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Exercise 8.4 · Q2

Q.Show that a⃗×(b⃗+c⃗)+b⃗×(c⃗+a⃗)+c⃗×(a⃗+b⃗)=0⃗\vec a\times(\vec b+\vec c)+\vec b\times(\vec c+\vec a)+\vec c\times(\vec a+\vec b)=\vec 0.

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✓ Free question

Step 1. Expand each term using distributivity of the cross product: a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗,\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c, b⃗×(c⃗+a⃗)=b⃗×c⃗+b⃗×a⃗,\vec b\times(\vec c+\vec a)=\vec b\times\vec c+\vec b\times\vec a, c⃗×(a⃗+b⃗)=c⃗×a⃗+c⃗×b⃗.\vec c\times(\vec a+\vec b)=\vec c\times\vec a+\vec c\times\vec b.

Step 2. Add all six terms: (a⃗×b⃗+b⃗×a⃗)+(a⃗×c⃗+c⃗×a⃗)+(b⃗×c⃗+c⃗×b⃗).(\vec a\times\vec b+\vec b\times\vec a)+(\vec a\times\vec c+\vec c\times\vec a)+(\vec b\times\vec c+\vec c\times\vec b).

Step 3. Since x⃗×y⃗=−(y⃗×x⃗)\vec x\times\vec y=-(\vec y\times\vec x) for any vectors, each bracket is 0⃗\vec 0: a⃗×b⃗+b⃗×a⃗=0⃗\vec a\times\vec b+\vec b\times\vec a=\vec0, and likewise for the other two brackets.

Step 4. The whole sum is therefore 0⃗+0⃗+0⃗=0⃗\vec 0+\vec0+\vec0=\vec0.

✓Final answer

The expression equals 0⃗\vec 0 for any vectors a⃗,b⃗,c⃗\vec a,\vec b,\vec c.

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