Problem 20 - Essential Calculus Early Transcendentals

Question

Find two unit vectors orthogonal to both \(\mathbf{j}-\mathbf{k}\) and \(\mathbf{i}+\mathbf{j}\).

Step-by-Step Solution

Answer

The unit vectors are \( \frac{1}{\sqrt{3}} \langle 1, -1, -1 \rangle \) and \( \frac{1}{\sqrt{3}} \langle -1, 1, 1 \rangle \).

Step 1: Understanding the Problem

We need two unit vectors that are orthogonal to both \( \mathbf{j} - \mathbf{k} \) and \( \mathbf{i} + \mathbf{j} \). To achieve this, we will use the cross product of these vectors to find a vector orthogonal to both.

Step 2: Find the Cross Product

Calculate the cross product \( \mathbf{v} = (\mathbf{j} - \mathbf{k}) \times (\mathbf{i} + \mathbf{j}) \). Given \( \mathbf{a} = \langle 0, 1, -1 \rangle \) and \( \mathbf{b} = \langle 1, 1, 0 \rangle \), the cross product is:\[\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & -1 \ 1 & 1 & 0 \end{vmatrix}\]Calculating the determinant, we get \( \mathbf{v} = \langle 1, -1, -1 \rangle \).

Step 3: Normalize the Cross Product to a Unit Vector

Normalize \( \mathbf{v} = \langle 1, -1, -1 \rangle \) to get a unit vector. The magnitude \( \| \mathbf{v} \| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} \).So, the unit vector \( \mathbf{u} = \frac{1}{\sqrt{3}} \langle 1, -1, -1 \rangle \).

Step 4: Find the Second Unit Vector

Since the vectors form a plane, and we need a second orthogonal vector, we can take the negative of the found unit vector as the second vector.Therefore, the second unit vector is \( -\frac{1}{\sqrt{3}} \langle 1, -1, -1 \rangle = \frac{1}{\sqrt{3}} \langle -1, 1, 1 \rangle \).

Step 5: Conclusion

The two unit vectors orthogonal to both \( \mathbf{j} - \mathbf{k} \) and \( \mathbf{i} + \mathbf{j} \) are \( \frac{1}{\sqrt{3}} \langle 1, -1, -1 \rangle \) and \( \frac{1}{\sqrt{3}} \langle -1, 1, 1 \rangle \).

Key Concepts

Cross Product

The cross product is a mathematical operation between two vectors in three-dimensional space, denoted as \( \mathbf{a} \times \mathbf{b} \). This operation results in a new vector that is perpendicular, or orthogonal, to both of the original vectors. This vector not only helps in finding directions but also plays a vital role in physics and engineering for determining torque and rotational effects.
To compute the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), you can use the determinant formula:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \\end{vmatrix}\]
This results in a vector \( \mathbf{c} = \langle c_1, c_2, c_3 \rangle \) where:

• \( c_1 = a_2b_3 - a_3b_2 \)
• \( c_2 = a_3b_1 - a_1b_3 \)
• \( c_3 = a_1b_2 - a_2b_1 \)

In our exercise, using the cross product, we found \( \mathbf{v} = \langle 1, -1, -1 \rangle \) by substituting vectors \( \mathbf{j} - \mathbf{k} \) and \( \mathbf{i} + \mathbf{j} \) into this equation.
This newly found vector \( \mathbf{v} \) is orthogonal to both \( \mathbf{j} - \mathbf{k} \) and \( \mathbf{i} + \mathbf{j} \). That's the magic of cross products!

Orthogonal Vectors

Orthogonal vectors are vectors that are at right angles to each other. This means their dot product is zero. Two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal if \( \mathbf{a} \cdot \mathbf{b} = 0 \). This concept is essential in various fields like geometry, physics, and computer graphics, where the need to understand relative directions and orientations of objects is crucial.
In simpler terms, imagine two lines intersecting at a 90-degree angle, they are orthogonal. In 3D space, it's like having one line going up, another going left-right and a third going forward. These lines are all orthogonal to each other.
In the original problem, the desired vectors are orthogonal to both \( \mathbf{j} - \mathbf{k} \) and \( \mathbf{i} + \mathbf{j} \). We accomplished this by using the cross product, which naturally creates orthogonal vectors. Don't forget, the orthogonal property ensures no overlap in direction, which can be particularly useful when analyzing forces or constructing coordinate systems.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector. A unit vector has a magnitude of one and maintains the direction of the original vector. This is crucial when you need to represent directions without concerning yourself about the length.
The process involves dividing each component of the vector by its magnitude. The magnitude of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is calculated by \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \). The normalized vector is then given by:

• \( \mathbf{u} = \frac{1}{\| \mathbf{v} \|} \langle x, y, z \rangle \)

In our exercise, after finding the cross product \( \mathbf{v} = \langle 1, -1, -1 \rangle \), we normalized it to create unit vectors. We calculated the magnitude as \( \sqrt{3} \) and divided each component by \( \sqrt{3} \), resulting in a unit vector \( \mathbf{u} = \frac{1}{\sqrt{3}} \langle 1, -1, -1 \rangle \).
Remember, by normalizing a vector, we scale it down to a size of one, allowing it to solely represent direction, making it a powerful tool in vector mathematics and physics.