Vector triple product

The product of three vectors $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}$ $\boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C})$ is called vector triple product and the following formula holds.．

$$\boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C}) = (\... ...oldsymbol{C})\boldsymbol{B} - (\boldsymbol{A}\cdot\boldsymbol{B})\boldsymbol{C}$$

Example 1..5  

Prove the above theorem Answer Let $\boldsymbol{A} = A_{1}\:\boldsymbol{i} + A_{2}\:\boldsymbol{j} + A_{3}\:\boldsymbol{k}$, $\boldsymbol{B} = B_{1}\:\boldsymbol{i}+ B_{2}\:\boldsymbol{j} + B_{3}\:\boldsymbol{k}$, $\boldsymbol{C} = C_{1}\:\boldsymbol{i} + C_{2}\:\boldsymbol{j} + C_{3}\:\boldsymbol{k}$. Then，

$$\boldsymbol{B} \times \boldsymbol{C} = (B_{2}C_{3} - B_{3}C_{2})\... ...C_{3} - B_{3}C_{1})\:\boldsymbol{j} + (B_{1}C_{2} - B_{2}C_{1})\:\boldsymbol{k}$$

Thus，the first component $V_{1}$ of ${\bf V} = \boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C})$ is

$$V_{1} = A_{2}(B_{1}C_{2} - B_{2}C_{1}) + A_{3}(B_{1}C_{3} - B_{3}C_{1})$$

$$= B_{1}(A_{2}C_{2} + A_{3}C_{3}) - C_{1}(A_{2}B_{2} + A_{3}B_{3})$$

$$= B_{1}(A_{1}C_{1} + A_{2}C_{2} + A_{3}C_{3}) - C_{1}(A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3})$$

$$= (\boldsymbol{A}\cdot\boldsymbol{C})B_{1}-(\boldsymbol{A}\cdot\boldsymbol{B})C_{1}$$

Similarly, we find， $V_{2}, V_{3}$.

$$V_{2} = (\boldsymbol{A}\cdot\boldsymbol{C})B_{2} - (\boldsymbol{A... ...ldsymbol{A}\cdot\boldsymbol{C})B_{3} - (\boldsymbol{A}\cdot\boldsymbol{C})C_{3}$$

Therefore,

$${\bf V} = (\boldsymbol{A}\cdot\boldsymbol{C})\boldsymbol{B} - (\boldsymbol{A}\cdot\boldsymbol{C})\boldsymbol{C}$$

Question 1..9  

Show that $(\boldsymbol{A}\times \boldsymbol{B})\cdot(\boldsymbol{C} \cdot{\bf d}) = \left... ...ymbol{B}\cdot\boldsymbol{C} & \boldsymbol{B}\cdot{\bf d} \end{array}\right\vert$．

Question 1..10  

Show that $\vert\boldsymbol{A}\times \boldsymbol{B}\vert^2 = (\boldsymbol{A}\cdot\boldsymbol{A})(\boldsymbol{B} \cdot\boldsymbol{B}) - (\boldsymbol{A}\cdot\boldsymbol{B})^2$．