Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=-3 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{b}=-\mathbf{i}+2 \mathbf{j}-4 \mathbf{k}\), and \(\mathbf{c}=7 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}\). Find each of the following: (a) \(\mathbf{a} \times \mathbf{b}\) (b) \(\mathbf{a} \times(\mathbf{b}+\mathbf{c})\) (c) \(\mathbf{a} \cdot(\mathbf{b}+\mathbf{c})\) (d) \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})\)

Step-by-Step Solution

Answer

(a) \(-4\mathbf{i} + 10\mathbf{j} - 4\mathbf{k}\). (b) \(-6\mathbf{i} - 36\mathbf{j} - 27\mathbf{k}\). (c) 8. (d) \(-98\mathbf{i} - 59\mathbf{j} + 88\mathbf{k}\).

Step 1: Compute \( \mathbf{a} \times \mathbf{b} \)

Using the cross product formula for vectors \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & -2 \ -1 & 2 & -4 \end{vmatrix} \), the determinant is calculated as: \( \mathbf{i} (2 \cdot -4 - (-2) \cdot 2) - \mathbf{j} (-3 \cdot -4 - (-2) \cdot -1) + \mathbf{k} (-3 \cdot 2 - 2 \cdot -1) \).Simplifying, we get:\( \mathbf{i} (-8 + 4) - \mathbf{j} (12 - 2) + \mathbf{k} (-6 + 2) \)\( = -4\mathbf{i} + 10\mathbf{j} - 4\mathbf{k} \).

Step 2: Compute \( \mathbf{b} + \mathbf{c} \)

Add vectors \( \mathbf{b} \) and \( \mathbf{c} \):\( \mathbf{b} + \mathbf{c} = (-1 + 7) \mathbf{i} + (2 + 3) \mathbf{j} + (-4 - 4) \mathbf{k} \).This simplifies to \( 6\mathbf{i} + 5\mathbf{j} - 8\mathbf{k} \).

Step 3: Compute \( \mathbf{a} \times (\mathbf{b}+\mathbf{c}) \)

Calculate \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) \) using \( \mathbf{a} = -3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \) and \( \mathbf{b} + \mathbf{c} = 6\mathbf{i} + 5\mathbf{j} - 8\mathbf{k} \):\( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & -2 \ 6 & 5 & -8 \end{vmatrix} \)Calculating the determinants gives:\( \mathbf{i} (2 \cdot -8 - (-2) \cdot 5) - \mathbf{j} (-3 \cdot -8 - (-2) \cdot 6) + \mathbf{k} (-3 \cdot 5 - 2 \cdot 6) \),which simplifies to:\( -16 + 10 \), \( 24 + 12 \), \( -15 - 12 \).Thus, \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = -6\mathbf{i} -36\mathbf{j} -27\mathbf{k} \).

Step 4: Compute \( \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) \)

Calculate the dot product using \( \mathbf{a} = -3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \) and \( \mathbf{b} + \mathbf{c} = 6\mathbf{i} + 5\mathbf{j} - 8\mathbf{k} \):\( \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = -3 \cdot 6 + 2 \cdot 5 + (-2) \cdot (-8) \).Simplifying gives:\( -18 + 10 + 16 \),which results in \( 8 \).

Step 5: Compute \( \mathbf{b} \times \mathbf{c} \)

First, use the cross product formula for \( \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & -4 \ 7 & 3 & -4 \end{vmatrix} \).So, \( \mathbf{i} (2 \cdot -4 - 3 \cdot -4) - \mathbf{j} (-1 \cdot -4 - 7 \cdot -4) + \mathbf{k} (-1 \cdot 3 - 2 \cdot 7) \).This yields the simplified terms:\( \mathbf{i}(-8 + 12) - \mathbf{j}(4 + 28) + \mathbf{k}(-3 - 14) \),which results in \( 4\mathbf{i} - 32\mathbf{j} - 17\mathbf{k} \).

Step 6: Compute \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \)

Using the result from Step 5, \( \mathbf{b} \times \mathbf{c} = 4\mathbf{i} - 32\mathbf{j} - 17\mathbf{k} \), and \( \mathbf{a} = -3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \), compute:\( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & -2 \ 4 & -32 & -17 \end{vmatrix} \).Calculating the determinant:\( \mathbf{i} (2 \cdot -17 - (-2) \cdot -32) - \mathbf{j} (-3 \cdot -17 - (-2) \cdot 4) + \mathbf{k} (-3 \cdot -32 - 2 \cdot 4) \).Solving gives:\( \mathbf{i}(-34 - 64) + \mathbf{j}(51 + 8) + \mathbf{k}(96 - 8) \),resulting in \( -98\mathbf{i} - 59\mathbf{j} + 88\mathbf{k} \).

Key Concepts

Cross Product

The cross product, also known as the vector product, is a way to multiply two 3D vectors. Unlike the dot product, which results in a scalar, the cross product yields another vector. This vector is perpendicular to both of the original vectors. The magnitude of the cross product represents the area of the parallelogram that the two vectors span.
To compute the cross product of vectors \(\mathbf{a}\) and \(\mathbf{b}\):

• Arrange the vectors in a determinant: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \]
• Calculate the determinant by finding the sum of the products of the diagonals.
• Express the result as \( c = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} \).

The direction of the resulting vector follows the right-hand rule: when you point your index finger along the first vector and your middle finger along the second, your thumb points in the direction of the cross product vector. Remember, the cross product is zero if the vectors are parallel or any of them are zero vectors.

Dot Product

The dot product, or scalar product, of two vectors, results in a scalar quantity. This makes the dot product useful for measuring magnitudes and the angle between vectors.
To calculate the dot product of \(\mathbf{a}\) and \(\mathbf{b}\):

• Multiply corresponding components together: \( a_1 \cdot b_1 \), \( a_2 \cdot b_2 \), \( a_3 \cdot b_3 \).
• Add those products up: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

This operation helps in understanding the angle between vectors:\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \]
where \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\), and \(\theta\) is the angle between them. If the dot product is zero, it indicates the vectors are orthogonal.

Determinants

Determinants are a fundamental concept in linear algebra and are used to solve systems of linear equations, among other applications. A determinant can be associated with a square matrix and provides specific information about the matrix, such as its invertibility.
Here’s how they work in a 3x3 matrix:

• Consider a 3x3 matrix: \( \begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} \)
• The determinant is calculated using: \( a(ei - fh) - b(di - fg) + c(dh - eg) \)

Each minor determinant is computed by omitting the row and column of each element and finding the determinant of the remaining 2x2 matrix.
In vector calculus, determinants are invaluable in computing cross products. This is because the cross product of two vectors is actually the determinant of a matrix that includes the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) as its first row, followed by the two vectors in its subsequent rows.