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Calculate the triple scalar products \(\mathbf{w} \cdot(\mathbf{v} \times \mathbf{u})\) and \(\mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}),\) where \(\mathbf{u}=\langle 4,2,-1\rangle,\) \(\mathbf{v}=\langle 2,5,-3\rangle,\) and \(\mathbf{w}=\langle 9,5,-10\rangle.\)

Short Answer

Expert verified
\( \mathbf{w} \cdot (\mathbf{v} \times \mathbf{u}) = 119 \) and \( \mathbf{u} \cdot (\mathbf{w} \times \mathbf{v}) = 71 \).

Step by step solution

01

Calculate Cross Product \( \mathbf{v} \times \mathbf{u} \)

To find the cross product \( \mathbf{v} \times \mathbf{u} \), use the determinant formula:\[\mathbf{v} \times \mathbf{u} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \2 & 5 & -3 \4 & 2 & -1 \\end{vmatrix}\]Expanding the determinant, we have:\[\mathbf{v} \times \mathbf{u} = (5(-1) - (-3)(2))\mathbf{i} - (2(-1) - (-3)(4))\mathbf{j} + (2 \cdot 2 - 5 \cdot 4)\mathbf{k}\]Simplifying, we find:\[\mathbf{v} \times \mathbf{u} = (1)\mathbf{i} + (-10)\mathbf{j} - (16)\mathbf{k} = \langle 1, -10, -16 \rangle \]
02

Calculate Dot Product \( \mathbf{w} \cdot (\mathbf{v} \times \mathbf{u}) \)

With \( \mathbf{v} \times \mathbf{u} = \langle 1, -10, -16 \rangle \), we calculate the dot product:\[\mathbf{w} \cdot (\mathbf{v} \times \mathbf{u}) = \langle 9, 5, -10 \rangle \cdot \langle 1, -10, -16 \rangle\]Using the formula \( a \cdot b = a_1b_1 + a_2b_2 + a_3b_3 \), we have:\[ 9 \times 1 + 5 \times (-10) + (-10) \times (-16) \]Simplifying, we obtain:\[ 9 - 50 + 160 = 119 \]
03

Calculate Cross Product \( \mathbf{w} \times \mathbf{v} \)

To find the cross product \( \mathbf{w} \times \mathbf{v} \), use the determinant formula:\[\mathbf{w} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \9 & 5 & -10 \2 & 5 & -3 \\end{vmatrix}\]Expanding the determinant, we have:\[\mathbf{w} \times \mathbf{v} = (5(-3) - (-10)(5))\mathbf{i} - (9(-3) - (-10)(2))\mathbf{j} + (9 \cdot 5 - 5 \cdot 2)\mathbf{k}\]Simplifying, we find:\[\mathbf{w} \times \mathbf{v} = (35)\mathbf{i} - (17)\mathbf{j} + (35)\mathbf{k} = \langle 35, -17, 35 \rangle \]
04

Calculate Dot Product \( \mathbf{u} \cdot (\mathbf{w} \times \mathbf{v}) \)

Now calculate the dot product using \( \mathbf{w} \times \mathbf{v} = \langle 35, -17, 35 \rangle \):\[\mathbf{u} \cdot (\mathbf{w} \times \mathbf{v}) = \langle 4, 2, -1 \rangle \cdot \langle 35, -17, 35 \rangle\]Using the dot product formula, we have:\[ 4 \times 35 + 2 \times (-17) + (-1) \times 35 \]Simplifying, we obtain:\[ 140 - 34 - 35 = 71 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product is denoted as \( \mathbf{a} \times \mathbf{b} \), resulting in a third vector \( \mathbf{c} \). This operation is particularly useful in physics and engineering to determine directions, such as torque or rotational force.

The cross product of vectors \( \mathbf{v} = \langle 2,5,-3 \rangle \) and \( \mathbf{u} = \langle 4,2,-1 \rangle \) is calculated using a determinant involving unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). The determinant is:
  • \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 5 & -3 \ 4 & 2 & -1 \end{vmatrix} \)
By expanding the determinant, the cross product yields another vector, calculated via:
  • \( (5(-1) - (-3)(2))\mathbf{i} \)
  • \( (2(-1) - (-3)(4)) \mathbf{j} \)
  • \( (2(2) - 5(4)) \mathbf{k} \)
This simplifies to \( \langle 1, -10, -16 \rangle \), which is used in subsequent calculations. Cross products are non-commutative, meaning that \( \mathbf{a} \times \mathbf{b} eq \mathbf{b} \times \mathbf{a} \) except in magnitude, where they are negatives of each other.
Dot Product
The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is represented by the symbol \( \cdot \). For two vectors \( \mathbf{a} = \langle a_1,a_2,a_3 \rangle \) and \( \mathbf{b} = \langle b_1,b_2,b_3 \rangle \), the dot product is computed as:
  • \( a_1b_1 + a_2b_2 + a_3b_3 \)
The result is a scalar, hence the name.

In the exercise, the dot product \( \mathbf{w} \cdot (\mathbf{v} \times \mathbf{u}) \) is calculated with \( \mathbf{w} = \langle 9,5,-10 \rangle \) and the cross product \( \langle 1, -10, -16 \rangle \) from earlier. Plugging into the dot product formula gives:
  • \( 9 \times 1 + 5 \times (-10) + (-10) \times (-16) \)
After computations, this results in \( 119 \). These products measure how much one vector extends in the direction of another, making it integral in projections and finding angles between vectors.
Determinant Calculation
Determinant calculation is crucial for finding cross products, as it helps determine the scalar multiplication of components in a matrix. In three-dimensional vector calculus, the determinant of a 3x3 matrix allows us to compute the cross product effectively.

The matrix for cross product is set up with the first row as unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), followed by the components of each vector in subsequent rows. This setup helps visualize and calculate the resultant vector perpendicular to the original vectors.
  • Example matrix for \( \mathbf{v} \times \mathbf{u} \):
  • \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 5 & -3 \ 4 & 2 & -1 \end{vmatrix} \)
Each expansion involves simple arithmetic computations of smaller 2x2 matrices formed by excluding the row and column of specific elements.

Through breaking down into minor matrices and using subtractions and additions, we get accurate vector components, simplifying to vectors like \( \langle 1, -10, -16 \rangle \). Understanding determinant calculation aids in properly evaluating vector orientations and transformations.

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