Question
Given three vectors a = − i − 4 j + 2 k a = -i - 4j + 2k a = − i − 4 j + 2 k , b = 3 i + 2 j − 2 k b = 3i + 2j - 2k b = 3 i + 2 j − 2 k , c = 2 i − 3 j + k c = 2i - 3j + k c = 2 i − 3 j + k , calculate a *(b × c b \times c b × c )
a.-6
b.6
c.9
d.-9
Solution
a = ( − 1 ; − 4 ; 2 ) ; b = ( 3 ; 2 ; − 2 ) ; c = ( 2 ; − 3 ; 1 ) ; ( a ∗ [ b × c ] ) − ? a = (-1; -4; 2); \quad b = (3; 2; -2); \quad c = (2; -3; 1); \quad (a^*[b \times c]) - ? a = ( − 1 ; − 4 ; 2 ) ; b = ( 3 ; 2 ; − 2 ) ; c = ( 2 ; − 3 ; 1 ) ; ( a ∗ [ b × c ]) − ?
Firstly we find [ b × c ] [b \times c] [ b × c ]
[ b × c ] = ( i j k 3 2 − 2 2 − 3 1 ) = i ( 2 ∗ 1 − ( − 3 ) ∗ ( − 2 ) ) + j ( 2 ∗ ( − 2 ) − 3 ∗ 1 ) + k ( 3 ∗ ( − 3 ) − 2 ∗ 2 ) \begin{array}{l}
[b \times c] = \left( \begin{array}{ccc} i & j & k \\ 3 & 2 & -2 \\ 2 & -3 & 1 \end{array} \right) \\
= i(2 * 1 - (-3) * (-2)) + j(2 * (-2) - 3 * 1) + k(3 * (-3) - 2 * 2)
\end{array} [ b × c ] = ⎝ ⎛ i 3 2 j 2 − 3 k − 2 1 ⎠ ⎞ = i ( 2 ∗ 1 − ( − 3 ) ∗ ( − 2 )) + j ( 2 ∗ ( − 2 ) − 3 ∗ 1 ) + k ( 3 ∗ ( − 3 ) − 2 ∗ 2 ) [ b × c ] = − 4 i − 5 j − 5 k ; [b \times c] = -4i - 5j - 5k; [ b × c ] = − 4 i − 5 j − 5 k ; ( a ∗ [ b × c ] ) = − 1 ∗ ( − 4 ) + ( − 4 ) ∗ ( − 5 ) + 2 ∗ ( − 5 ) = 14 (a^*[b \times c]) = -1 * (-4) + (-4) * (-5) + 2 * (-5) = 14 ( a ∗ [ b × c ]) = − 1 ∗ ( − 4 ) + ( − 4 ) ∗ ( − 5 ) + 2 ∗ ( − 5 ) = 14
Answer: ( a ∗ [ b × c ] ) = 14 (a^*[b \times c]) = 14 ( a ∗ [ b × c ]) = 14 . We don't have correct answer in given multiple choice answers.
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