Q.If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, show that $\vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}$. Interpret the result geometrically?

Cross the given relation with each vector and use $\vec{x}\times\vec{x}=\vec{0}$ and $\vec{x}\times\vec{y}=-\vec{y}\times\vec{x}$.

Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.

Cross with $\vec{a}$: $\vec{a}\times\vec{a}+\vec{a}\times\vec{b}+\vec{a}\times\vec{c}=\vec{0}\Rightarrow\vec{a}\times\vec{b}=-\vec{a}\times\vec{c}=\vec{c}\times\vec{a}$.

Cross with $\vec{b}$: $\vec{b}\times\vec{a}+\vec{b}\times\vec{b}+\vec{b}\times\vec{c}=\vec{0}\Rightarrow\vec{b}\times\vec{c}=-\vec{b}\times\vec{a}=\vec{a}\times\vec{b}$.

Hence $\vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}$. …

Tools we use

Two cross-product facts do all the work:

• $\vec{x}\times\vec{x}=\vec{0}$ (a vector crossed with itself is zero),
• $\vec{x}\times\vec{y}=-(\vec{y}\times\vec{x})$ (anti-commutativity).

Step 1: cross the relation with $\vec{a}$

Start from $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and take the cross product of both sides with $\vec{a}$ on the left:

Since $\vec{a}\times\vec{a}=\vec{0}$,

Step 2: cross the relation with $\vec{b}$

Step 3: combine

From Steps 1 and 2,

Geometric meaning …

Method: Cross-multiplying a vector identity to expose equal cross products

Use this reasoning pattern when a relation among vectors (like $\vec{a}+\vec{b}+\vec{c}=\vec{0}$) must be converted into statements about cross products.

Steps

Step 1: Recall the two cross-product facts that do the work.

The first kills self-terms; the second lets you flip an order at the cost of a sign.

Step 2: Cross the whole relation by one vector at a time.

Taking $\times\vec{a}$ of $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ removes $\vec{a}\times\vec{a}$ and yields one equality; taking $\times\vec{b}$ yields another. Use anti-commutativity to rewrite each term in a consistent order. …

Common Mistakes

Mistake 1: Ignoring anti-commutativity when reordering cross products.

Why it's wrong: $\vec{a}\times\vec{c}=-\,\vec{c}\times\vec{a}$, so the sign must flip when you rewrite a term; dropping it produces a false equality. Correct approach: apply $\vec{x}\times\vec{y}=-(\vec{y}\times\vec{x})$ every time you swap.

Mistake 2: Forgetting that $\vec{a}\times\vec{a}=\vec{0}$.

Why it's wrong: the whole method relies on the self-cross term vanishing when you cross the relation with $\vec{a}$ or $\vec{b}$. Correct approach: cancel $\vec{a}\times\vec{a}$ and $\vec{b}\times\vec{b}$ to zero. …