3-37 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a = 3.0\i + 3.0\j +(- 2.0)\k\\ \vec b =(-1.0)\i +(-4.0)\j +(-2.0)\k\\ \vec c = 2.0\i + 2.0\j + 1.0\k\\ \end{cases} $$
(a)$\vec a\cdot(\vec b \times \vec c)$ $$ \begin{aligned} \Ans=&\vec a\cdot(\vec b \times \vec c)\\ =&\vec a\cdot \begin{vmatrix} \i & \j & \k\\ -1&-4&-2\\ 2&2&1\\ \end{vmatrix}\\ =&\vec a\cdot \( \begin{vmatrix} -4&-2\\ 2&1\\ \end{vmatrix}\i+ \begin{vmatrix} -2&-1\\ 1&2\\ \end{vmatrix}\j+ \begin{vmatrix} -1&-4\\ 2&2\\ \end{vmatrix}\k\)\\ =&\(3.0\i + 3.0\j +(- 2.0)\k\)\cdot \(-3\hat j + 6\hat k\)\\ =&-21 \end{aligned} $$
(b)$\vec a\cdot(\vec b + \vec c)$ $$ \begin{aligned} \Ans=&\vec a\cdot(\vec b + \vec c)\\ =&\(3.0\i + 3.0\j - 2.0\k\)\cdot\(\hat i -2\hat j -\hat k\)\\ =&-1 \end{aligned} $$
(c)$\vec a\times(\vec b + \vec c)$ $$ \begin{aligned} \Ans=&\vec a\times(\vec b + \vec c)\\ =&\(3.0\i + 3.0\j - 2.0\k\)\times\(\hat i -2\hat j -\hat k\)\\ =&\begin{vmatrix} \i & \j & \k\\ 3&3&-2\\ 1&-2&-1\\ \end{vmatrix}\\ =&\begin{vmatrix} 3&-2\\ -2&-1\\ \end{vmatrix}\i+ \begin{vmatrix} -2&3\\ -1&1\\ \end{vmatrix}\j+ \begin{vmatrix} 3&3\\ 1&-2\\ \end{vmatrix}\k\\ =&(-7)\i+\j+(-9)\k \end{aligned} $$