Answer on Question #59204 – Math – Linear Algebra
Question
6. If A = 2 i − 3 j − k A = 2i - 3j - k A = 2 i − 3 j − k B = i + 4 j − 2 k B = i + 4j - 2k B = i + 4 j − 2 k ( A + B + × ( A − B ) ) (A + B + \times (A - B)) ( A + B + × ( A − B ))
Solution
Remark: Statement may contain a bug and perhaps we have to find ( A + B ) × ( A − B ) (A + B) \times (A - B) ( A + B ) × ( A − B )
If A = 2 i − 3 j − k A = 2i - 3j - k A = 2 i − 3 j − k B = i + 4 j − 2 k B = i + 4j - 2k B = i + 4 j − 2 k
( A + B ) = 3 i + j − 3 k (A + B) = 3i + j - 3k ( A + B ) = 3 i + j − 3 k
and
( A − B ) = i − 7 j + k . (A - B) = i - 7j + k. ( A − B ) = i − 7 j + k .
So we have the following cross product:
( A + B ) × ( A − B ) = ∣ i j k 3 1 − 3 1 − 7 1 ∣ = i ∣ 1 − 3 − 7 1 ∣ − j ∣ 3 − 3 1 1 ∣ + k ∣ 3 1 1 − 7 ∣ = ( 1 − 21 ) i − ( 3 + 3 ) j + ( − 21 − 1 ) k = − 20 i − 6 j − 22 k . \begin{array}{l}
(A + B) \times (A - B) = \left| \begin{array}{ccc} i & j & k \\ 3 & 1 & -3 \\ 1 & -7 & 1 \end{array} \right| = i \left| \begin{array}{cc} 1 & -3 \\ -7 & 1 \end{array} \right| - j \left| \begin{array}{cc} 3 & -3 \\ 1 & 1 \end{array} \right| + k \left| \begin{array}{cc} 3 & 1 \\ 1 & -7 \end{array} \right| \\
= (1 - 21)i - (3 + 3)j + (-21 - 1)k = -20i - 6j - 22k.
\end{array} ( A + B ) × ( A − B ) = ∣ ∣ i 3 1 j 1 − 7 k − 3 1 ∣ ∣ = i ∣ ∣ 1 − 7 − 3 1 ∣ ∣ − j ∣ ∣ 3 1 − 3 1 ∣ ∣ + k ∣ ∣ 3 1 1 − 7 ∣ ∣ = ( 1 − 21 ) i − ( 3 + 3 ) j + ( − 21 − 1 ) k = − 20 i − 6 j − 22 k .
Answer: − 20 i − 6 j − 22 k -20i - 6j - 22k − 20 i − 6 j − 22 k
Question
7. If A = 3 i − j + 2 k A = 3i - j + 2k A = 3 i − j + 2 k B = 2 i + j − k B = 2i + j - k B = 2 i + j − k C = i − 2 j + 2 k C = i - 2j + 2k C = i − 2 j + 2 k ( A × B ) × C (A \times B) \times C ( A × B ) × C
Solution
If A = 3 i − j + 2 k A = 3i - j + 2k A = 3 i − j + 2 k B = 2 i + j − k B = 2i + j - k B = 2 i + j − k C = i − 2 j + 2 k C = i - 2j + 2k C = i − 2 j + 2 k
( A × B ) = ∣ i j k 3 − 1 2 2 1 − 1 ∣ = i ∣ − 1 2 1 − 1 ∣ − j ∣ 3 2 2 − 1 ∣ + k ∣ 3 − 1 2 1 ∣ = ( 1 − 2 ) i − ( − 3 − 4 ) j + ( 3 + 2 ) k = − i + 7 j + 5 k \begin{array}{l}
(A \times B) = \left| \begin{array}{ccc} i & j & k \\ 3 & -1 & 2 \\ 2 & 1 & -1 \end{array} \right| = i \left| \begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array} \right| - j \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \end{array} \right| + k \left| \begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array} \right| \\
= (1 - 2)i - (-3 - 4)j + (3 + 2)k = -i + 7j + 5k
\end{array} ( A × B ) = ∣ ∣ i 3 2 j − 1 1 k 2 − 1 ∣ ∣ = i ∣ ∣ − 1 1 2 − 1 ∣ ∣ − j ∣ ∣ 3 2 2 − 1 ∣ ∣ + k ∣ ∣ 3 2 − 1 1 ∣ ∣ = ( 1 − 2 ) i − ( − 3 − 4 ) j + ( 3 + 2 ) k = − i + 7 j + 5 k
and the expression of the vector triple product
( A × B ) × C = ∣ i j k − 1 7 5 1 − 2 2 ∣ = i ∣ 7 5 − 2 2 ∣ − j ∣ − 1 5 1 2 ∣ + k ∣ − 1 7 1 − 2 ∣ = ( 14 + 10 ) i − ( − 2 − 5 ) j + ( 2 − 7 ) k = 24 i + 7 j − 5 k \begin{array}{l}
(A \times B) \times C = \left| \begin{array}{ccc} i & j & k \\ -1 & 7 & 5 \\ 1 & -2 & 2 \end{array} \right| = i \left| \begin{array}{cc} 7 & 5 \\ -2 & 2 \end{array} \right| - j \left| \begin{array}{cc} -1 & 5 \\ 1 & 2 \end{array} \right| + k \left| \begin{array}{cc} -1 & 7 \\ 1 & -2 \end{array} \right| \\
= (14 + 10)i - (-2 - 5)j + (2 - 7)k = 24i + 7j - 5k
\end{array} ( A × B ) × C = ∣ ∣ i − 1 1 j 7 − 2 k 5 2 ∣ ∣ = i ∣ ∣ 7 − 2 5 2 ∣ ∣ − j ∣ ∣ − 1 1 5 2 ∣ ∣ + k ∣ ∣ − 1 1 7 − 2 ∣ ∣ = ( 14 + 10 ) i − ( − 2 − 5 ) j + ( 2 − 7 ) k = 24 i + 7 j − 5 k
Answer: 24 i + 7 j − 5 k 24i + 7j - 5k 24 i + 7 j − 5 k
Question
8. Determine a unit vector perpendicular to the plane of A = 2 i − 6 j − 3 k A = 2i - 6j - 3k A = 2 i − 6 j − 3 k B = 4 i + 3 j − k B = 4i + 3j - k B = 4 i + 3 j − k
Solution
If A = 2 i − 6 j − 3 k A = 2i - 6j - 3k A = 2 i − 6 j − 3 k B = 4 i + 3 j − k B = 4i + 3j - k B = 4 i + 3 j − k
A × B = ∣ i j k 2 − 6 − 3 4 3 − 1 ∣ = i ∣ − 6 − 3 3 − 1 ∣ − j ∣ 2 − 3 4 − 1 ∣ + k ∣ 2 − 6 4 3 ∣ = ( 6 + 9 ) i − ( − 2 + 12 ) j + ( 6 + 24 ) k = 15 i − 10 j + 30 k \begin{array}{l}
A \times B = \left| \begin{array}{ccc} i & j & k \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{array} \right| = i \left| \begin{array}{cc} -6 & -3 \\ 3 & -1 \end{array} \right| - j \left| \begin{array}{cc} 2 & -3 \\ 4 & -1 \end{array} \right| + k \left| \begin{array}{cc} 2 & -6 \\ 4 & 3 \end{array} \right| \\
= (6 + 9)i - (-2 + 12)j + (6 + 24)k = 15i - 10j + 30k
\end{array} A × B = ∣ ∣ i 2 4 j − 6 3 k − 3 − 1 ∣ ∣ = i ∣ ∣ − 6 3 − 3 − 1 ∣ ∣ − j ∣ ∣ 2 4 − 3 − 1 ∣ ∣ + k ∣ ∣ 2 4 − 6 3 ∣ ∣ = ( 6 + 9 ) i − ( − 2 + 12 ) j + ( 6 + 24 ) k = 15 i − 10 j + 30 k
and the unit vector perpendicular to the plane of A A A B B B
n = A × B ∣ A × B ∣ = 15 i − 10 j + 30 k 1 5 2 + 1 0 2 + 3 0 2 = 15 i − 10 j + 30 k 1225 = 15 i − 10 j + 30 k 35 = 3 i 7 − 2 j 7 + 6 k 7 n = \frac{A \times B}{|A \times B|} = \frac{15i - 10j + 30k}{\sqrt{15^2 + 10^2 + 30^2}} = \frac{15i - 10j + 30k}{\sqrt{1225}} = \frac{15i - 10j + 30k}{35} = \frac{3i}{7} - \frac{2j}{7} + \frac{6k}{7} n = ∣ A × B ∣ A × B = 1 5 2 + 1 0 2 + 3 0 2 15 i − 10 j + 30 k = 1225 15 i − 10 j + 30 k = 35 15 i − 10 j + 30 k = 7 3 i − 7 2 j + 7 6 k
Answer: \frac{3}{7} i - \frac{2}{7} j + \frac{6}{7} k.
Question
9. Evaluate ( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] (2i - 3j) \cdot [(i + j - k) \times (3i - k)] ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )]
Solution
The cross product is
( i + j − k ) × ( 3 i − k ) = ∣ i j k 1 1 − 1 3 0 − 1 ∣ = i ∣ 1 − 1 0 − 1 ∣ − j ∣ 1 − 1 3 − 1 ∣ + k ∣ 1 1 3 0 ∣ = ( − 1 − 0 ) i − ( − 1 + 3 ) j + ( 0 − 3 ) k = − i − 2 j − 3 k . (i + j - k) \times (3i - k) = \left| \begin{array}{ccc} i & j & k \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{array} \right| = i \left| \begin{array}{ccc} 1 & -1 \\ 0 & -1 \end{array} \right| - j \left| \begin{array}{ccc} 1 & -1 \\ 3 & -1 \end{array} \right| + k \left| \begin{array}{cc} 1 & 1 \\ 3 & 0 \end{array} \right|
= (-1 - 0)i - (-1 + 3)j + (0 - 3)k = -i - 2j - 3k. ( i + j − k ) × ( 3 i − k ) = ∣ ∣ i 1 3 j 1 0 k − 1 − 1 ∣ ∣ = i ∣ ∣ 1 0 − 1 − 1 ∣ ∣ − j ∣ ∣ 1 3 − 1 − 1 ∣ ∣ + k ∣ ∣ 1 3 1 0 ∣ ∣ = ( − 1 − 0 ) i − ( − 1 + 3 ) j + ( 0 − 3 ) k = − i − 2 j − 3 k .
The scalar triple product is
( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) + 0 ⋅ ( − 3 ) = − 2 + 6 = 4 (2i - 3j) \cdot [(i + j - k) \times (3i - k)] = (2i - 3j) \cdot (-i - 2j - 3k) = 2 \cdot (-1) + (-3) \cdot (-2) + 0 \cdot (-3) = -2 + 6 = 4 ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) + 0 ⋅ ( − 3 ) = − 2 + 6 = 4 Question
10. If A = i − 2 j − 3 k A = i - 2j - 3k A = i − 2 j − 3 k B = 2 i + j − k B = 2i + j - k B = 2 i + j − k C = i + 3 j − 2 k C = i + 3j - 2k C = i + 3 j − 2 k ( A × B ) ⋅ C (A \times B) \cdot C ( A × B ) ⋅ C
Solution
The scalar triple product is
( A × B ) ⋅ C = ∣ 1 − 2 − 3 2 1 − 1 1 3 − 2 ∣ = 1 ∣ 1 − 1 3 − 2 ∣ − ( − 2 ) ∣ 2 − 1 1 − 2 ∣ + ( − 3 ) ∣ 2 1 1 3 ∣ = ( − 2 + 3 ) + 2 ( − 4 + 1 ) − 3 ( 6 − 1 ) = 1 − 6 − 15 = − 20. (A \times B) \cdot C = \left| \begin{array}{ccc} 1 & -2 & -3 \\ 2 & 1 & -1 \\ 1 & 3 & -2 \end{array} \right| = 1 \left| \begin{array}{ccc} 1 & -1 \\ 3 & -2 \end{array} \right| - (-2) \left| \begin{array}{ccc} 2 & -1 \\ 1 & -2 \end{array} \right| + (-3) \left| \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right|
= (-2 + 3) + 2(-4 + 1) - 3(6 - 1) = 1 - 6 - 15 = -20. ( A × B ) ⋅ C = ∣ ∣ 1 2 1 − 2 1 3 − 3 − 1 − 2 ∣ ∣ = 1 ∣ ∣ 1 3 − 1 − 2 ∣ ∣ − ( − 2 ) ∣ ∣ 2 1 − 1 − 2 ∣ ∣ + ( − 3 ) ∣ ∣ 2 1 1 3 ∣ ∣ = ( − 2 + 3 ) + 2 ( − 4 + 1 ) − 3 ( 6 − 1 ) = 1 − 6 − 15 = − 20. Answer: -20.
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