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Chapter 1: Problem 10

Three vectors \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) are given by $$ \begin{array}{l} \mathbf{A}=3 \hat{\mathbf{x}}-2 \hat{\mathbf{y}}+2 \hat{z} \\ \mathbf{B}=6 \hat{\mathbf{x}}+4 \hat{\mathbf{y}}-2 \hat{\mathbf{z}} \\ \mathbf{C}=-3 \hat{\mathbf{x}}-2 \hat{\mathbf{y}}-4 \hat{\mathbf{z}} \end{array} $$ Compute the values of \(\mathbf{A} \cdot \mathbf{B} \times \mathbf{C}\) and \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C}), \mathbf{C} \times(\mathbf{A} \times \mathbf{B})\), and \(\mathbf{B} \times(\mathbf{C} \times \mathbf{A})\).

Short Answer

Expert verified
The results are: \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \), \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = 60 \hat{\mathbf{x}} + 40 \hat{\mathbf{y}} - 130 \hat{\mathbf{z}} \), \( \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = 24 \hat{\mathbf{x}} - 56 \hat{\mathbf{y}} - 46 \hat{\mathbf{z}} \), \( \mathbf{B} \times (\mathbf{C} \times \mathbf{A}) = 36 \hat{\mathbf{x}} - 24 \hat{\mathbf{y}} + 156 \hat{\mathbf{z}} \).

Step by step solution

01

Calculate \( \mathbf{B} \times \mathbf{C} \)

To find the cross product \( \mathbf{B} \times \mathbf{C} \), use the determinant method for vectors:\[\mathbf{B} \times \mathbf{C} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \6 & 4 & -2 \-3 & -2 & -4 \end{vmatrix} = \hat{\mathbf{x}}(4(-4) - (-2)(-2)) - \hat{\mathbf{y}}(6(-4) - (-2)(-3)) + \hat{\mathbf{z}}(6(-2) - 4(-3))\]Calculating each component gives:\[=-16 - 4 = -20,\quad 24 + 6 = 30,\quad -12 + 12 = 0\]Thus, \( \mathbf{B} \times \mathbf{C} = -20 \hat{\mathbf{x}} - 30 \hat{\mathbf{y}} \).
02

Compute \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \)

Calculate the dot product \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \):\[\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (3 \hat{\mathbf{x}} - 2 \hat{\mathbf{y}} + 2 \hat{\mathbf{z}}) \cdot (-20 \hat{\mathbf{x}} - 30 \hat{\mathbf{y}})\]This gives\[= 3(-20) + (-2)(-30) + 2(0) = -60 + 60 + 0 = 0\]Thus, \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \).
03

Calculate \( \mathbf{B} \times (\mathbf{C} \times \mathbf{A}) \)

First, find \( \mathbf{C} \times \mathbf{A} \) using the determinant method:\[\mathbf{C} \times \mathbf{A} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \-3 & -2 & -4 \3 & -2 & 2 \end{vmatrix} = \hat{\mathbf{x}}((-2)(2) - (-4)(-2)) - \hat{\mathbf{y}}((-3)(2) - (-4)(3)) + \hat{\mathbf{z}}((-3)(-2) - (-2)(3))\]Calculating each gives\[-4 - 8 = -12 \quad + 6 + 12 = 18 \quad + 6 - 6 = 0\]So, \( \mathbf{C} \times \mathbf{A} = -12 \hat{\mathbf{x}} + 18 \hat{\mathbf{y}} \).Now, find \( \mathbf{B} \times (\mathbf{C} \times \mathbf{A}) \):\[\mathbf{B} \times (\mathbf{C} \times \mathbf{A}) = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \6 & 4 & -2 \-12 & 18 & 0\end{vmatrix} = \hat{\mathbf{x}}(4 \cdot 0 - 18(-2)) - \hat{\mathbf{y}}(6 \cdot 0 - (-2)(-12)) + \hat{\mathbf{z}}(6(18) - 4(-12))\]Calculating gives\[= 0 + 36 \quad = 0 - 24 \quad = 108 + 48\]Thus, \( \mathbf{B} \times (\mathbf{C} \times \mathbf{A}) = 36 \hat{\mathbf{x}} - 24 \hat{\mathbf{y}} + 156 \hat{\mathbf{z}} \).
04

Compute \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \)

Use the result of Step 1, \( \mathbf{B} \times \mathbf{C} = -20 \hat{\mathbf{x}} - 30 \hat{\mathbf{y}} \):\[\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \3 & -2 & 2 \-20 & -30 & 0\end{vmatrix} = \hat{\mathbf{x}}(-2 \cdot 0 - 2(-30)) - \hat{\mathbf{y}}(3 \cdot 0 - 2(-20)) + \hat{\mathbf{z}}(3(-30) - (-2)(-20))\]Calculating gives\[= 60 \quad = 40 \quad = -90 - 40\]Thus, \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = 60 \hat{\mathbf{x}} + 40 \hat{\mathbf{y}} - 130 \hat{\mathbf{z}} \).
05

Calculate \( \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) \)

First, find \( \mathbf{A} \times \mathbf{B} \):\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \3 & -2 & 2 \6 & 4 & -2\end{vmatrix} = \hat{\mathbf{x}}((-2)(-2) - (2)(4)) - \hat{\mathbf{y}}((3)(-2) - (2)(6)) + \hat{\mathbf{z}}((3)(4) - (-2)(6))\]Calculating gives\[4 - 8 = -4 \quad + 6 + 12 = 18 \quad + 12 + 12\]Thus, \( \mathbf{A} \times \mathbf{B} = -4 \hat{\mathbf{x}} + 18 \hat{\mathbf{y}} + 24 \hat{\mathbf{z}} \).Now, compute \( \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) \):\[\mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \-3 & -2 & -4 \-4 & 18 & 24\end{vmatrix} = \hat{\mathbf{x}}((-2)(24) - (18)(-4)) - \hat{\mathbf{y}}((-3)(24) - (-4)(-4)) + \hat{\mathbf{z}}((-3)(18) - (-2)(-4))\]Calculating gives\[-48 + 72 \quad -72 + 16 \quad -54 + 8\]Thus, \( \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = 24 \hat{\mathbf{x}} - 56 \hat{\mathbf{y}} - 46 \hat{\mathbf{z}} \).
06

Final Result Verification

Each vector triple product and scalar triple product should be reviewed for consistency and calculations ensuring non-zero or zero results for accuracy. Confirm calculations are consistent with vector cross product identities.

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Key Concepts

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

Vector Cross Product
The vector cross product is a fundamental operation in vector calculus. It involves two vectors and results in a new vector that is perpendicular to both original vectors. This is particularly useful in physics and engineering to determine the orientation of objects in three-dimensional space. Notably, the magnitude of the cross product is given by multiplying the magnitudes of the two vectors and the sine of the angle between them:\[|\mathbf{A} \times \mathbf{B}|=|\mathbf{A}|\times|\mathbf{B}|\times \sin(\theta)\]This product is non-commutative, meaning \( \mathbf{A} \times \mathbf{B} eq \mathbf{B} \times \mathbf{A} \). The direction is given by the right-hand rule. In practical scenarios, the determinant method is often used to calculate the cross product more efficiently when the vectors are expressed in terms of their components.
Scalar Triple Product
The scalar triple product combines three vectors to produce a scalar value. This result provides insight into the volume of a parallelepiped, which is the three-dimensional shape formed by these vectors. The scalar triple product is given by:\[\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\]This can also be expressed as a determinant, using the components of the three vectors. If the scalar triple product is zero, it indicates the vectors are coplanar, meaning they lie on the same plane. It's essential in determining properties of geometrical configurations in vector spaces and is widely used for simplification in physical problems.
Vector Dot Product
The vector dot product, also known as the scalar product, involves multiplying two vectors to produce a scalar. It is computed as:\[\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| \times |\mathbf{B}|\times \cos(\theta)\]where \( \theta \) is the angle between the two vectors. It's a quick way to understand if two vectors are orthogonal鈥攊f the dot product is zero, the vectors are perpendicular. This operation helps in projections and finding angles between vectors. Additionally, it plays a critical role in physics, such as in computing work done, where force and displacement vectors are considered.
Determinant Method
The determinant method is an efficient technique for computing vector cross products. This method utilizes matrices where each row corresponds to a vector, and the first row comprises unit vectors along the axes \( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \). To find \( \mathbf{A} \times \mathbf{B} \), set up the matrix as:\[\begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \A_x & A_y & A_z \B_x & B_y & B_z\end{vmatrix}\]By solving this determinant, each component of the resultant vector can be derived. It's a helpful and standardized method, reducing the complexity of handling calculations in vector algebra.
Vector Identities
Vector identities are crucial in simplifying complex vector calculus problems. Some fundamental identities involve simplifications using the distributive, associative, and commutative properties of vectors. Key vector identities include:
  • The anti-commutative property of the cross product: \( \mathbf{A} \times \mathbf{B} = -\mathbf{B} \times \mathbf{A} \)
  • The distributive law: \( \mathbf{A} \times (\mathbf{B} + \mathbf{C}) = \mathbf{A} \times \mathbf{B} + \mathbf{A} \times \mathbf{C} \)
  • The Jacobi identity or the vector triple product identity, which states: \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C})\mathbf{B} - (\mathbf{A} \cdot \mathbf{B})\mathbf{C} \)
These identities facilitate vector calculus by reducing algebraic expressions and proving helpful in the simplification of vector operations.

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Most popular questions from this chapter

The vertices of parallelogram \(A B C D\) are \((1,0,0),(2,-1,0),(0,-1,1)\), and \((-1,0,1)\) in order. Calculate the vector areas of triangle \(A B D\) and of triangle \(B C D\). Are the two vector areas equal? $$ \text { ANS. } \operatorname{Area}_{A B D}=-\frac{1}{2}(\hat{\mathbf{x}}+\hat{\mathbf{y}}+2 \hat{\mathbf{z}}) \text { . } $$

Given that: (1) the dot product of a unit vector with itself is unity, and (2) this relation is valid in all (rotated) coordinate systems, show that \(\hat{\mathbf{x}}^{\prime} \cdot \hat{\mathbf{x}}^{\prime}=1\) (with the primed system rotated \(45^{\circ}\) about the \(z\) -axis relative to the unprimed) implies that \(\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}=0\).

A triangle is defined by the vertices of three vectors \(\mathbf{A}, \mathrm{B}\), and \(\mathrm{C}\) that extend from the origin. In terms of \(A, B\), and \(C\) show that the vector sum of the successive sides of the triangle \((A B+B C+C A)\) is zero.

Prove \(\nabla \cdot(\mathbf{a} \times \mathbf{b})=\mathbf{b} \cdot(\nabla \times \mathbf{a})-\mathbf{a} \cdot(\nabla \times \mathbf{b})\). Hint. Treat as a triple scalar product.

Prove that \((\mathrm{A} \times \mathbf{B}) \cdot(\mathbf{A} \times \mathbf{B})=(A B)^{2}-(\mathbf{A} \cdot \mathbf{B})^{2}\).
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