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Question 154 of 162

Q.Prove that [a⃗−b⃗, b⃗−c⃗, c⃗−a⃗]=0\left[\vec{a}-\vec{b},\ \vec{b}-\vec{c},\ \vec{c}-\vec{a}\right]=0.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2024Subjective· 3mImportance★★★★★
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Notices the three difference-vectors sum to zero, so the third is minus the sum of the first two — making the scalar triple product automatically zero.

  1. Let u⃗=a⃗−b⃗\vec u=\vec a-\vec b, v⃗=b⃗−c⃗\vec v=\vec b-\vec c, w⃗=c⃗−a⃗\vec w=\vec c-\vec a.
  2. Adding: u⃗+v⃗+w⃗=(a⃗−b⃗)+(b⃗−c⃗)+(c⃗−a⃗)=0⃗\vec u+\vec v+\vec w=(\vec a-\vec b)+(\vec b-\vec c)+(\vec c-\vec a)=\vec0, so w⃗=−(u⃗+v⃗)\vec w=-(\vec u+\vec v).
  3. Scalar triple product: [u⃗,v⃗,w⃗]=[u⃗,v⃗,−(u⃗+v⃗)]=−[u⃗,v⃗,u⃗]−[u⃗,v⃗,v⃗][\vec u,\vec v,\vec w]=[\vec u,\vec v,-(\vec u+\vec v)]=-[\vec u,\vec v,\vec u]-[\vec u,\vec v,\vec v]. …
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