10판/3. 벡터

3-36 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 10. 20:37

{p1=4i^3j^p2=6i^+3j^2k^\begin{cases} \vec p_1 = 4\hat i - 3 \hat j\\ \vec p_2= -6\hat i + 3\hat j - 2\hat k\\ \end{cases} Ans=(p1+p2)(p1×p2)={(2)i^+(2)k^}i^j^k^430632={(2)i^+(2)k^}(3032i^+0426j^+3032k^)={(2)i^+(2)k^}{6i^+8j^+(6)k^}=0 \begin{aligned} \Ans =& (\vec p_1 + \vec p_2)\cdot(\vec p_1\times \vec p_2)\\ =&\bra{(-2)\i+(-2)\k}\cdot \begin{vmatrix} \i & \j & \k\\ 4&-3&0\\ -6&3&-2\\ \end{vmatrix}\\ =&\bra{(-2)\i+(-2)\k}\\ &\cdot\( \begin{vmatrix} -3&0\\ 3&-2\\ \end{vmatrix}\i+ \begin{vmatrix} 0&4\\ -2&-6\\ \end{vmatrix}\j+ \begin{vmatrix} -3&0\\ 3&-2\\ \end{vmatrix}\k\)\\ =&\bra{(-2)\i+(-2)\k}\cdot\bra{6\i+8\j+(-6)\k}\\ =&0 \end{aligned}