ベクトル3重積(Vector triple product)

3つのベクトル $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}$の積 $\boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C})$をベクトル3重積といい，次の公式が成り立ちます．

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

例題 1..5  

上の公式を証明しよう． 解 $\boldsymbol{A} = A_{1}\:\boldsymbol{i} + A_{2}\:\boldsymbol{j} + A_{3}\:\boldsymbol{k}$, $\boldsymbol{B} = B_{1}\:\boldsymbol{i}+ B_{2}\:\boldsymbol{j} + B_{3}\:\boldsymbol{k}$, $\boldsymbol{C} = C_{1}\:\boldsymbol{i} + C_{2}\:\boldsymbol{j} + C_{3}\:\boldsymbol{k}$とすると，

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

より， ${\bf V} = \boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C})$の第1成分$V_{1}$は

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

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

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

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

同様に， $V_{2}, V_{3}$を求めると，

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

したがって，

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

問 1..9  

$(\boldsymbol{A}\times \boldsymbol{B})\cdot(\boldsymbol{C} \cdot{\bf d}) = \left... ...ymbol{B}\cdot\boldsymbol{C} & \boldsymbol{B}\cdot{\bf d} \end{array}\right\vert$を証明せよ．

問 1..10  

$\vert\boldsymbol{A}\times \boldsymbol{B}\vert^2 = (\boldsymbol{A}\cdot\boldsymbol{A})(\boldsymbol{B} \cdot\boldsymbol{B}) - (\boldsymbol{A}\cdot\boldsymbol{B})^2$を証明せよ．