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 -3.
Step by step solution
01
Understanding Dot and Cross Products
The goal is to find the scalar triple product \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\), which involves calculating the cross product of \(\mathbf{v}\) and \(\mathbf{w}\), and then taking the dot product of the result with \(\mathbf{u}\).
02
Calculate the Cross Product \(\mathbf{v} \times \mathbf{w}\)
To find \(\mathbf{v} \times \mathbf{w}\), use the determinant formula: \[\mathbf{v} \times \mathbf{w} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -3 & 1 \4 & 0 & 1\end{vmatrix}\]This results in:\[\mathbf{v} \times \mathbf{w} = ((-3)(1) - (1)(0))\mathbf{i} - (1\cdot 1 - 4 \cdot 1)\mathbf{j} + (1 \cdot 0 - (-3)\cdot 4)\mathbf{k} = -3\mathbf{i} + 3\mathbf{j} + 12\mathbf{k}\] Thus, \(\mathbf{v} \times \mathbf{w} = \langle -3, 3, 12 \rangle\).
03
Calculate the Dot Product \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\)
Now, compute \(\mathbf{u} \cdot \langle -3, 3, 12 \rangle\):\[\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \langle 2, 1, 0 \rangle \cdot \langle -3, 3, 12 \rangle = (2 \times -3) + (1 \times 3) + (0 \times 12)\]This simplifies to:\[= -6 + 3 + 0 = -3\]
04
Conclusion
The scalar triple product \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) results in the scalar value \(-3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Cross Product
To understand the vector cross product, think of it as a way to combine two vectors in three-dimensional space to create a new vector, which is perpendicular to both of the original vectors. This operation is essential for determining the area of parallelograms and the orientation of planes.
- The cross product of two vectors \(\mathbf{v}\) and \(\mathbf{w}\) is denoted \(\mathbf{v} \times \mathbf{w}\).
- The magnitude of \(\mathbf{v} \times \mathbf{w}\) is given by \(|\mathbf{v}| |\mathbf{w}| \sin(\theta)\), where \(\theta\) is the angle between \(\mathbf{v}\) and \(\mathbf{w}\).
- The direction of the resulting vector follows the right-hand rule.
Dot Product
The dot product is another fundamental operation involving vectors. Unlike the cross product, the dot product results in a scalar quantity, which is why it's often called the scalar product. It measures how much one vector goes in the direction of another.
- The dot product of two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\) is given by the formula \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\).
- It represents the product of the magnitudes of the vectors and the cosine of the angle \(\theta\) between them: \(|\mathbf{a}| |\mathbf{b}| \cos(\theta)\).
- Two vectors are orthogonal if their dot product is zero.
Determinants
Determinants are a mathematical tool used to solve systems of linear equations, compute volumes, and find cross products among others. They transform matrices into a single scalar value that provides a lot of information about the matrix. In the context of vectors, the determinant helps in finding the cross product by organizing it in a structured format.
- A 2x2 determinant is quite simple. Given a matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is \(ad - bc\).
- A 3x3 determinant, such as the one used in the exercise to compute the cross product, expands to a series of 2x2 determinants.