 2007]

JEE Main 2007 Mathematics question on Vector Algebra