Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all vectors perpendicular to both \(\langle 1,-2,-3\rangle\) and \(\langle-3,2,0\rangle\).

Step-by-Step Solution

Answer

The vector \( \langle 6, -9, -4 \rangle \) is perpendicular to both given vectors.

Step 1: Understand the Perpendicular Requirement

A vector is perpendicular to two other vectors if it is the cross product of those vectors. We need to find the cross product of \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \).

Step 2: Set Up the Cross Product Matrix

To find the cross product \( \mathbf{a} \times \mathbf{b} \) where \( \mathbf{a} = \langle 1, -2, -3 \rangle \) and \( \mathbf{b} = \langle -3, 2, 0 \rangle \), we set up a matrix using unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -2 & -3 \-3 & 2 & 0 \\end{vmatrix}\]

Step 3: Calculate the Determinant for Each Component

Compute the determinant for each unit vector component:- \( \mathbf{i} \)-component: \( (-2)(0) - (-3)(2) = 6 \)- \( \mathbf{j} \)-component: \(-\left( (1)(0) - (-3)(-3) \right) = -9 \)- \( \mathbf{k} \)-component: \( (1)(2) - (-3)(-2) = 2 - 6 = -4 \)

Step 4: Form the Resulting Vector

Using the components obtained from the determinant, form the vector:\[ \mathbf{v} = \langle 6, -9, -4 \rangle \]

Step 5: Confirm the Result

Verify that \( \mathbf{v} = \langle 6, -9, -4 \rangle \) is indeed perpendicular to both original vectors by checking dot products:- \( \langle 6, -9, -4 \rangle \cdot \langle 1, -2, -3 \rangle = 6(1) + (-9)(-2) + (-4)(-3) = 0 \)- \( \langle 6, -9, -4 \rangle \cdot \langle -3, 2, 0 \rangle = 6(-3) + (-9)(2) + (-4)(0) = 0 \)Since both dot products are zero, \( \langle 6, -9, -4 \rangle \) is perpendicular to both given vectors.

Key Concepts

Understanding Cross Product

The cross product, denoted as \( \mathbf{a} \times \mathbf{b} \), is a crucial operation in vector calculus when dealing with three-dimensional vectors. Its primary function is to find a vector that is perpendicular to the plane formed by the input vectors. This is particularly useful in physics and engineering to determine forces and torques.

For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the resultant cross product is a new vector expressed as:

• \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \)

This process involves setting up and solving a determinant, capturing the essence of crossing vectors in space. It's noteworthy that if \( \mathbf{a} \times \mathbf{b} = \mathbf{0} \), the vectors are parallel.

Role of Perpendicular Vectors

In vector mathematics, perpendicular vectors (or orthogonal vectors) have a dot product of zero. This means that they are at a perfect 90-degree angle to each other. The significance of finding a vector that is perpendicular to two other vectors lies in various applications.

Consider a scenario in computer graphics, where you need to find a normal vector to shade a surface correctly. By calculating the cross product of two known edges of a surface, you obtain a vector perpendicular to that surface, aiding in accurate rendering.

To ensure a vector is perpendicular to the given vectors \( \mathbf{a} \) and \( \mathbf{b} \):

• Compute \( \mathbf{a} \times \mathbf{b} \).
• Check if the dot product with both vectors is zero.

It's also important in electrical engineering, where perpendicular vectors can represent different phases of an AC system.

Determinant Calculation in Cross Product

The calculation of cross products involves determinants, which are structured as a 3x3 matrix in this context. The determinant is essentially a scalar value that helps find the orientation and area (magnitude of the vector) in 3D space.
The setup for finding \( \mathbf{a} \times \mathbf{b} \) is:

• \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} ewline 1 & -2 & -3 ewline -3 & 2 & 0 \end{vmatrix} \]

Here, each symbol \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) represents a unit vector in the x, y, and z directions respectively. By expanding this determinant, you calculate each of the components of the resulting vector:

• The \( \mathbf{i} \)-component uses the 2x2 minors of the remaining elements after excluding the row and column of \( \mathbf{i} \).
• The \( \mathbf{j} \)-component includes adjusting signs as per the cofactor expansion method.
• The \( \mathbf{k} \)-component is similar to the \( \mathbf{i} \)-component but uses different elements.

This systematic method ensures accuracy in finding the vector that is orthogonal to the initial pair of vectors.