Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all vectors perpendicular to both of the vectors \(\mathbf{a}=-2 \mathbf{i}+5 \mathbf{j}-2 \mathbf{k}\) and \(\mathbf{b}=3 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}\).

Step-by-Step Solution

Answer

The vector perpendicular to both is \( 16\mathbf{i} + 2\mathbf{j} - 11\mathbf{k} \).

Step 1: Understanding the Problem

To find a vector perpendicular to two given vectors \( \mathbf{a} = -2 \mathbf{i} + 5 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{b} = 3 \mathbf{i} - 2 \mathbf{j} + 4 \mathbf{k} \), we need to compute their cross product \( \mathbf{a} \times \mathbf{b} \). The cross product of two vectors results in a vector that is perpendicular to both of the original vectors.

Step 2: Setup the Cross Product Formula

The cross product of two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \) is given by the determinant:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix}\]

Step 3: Calculate the Determinant

For our vectors \( \mathbf{a} = -2 \mathbf{i} + 5 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{b} = 3 \mathbf{i} - 2 \mathbf{j} + 4 \mathbf{k} \), setup the determinant:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-2 & 5 & -2 \3 & -2 & 4\end{vmatrix}\]Now expand the determinant.

Step 4: Expand the Determinant

Expanding the determinant, we have:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(5 \times 4 - (-2) \times (-2)) - \mathbf{j}((-2) \times 4 - (-2) \times 3) + \mathbf{k}((-2) \times (-2) - 3 \times 5)\]Simplify each term.

Step 5: Simplify Each Component

Compute each component:1. \( \mathbf{i} \) component: \( 5 \times 4 - (-2) \times (-2) = 20 - 4 = 16 \)2. \( \mathbf{j} \) component: \(-((-2) \times 4 - (-2) \times 3) = -(-8 + 6) = 2 \)3. \( \mathbf{k} \) component: \( (-2) \times (-2) - 3 \times 5 = 4 - 15 = -11 \)Thus, \( \mathbf{a} \times \mathbf{b} = 16\mathbf{i} + 2\mathbf{j} - 11\mathbf{k} \).

Step 6: Result of the Cross Product

The vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is \( 16 \mathbf{i} + 2 \mathbf{j} - 11 \mathbf{k} \). This is the only vector, apart from its scalar multiples, that satisfies the condition of being perpendicular to both given vectors.

Key Concepts

Perpendicular Vectors

Perpendicular vectors are vectors that intersect at a 90-degree angle. In the mathematical realm, these vectors are considered orthogonal. A unique property of the cross product is that it generates a third vector that is perpendicular to the two original vectors. This means if you want to find a vector

• that is perpendicular to two other given vectors,
• the cross product is your go-to operation.

The problem we are tackling involves finding such a vector that is perpendicular to

• \(-2\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}\) (denoted as \(\mathbf{a}\)), and
• \(3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\) (denoted as \(\mathbf{b}\)).

Using the cross product, we ensure that the resulting vector not only meets the conditions of perpendicularity but also maintains the orientation and magnitude dictated by the original vectors.

Determinant

The concept of a determinant is fundamental in finding the cross product of two vectors, especially in three-dimensional space. A determinant is essentially a scalar value derived from a square matrix. When you calculate the cross product \(\ \mathbf{a} \times \mathbf{b}\ \),

• you use a special 3x3 matrix filled with unit vectors \(\mathbf{i}, \mathbf{j}, \) and \(\mathbf{k}\) in the first row,
• followed by the components of vector \(\mathbf{a}\) in the second row, and
• the components of vector \(\mathbf{b}\) in the third row.

To solve the problem, you'd write:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-2 & 5 & -2 \3 & -2 & 4 \end{vmatrix}\]This determinant helps compute each component—\(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\)—of the resulting perpendicular vector. Each minor of this determinant expresses the product of differences between the vector components.

Vector Operations

Vector operations are crucial manipulations defined on vectors, helping us to understand and solve geometrical and physical problems. The cross product is one such operation. Unlike the dot product, which returns a scalar, the cross product returns another vector. This new vector embodies the following characteristics:

• It is perpendicular to the two input vectors.
• Its direction is determined by the right-hand rule.
• The magnitude is equal to the area of the parallelogram spanned by the original vectors.

When solving for perpendicular vectors in our example, we computed the cross product of \(\mathbf{a}\) and \(\mathbf{b}\), resulting in a vector represented as a linear combination of \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). This process allows the identification of a unique perpendicular solution vector characterized by its calculated components.

Linear Algebra Concepts

Linear algebra offers a powerful framework for working with vectors and matrices, central to which are operations like addition, scalar multiplication, and cross products. Here's how a few critical concepts are applied:

• Vector Representation: Vectors are expressed as linear combinations of the standard basis vectors \(\mathbf{i}, \mathbf{j}, \and \mathbf{k}\), enabling easy manipulation and computation.
• Matrix Determinants: These are effective tools for operations involving vectors, such as finding perpendicular vectors through cross products.
• Cross Product: This operation helps derive vectors perpendicular to a given pair, utilizing determinants and orientation rules like the right-hand rule.

Linear algebra thus simplifies complex problems involving direction and space, especially when visualizing vector behaviors in three dimensions. The idea of perpendicular vectors, for instance, becomes more intuitive when viewed through the lens of cross products and their accompanying determinants.