题目
已知a =3i+2j-k, b =i-j+2k,求:(1) a×b; (2) 2a×7b;…(3) 7b×2a; (4) a×a.
已知a =3i+2j-k, b =i-j+2k,求:
(1) a×b; (2) 2a×7b;
…
(3) 7b×2a; (4) a×a.
题目解答
答案
解:(1) 
(2) 
(3) 
(4)
.
解析
步骤 1:计算向量a和b的叉乘a×b
向量a和b的叉乘可以通过行列式计算,即
\[ a \times b = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]
步骤 2:计算行列式
\[ a \times b = i(2 \times 2 - (-1) \times (-1)) - j(3 \times 2 - (-1) \times 1) + k(3 \times (-1) - 2 \times 1) \]
\[ = i(4 - 1) - j(6 + 1) + k(-3 - 2) \]
\[ = 3i - 7j - 5k \]
步骤 3:计算2a×7b
\[ 2a \times 7b = 14(a \times b) = 14(3i - 7j - 5k) = 42i - 98j - 70k \]
步骤 4:计算7b×2a
\[ 7b \times 2a = 14(b \times a) = -14(a \times b) = -14(3i - 7j - 5k) = -42i + 98j + 70k \]
步骤 5:计算a×a
\[ a \times a = 0 \]
向量a和b的叉乘可以通过行列式计算,即
\[ a \times b = \begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]
步骤 2:计算行列式
\[ a \times b = i(2 \times 2 - (-1) \times (-1)) - j(3 \times 2 - (-1) \times 1) + k(3 \times (-1) - 2 \times 1) \]
\[ = i(4 - 1) - j(6 + 1) + k(-3 - 2) \]
\[ = 3i - 7j - 5k \]
步骤 3:计算2a×7b
\[ 2a \times 7b = 14(a \times b) = 14(3i - 7j - 5k) = 42i - 98j - 70k \]
步骤 4:计算7b×2a
\[ 7b \times 2a = 14(b \times a) = -14(a \times b) = -14(3i - 7j - 5k) = -42i + 98j + 70k \]
步骤 5:计算a×a
\[ a \times a = 0 \]