Chapter 11: Problem 23
Find \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})\) $$ \mathbf{u}=\langle 2,1,0\rangle, \mathbf{v}=\langle 1,-3,1\rangle, \mathbf{w}=\langle 4,0,1\rangle $$
Short Answer
Expert verified
The scalar triple product is -9.
Step by step solution
01
Set Up the Problem
Our goal is to find the scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). This requires first calculating the cross product \( \mathbf{v} \times \mathbf{w} \).
02
Calculate the Cross Product \( \mathbf{v} \times \mathbf{w} \)
To find the cross product \( \mathbf{v} \times \mathbf{w} \), use the determinant formula:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -3 & 1 \4 & 0 & 1 \\end{vmatrix}\]Calculating this determinant, we get:\[\mathbf{v} \times \mathbf{w} = \langle (-3)(1) - (1)(0), (1)(1) - (4)(1), (1)(0) - (-3)(4) \rangle = \langle -3, -3, 12 \rangle\]
03
Calculate the Dot Product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \)
Now, calculate the dot product of \( \mathbf{u} = \langle 2, 1, 0 \rangle \) and \( \mathbf{v} \times \mathbf{w} = \langle -3, -3, 12 \rangle \):\[\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (2)(-3) + (1)(-3) + (0)(12) = -6 - 3 + 0 = -9\]
04
Final Conclusion
The scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) has been found to be \(-9\). This indicates the volume of the parallelepiped formed by \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) is \(9\), with negative indicating orientation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Cross Product
The cross product, also known as the vector product, is a fundamental operation in vector algebra. It's used primarily in three-dimensional space and results in a vector that is perpendicular to two input vectors. If you have two vectors, \( \mathbf{A} \) and \( \mathbf{B} \), the cross product is denoted by \( \mathbf{A} \times \mathbf{B} \). The resulting vector's magnitude corresponds to the area of the parallelogram formed by \( \mathbf{A} \) and \( \mathbf{B} \).
The formula for a cross product involves a determinant of a 3x3 matrix and can be written as:
The formula for a cross product involves a determinant of a 3x3 matrix and can be written as:
- \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_1 & A_2 & A_3 \ B_1 & B_2 & B_3 \end{vmatrix} \]
- The operation is anti-commutative, meaning \( \mathbf{A} \times \mathbf{B} = - (\mathbf{B} \times \mathbf{A}) \)
- The cross product of parallel vectors is a zero-vector \( \mathbf{A} \times \mathbf{A} = \mathbf{0} \)
Determinants
Determinants are an essential mathematical tool used in various fields, particularly in linear algebra. They are a special number calculated from a square matrix. For our purposes in vector algebra, determinants are used to calculate the cross product of vectors.
For a 3x3 matrix:
For a 3x3 matrix:
- \[\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)\]
- Determinants are crucial for solving linear equations, finding eigenvalues, and working with vector spaces.
- In three dimensions, they assist in finding vector products and assessing vector independence.
Dot Product
The dot product, or scalar product, is a way to multiply two vectors resulting in a scalar, not a vector. It measures the extent to which two vectors are parallel. 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 calculated as:
- \( \mathbf{A} \cdot \mathbf{B} = A_1B_1 + A_2B_2 + A_3B_3 \)
- If the dot product is zero, the vectors are orthogonal or perpendicular.
- The larger the dot product's magnitude, the more parallel the vectors are.
- \( \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{\| \mathbf{A} \| \| \mathbf{B} \|} \)