Given that A×(B×C) = (A.C)B - ( A.B)C , deduce that (A×B)×C = (A.C)B - (B.C)A

To deduce the expression $(A\times B)\times C=(A\cdot C)B-(B\cdot C)A$ from the given identity $A\times (B\times C)=(A\cdot C)B-(A\cdot B)C$, we can use properties of the vector cross product and the scalar dot product. Let’s break this down step by step.

Understanding the Cross Product

The cross product of two vectors results in a vector that is perpendicular to the plane formed by the two original vectors. It has several important properties, including:

• Anticommutativity: $A\times B=-(B\times A)$
• Distributivity: $A\times (B+C)=A\times B+A\times C$
• Scalar Multiplication: $k(A\times B)=(kA)\times B=A\times (kB)$ for any scalar $k$

Applying the Given Identity

We start with the identity:

1. $A\times (B\times C)=(A\cdot C)B-(A\cdot B)C$

Now, we want to manipulate this to find $(A\times B)\times C$. We can use the anticommutative property of the cross product.

Rearranging the Expression

Using the anticommutativity property, we can rewrite $(A\times B)\times C$ as:

2. $(A\times B)\times C=-(C\times (A\times B))$

Now, we can apply the vector triple product identity, which states that:

3. $X\times (Y\times Z)=(X\cdot Z)Y-(X\cdot Y)Z$

In our case, let $X=C$, $Y=A$, and $Z=B$. Applying this identity gives us:

4. $C\times (A\times B)=(C\cdot B)A-(C\cdot A)B$

Substituting Back

Now substituting this back into our expression from step 2, we have:

5. $(A\times B)\times C=-((C\cdot B)A-(C\cdot A)B)$

Distributing the negative sign results in:

6. $(A\times B)\times C=-(C\cdot B)A+(C\cdot A)B$

Final Rearrangement

Now, we can rearrange this to match the desired form:

7. $(A\times B)\times C=(C\cdot A)B-(C\cdot B)A$

Notice that we can rename $C$ to $A$ and $B$ to $B$ in our final expression, leading us to:

8. $(A\times B)\times C=(A\cdot C)B-(B\cdot C)A$

Conclusion

Thus, we have successfully deduced that $(A\times B)\times C=(A\cdot C)B-(B\cdot C)A$ using the properties of the cross product and the vector triple product identity. This demonstrates the interconnectedness of vector operations and how they can be manipulated to derive new relationships.