Answer on Question #56113 – Math – Vector Calculus
If A = i − 2 j − 3 k A = i - 2j - 3k A = i − 2 j − 3 k , B = 2 i + 3 j + k B = 2i + 3j + k B = 2 i + 3 j + k and C = i + 3 j − 2 k C = i + 3j - 2k C = i + 3 j − 2 k , compute ∣ ( A × B ) × C ∣ |(A \times B) \times C| ∣ ( A × B ) × C ∣
2(√2)
5(√26)
3(√21)
4(√11)
Solution
A × B = [ i j k 1 − 2 − 3 2 3 1 ] = i ( − 2 ∗ 1 − ( − 3 ) ∗ 3 ) − j ( 1 ∗ 1 − ( − 3 ) ∗ 2 ) + k ( 1 ∗ 3 − ( − 2 ) ∗ 2 ) = = i ( − 2 + 9 ) − j ( 1 + 6 ) + k ( 3 + 4 ) = 7 i − 7 j + 7 k . A \times B = \begin{bmatrix} i & j & k \\ 1 & -2 & -3 \\ 2 & 3 & 1 \end{bmatrix} = i(-2 * 1 - (-3) * 3) - j(1 * 1 - (-3) * 2) + k(1 * 3 - (-2) * 2) = \\
= i(-2 + 9) - j(1 + 6) + k(3 + 4) = 7i - 7j + 7k. A × B = ⎣ ⎡ i 1 2 j − 2 3 k − 3 1 ⎦ ⎤ = i ( − 2 ∗ 1 − ( − 3 ) ∗ 3 ) − j ( 1 ∗ 1 − ( − 3 ) ∗ 2 ) + k ( 1 ∗ 3 − ( − 2 ) ∗ 2 ) = = i ( − 2 + 9 ) − j ( 1 + 6 ) + k ( 3 + 4 ) = 7 i − 7 j + 7 k . ( A × B ) × C = [ i j k 7 − 7 7 1 3 − 2 ] = i ( − 7 ∗ ( − 2 ) − 7 ∗ 3 ) − j ( 7 ∗ ( − 2 ) − 7 ∗ 1 ) + k ( 7 ∗ 3 − ( − 7 ) 1 ) = = i ( 14 − 21 ) − j ( − 14 − 7 ) + k ( 21 + 7 ) = − 7 i + 21 j + 28 k . (A \times B) \times C = \begin{bmatrix} i & j & k \\ 7 & -7 & 7 \\ 1 & 3 & -2 \end{bmatrix} = i(-7 * (-2) - 7 * 3) - j(7 * (-2) - 7 * 1) + k(7 * 3 - (-7)1) = \\
= i(14 - 21) - j(-14 - 7) + k(21 + 7) = -7i + 21j + 28k. ( A × B ) × C = ⎣ ⎡ i 7 1 j − 7 3 k 7 − 2 ⎦ ⎤ = i ( − 7 ∗ ( − 2 ) − 7 ∗ 3 ) − j ( 7 ∗ ( − 2 ) − 7 ∗ 1 ) + k ( 7 ∗ 3 − ( − 7 ) 1 ) = = i ( 14 − 21 ) − j ( − 14 − 7 ) + k ( 21 + 7 ) = − 7 i + 21 j + 28 k . ∣ ( A × B ) × C ∣ = ( − 7 ) 2 + 2 1 2 + 2 8 2 = 49 + 441 + 784 = 1274 = 7 ∗ 7 ∗ 26 = 7 26 . |(A \times B) \times C| = \sqrt{(-7)^2 + 21^2 + 28^2} = \sqrt{49 + 441 + 784} = \sqrt{1274} = \sqrt{7 * 7 * 26} = 7\sqrt{26}. ∣ ( A × B ) × C ∣ = ( − 7 ) 2 + 2 1 2 + 2 8 2 = 49 + 441 + 784 = 1274 = 7 ∗ 7 ∗ 26 = 7 26 .
Answer: 7 26 7\sqrt{26} 7 26 .
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