Answer on Question #46931 – Math – Vector Calculus
Evaluate the vectors ( 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 )]
7
4
6
2
Solution
Let's calculate the cross product:
[ ( i + j − k ) × ( 3 i − k ) ] = ∣ i j k 1 1 − 1 3 0 − 1 ∣ = 1 ⋅ ( − 1 ) ⋅ i + ( − 1 ) ⋅ 3 ⋅ j − 3 ⋅ 1 ⋅ k − 1 ⋅ ( − 1 ) ⋅ j = − i − 2 j − 3 k \begin{array}{l}
\left[ (i + j - k) \times (3i - k) \right] = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{array} \right| = 1 \cdot (-1) \cdot \mathbf{i} + (-1) \cdot 3 \cdot \mathbf{j} - 3 \cdot 1 \cdot \mathbf{k} - 1 \cdot (-1) \cdot \mathbf{j} \\
= - \mathbf{i} - 2 \mathbf{j} - 3 \mathbf{k}
\end{array} [ ( i + j − k ) × ( 3 i − k ) ] = ∣ ∣ i 1 3 j 1 0 k − 1 − 1 ∣ ∣ = 1 ⋅ ( − 1 ) ⋅ i + ( − 1 ) ⋅ 3 ⋅ j − 3 ⋅ 1 ⋅ k − 1 ⋅ ( − 1 ) ⋅ j = − i − 2 j − 3 k
Let's calculate the product:
( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = − 2 + 6 = 4 (2\mathbf{i} - 3\mathbf{j}) \cdot [(\mathbf{i} + \mathbf{j} - \mathbf{k}) \times (3\mathbf{i} - \mathbf{k})] = (2\mathbf{i} - 3\mathbf{j}) \cdot (-\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}) = -2 + 6 = 4 ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = − 2 + 6 = 4
Answer: 4
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#339914 on Dec 2023
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