Problem 23 - Essential Calculus Early Transcendentals

Question

Find a unit vector that is orthogonal to both \(\mathbf{i}+\mathbf{j}\) and \(\mathbf{i}+\mathbf{k}\)

Step-by-Step Solution

Answer

The unit vector orthogonal to both is \( \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k}) \).

Step 1: Identify the vectors

The vectors we have are \( \mathbf{a} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{k} \).

Step 2: Use the cross product to find an orthogonal vector

The cross product \( \mathbf{a} \times \mathbf{b} \) will give us a vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). Compute \( \mathbf{a} \times \mathbf{b} \): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 1 & 0 & 1 \end{vmatrix} = \mathbf{i}(1 \cdot 1 - 0 \cdot 0) - \mathbf{j}(1 \cdot 1 - 0 \cdot 1) + \mathbf{k}(1 \cdot 0 - 1 \cdot 1) = \mathbf{i} - \mathbf{j} - \mathbf{k} \] Hence, \( \mathbf{a} \times \mathbf{b} = \mathbf{i} - \mathbf{j} - \mathbf{k} \).

Step 3: Calculate the magnitude of the resulting vector

To make \( \mathbf{i} - \mathbf{j} - \mathbf{k} \) a unit vector, find its magnitude: \[ \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

Step 4: Normalize the vector

Divide each component of the vector \( \mathbf{i} - \mathbf{j} - \mathbf{k} \) by its magnitude \( \sqrt{3} \): \[ \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \] This gives us the unit vector \( \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k}) \), which is orthogonal to both given vectors.

Key Concepts

Cross Product

The cross product is a useful operation for finding a vector that is perpendicular, or orthogonal, to two given vectors. If you have two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), in a three-dimensional space, their cross product \( \mathbf{a} \times \mathbf{b} \) will result in a new vector that is perpendicular to both original vectors. This property is particularly helpful in geometry and physics.

To compute the cross product, you can use the determinant formula involving the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). For vectors \( \mathbf{a} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{k} \), the calculation is:

• Set up the determinant with \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) across the top.
• Fill in the components from the vectors \( \mathbf{a} \) and \( \mathbf{b} \).
• Calculate the determinant to get the new orthogonal vector.

The result is a vector like \( \mathbf{i} - \mathbf{j} - \mathbf{k} \), which is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. In mathematical terms, two vectors are orthogonal if their dot product equals zero. This means there is no component of one vector along the direction of the other.

This property is essential in many applications, such as computer graphics, physics, and engineering, where it might be necessary to find a direction that is completely independent of others. In our exercise, we wanted to find a vector that is orthogonal to both \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{i} + \mathbf{k} \).

The cross product of these vectors automatically gives us an orthogonal vector, satisfying this requirement. The resulting vector, \( \mathbf{i} - \mathbf{j} - \mathbf{k} \), is perpendicular to both original vectors and thus orthogonal.

Vector Normalization

Vector normalization is the process of converting any vector into a unit vector. A unit vector has a magnitude of 1 and simply represents direction. This process is crucial when you need to preserve the direction of a vector while adjusting its magnitude.

After calculating the orthogonal vector \( \mathbf{i} - \mathbf{j} - \mathbf{k} \) from the cross product, we need to normalize it. Normalization involves dividing each component of the vector by its magnitude.

For our vector:

• Calculate the magnitude, which is \( \sqrt{3} \).
• Divide each component by \( \sqrt{3} \).

This gives us the unit vector \( \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k}) \), maintaining the direction but with a magnitude of 1.

Vector Magnitude

The magnitude of a vector is a measure of its length. It tells you how long the vector is, starting from its initial point and stretching along its direction to its terminal point. The formula to find the magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is:

\[\sqrt{a^2 + b^2 + c^2}\]
In the context of our exercise, once we have the orthogonal vector \( \mathbf{i} - \mathbf{j} - \mathbf{k} \), we calculate its magnitude:

• Square each component: \( 1^2, (-1)^2, (-1)^2 \)
• Add these squares together: \( 1 + 1 + 1 \)
• Take the square root of the sum: \( \sqrt{3} \)

This gives us the length of the vector, which is essential for further steps such as normalization, allowing us to scale the vector to a unit vector without changing its direction.