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Chapter 11: Problem 2

If \(\mathbf{a}=\langle 3,3,1\rangle, \mathbf{b}=\langle-2,-1,0\rangle\), and \(\mathbf{c}=\langle-2,-3,-1\rangle\), find each of the following: (a) \(\mathbf{a} \times \mathbf{b}\) (b) \(\mathbf{a} \times(\mathbf{b}+\mathbf{c})\) (c) \(\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})\) (d) \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})\)

Short Answer

Expert verified
(a) \( \langle 1, -2, 3 \rangle \), (b) \( \langle 1, -1, 0 \rangle \), (c) 1, (d) \( \langle 14, -11, -9 \rangle \).\)

Step by step solution

01

Calculate Cross Product: \( \mathbf{a} \times \mathbf{b} \)

We use the formula for the cross product of two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \): \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).\For \( \mathbf{a} = \langle 3, 3, 1 \rangle \) and \( \mathbf{b} = \langle -2, -1, 0 \rangle \):- First component: \( 3 \cdot 0 - 1 \cdot (-1) = 0 + 1 = 1 \)- Second component: \( 1 \cdot (-2) - 3 \cdot 0 = -2 - 0 = -2 \)- Third component: \( 3 \cdot (-1) - 3 \cdot (-2) = -3 + 6 = 3 \)Thus, \( \mathbf{a} \times \mathbf{b} = \langle 1, -2, 3 \rangle \).
02

Calculate Vector Sum: \( \mathbf{b} + \mathbf{c} \)

Vector addition is performed component-wise. For \( \mathbf{b} = \langle -2, -1, 0 \rangle \) and \( \mathbf{c} = \langle -2, -3, -1 \rangle \), we get:- First component: \( -2 + (-2) = -4 \)- Second component: \( -1 + (-3) = -4 \)- Third component: \( 0 + (-1) = -1 \)Thus, \( \mathbf{b} + \mathbf{c} = \langle -4, -4, -1 \rangle \).
03

Calculate Cross Product: \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) \)

Now we calculate the cross product \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) \) using \( \mathbf{a} = \langle 3, 3, 1 \rangle \) and \( \mathbf{b} + \mathbf{c} = \langle -4, -4, -1 \rangle \):- First component: \( 3 \cdot (-1) - 1 \cdot (-4) = -3 + 4 = 1 \)- Second component: \( 1 \cdot (-4) - 3 \cdot (-1) = -4 + 3 = -1 \)- Third component: \( 3 \cdot (-4) - 3 \cdot (-4) = -12 + 12 = 0 \)Thus, \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \langle 1, -1, 0 \rangle \).
04

Calculate Cross Product: \( \mathbf{b} \times \mathbf{c} \)

Using the formula for the cross product again, for \( \mathbf{b} = \langle -2, -1, 0 \rangle \) and \( \mathbf{c} = \langle -2, -3, -1 \rangle \):- First component: \( -1 \cdot (-1) - 0 \cdot (-3) = 1 - 0 = 1 \)- Second component: \( 0 \cdot (-2) - (-2) \cdot (-1) = 0 - 2 = -2 \)- Third component: \( -2 \cdot (-3) - (-1) \cdot (-2) = 6 - 2 = 4 \)Thus, \( \mathbf{b} \times \mathbf{c} = \langle 1, -2, 4 \rangle \).
05

Calculate Dot Product: \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)

The dot product between two vectors is calculated using the formula: \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \). Using \( \mathbf{a} = \langle 3, 3, 1 \rangle \) and \( \mathbf{b} \times \mathbf{c} = \langle 1, -2, 4 \rangle \), compute:- \( 3 \cdot 1 = 3 \)- \( 3 \cdot (-2) = -6 \)- \( 1 \cdot 4 = 4 \)So, \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 3 - 6 + 4 = 1 \).
06

Calculate Triple Product: \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \)

Using the vector triple product identity: \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} \). First, determine the dot products:- \( \mathbf{a} \cdot \mathbf{b} = 3 \cdot (-2) + 3 \cdot (-1) + 1 \cdot 0 = -6 - 3 + 0 = -9 \)- \( \mathbf{a} \cdot \mathbf{c} = 3 \cdot (-2) + 3 \cdot (-3) + 1 \cdot (-1) = -6 - 9 - 1 = -16 \)Now calculate: - \( (\mathbf{a} \cdot \mathbf{c})\mathbf{b} = -16 \cdot \langle -2, -1, 0 \rangle = \langle 32, 16, 0 \rangle \)- \( (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = -9 \cdot \langle -2, -3, -1 \rangle = \langle 18, 27, 9 \rangle \)Thus, \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \langle 32, 16, 0 \rangle - \langle 18, 27, 9 \rangle = \langle 14, -11, -9 \rangle \).

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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 between two vectors in three-dimensional space, resulting in another vector perpendicular to the plane containing the original vectors. It's commonly denoted by \( \times \). In our exercise, we've calculated the cross product \( \mathbf{a} \times \mathbf{b} \) for vectors \( \mathbf{a} = \langle 3, 3, 1 \rangle \) and \( \mathbf{b} = \langle -2, -1, 0 \rangle \). To find \( \mathbf{a} \times \mathbf{b} \), we use these steps:
  • First component: \( a_2 \cdot b_3 - a_3 \cdot b_2 = 3 \cdot 0 - 1 \cdot (-1) = 1 \)
  • Second component: \( a_3 \cdot b_1 - a_1 \cdot b_3 = 1 \cdot (-2) - 3 \cdot 0 = -2 \)
  • Third component: \( a_1 \cdot b_2 - a_2 \cdot b_1 = 3 \cdot (-1) - 3 \cdot (-2) = 3 \)
The result is \( \mathbf{a} \times \mathbf{b} = \langle 1, -2, 3 \rangle \). The cross product is particularly useful in physics and engineering to determine torque, rotational forces, and directions orthogonal to surfaces.
Dot Product
The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers and returns a single number. It's useful in finding the angle between two vectors, determining vector projections, or calculating work done by a force.
In our exercise, one of the computations involved is the dot product of \( \mathbf{a} = \langle 3, 3, 1 \rangle \) and \( \mathbf{b} \times \mathbf{c} = \langle 1, -2, 4 \rangle \). The formula used is:
  • \( \mathbf{a} \cdot \mathbf{v} = a_1 \cdot v_1 + a_2 \cdot v_2 + a_3 \cdot v_3 \)
Substituting in the values gives us:
  • \( 3 \cdot 1 + 3 \cdot (-2) + 1 \cdot 4 = 3 - 6 + 4 \)
The result is 1. Such results are used in various applications including projections and checking orthogonality.
Vector Addition
Vector addition involves adding corresponding components of given vectors, resulting in a new vector. It is an essential operation used widely across various fields such as physics, computer graphics, and navigation.
In the exercise, we see the addition of vectors \( \mathbf{b} \) and \( \mathbf{c} \) resulting in \( \mathbf{b} + \mathbf{c} \):
  • First component: \( -2 + (-2) = -4 \)
  • Second component: \( -1 + (-3) = -4 \)
  • Third component: \( 0 + (-1) = -1 \)
Thus, the resulting vector is \( \langle -4, -4, -1 \rangle \). This basic component-wise addition forms the basis of more complex vector manipulations and is crucial in operations requiring vector sums.
Triple Product Identity
The triple product identity relates to the cross and dot products of three vectors. It is denoted as the vector triple product identity and is often used to simplify vector equations.
In our exercise, we used the identity to calculate \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \), involving vectors \( \mathbf{a} = \langle 3, 3, 1 \rangle \), \( \mathbf{b} = \langle -2, -1, 0 \rangle \), and \( \mathbf{c} = \langle -2, -3, -1 \rangle \).
The formula is:
  • \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} \)
First, calculate the dot products:
  • \( \mathbf{a} \cdot \mathbf{b} = -9 \)
  • \( \mathbf{a} \cdot \mathbf{c} = -16 \)
Then compute:
  • \( (\mathbf{a} \cdot \mathbf{c})\mathbf{b} = \langle 32, 16, 0 \rangle \)
  • \( (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = \langle 18, 27, 9 \rangle \)
Subtracting these gives \( \langle 14, -11, -9 \rangle \). This identity is notably useful in simplifying expressions in vector calculus and physics.

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