Problem 17 - Calculus: Early Transcendentals

Question

Let \(x=\langle 1,1,0\rangle\) and \(y=\langle 2,4,2\rangle .\) Find a unit vector that is perpendicular to both \(x\) and \(y\)

Step-by-Step Solution

Answer

A unit vector perpendicular to both is \( \langle \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle \).

Step 1: Compute cross product

\(\mathbf{x} \times \mathbf{y} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 2 & 4 & 2 \end{vmatrix} = (2-0)\mathbf{i} - (2-0)\mathbf{j} + (4-2)\mathbf{k} = (2,-2,2)\)

Step 2: Normalize

\(|\mathbf{x} \times \mathbf{y}| = \sqrt{4+4+4} = 2\sqrt{3}\)
Unit vector: \(\frac{1}{2\sqrt{3}}(2,-2,2) = \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)

Key Concepts

Cross Product

In vector calculus, the cross product is an important operation used to find a vector that is perpendicular to two given vectors. When given two vectors, like \(x = \langle 1, 1, 0 \rangle\) and \(y = \langle 2, 4, 2 \rangle\) in the exercise, the goal is to determine a new vector that is perpendicular to both. The cross product, denoted as \(x \times y\), is defined only in three-dimensional space and is calculated using the formula: \[x \times y = \langle b_1c_2 - c_1b_2, c_1a_2 - a_1c_2, a_1b_2 - b_1a_2 \rangle\]

• The first component involves the subtraction \(b_1c_2 - c_1b_2\),
• The second component \(c_1a_2 - a_1c_2\),
• The third component \(a_1b_2 - b_1a_2\).

For vectors \(x\) and \(y\) in the example, substitute the components into the formula:

• \((1 \times 2) - (0 \times 4) = 2 - 0 = 2\)
• \((0 \times 2) - (1 \times 2) = 0 - 2 = -2\)
• \((1 \times 4) - (1 \times 2) = 4 - 2 = 2\)

Thus, the cross product yields \(\langle 2, -2, 2 \rangle\), aligning with the concept of perpendicular vectors.

Unit Vector

A unit vector is a vector that has a magnitude, or length, of exactly one. The idea of a unit vector is particularly important in normalization, where we convert a given vector to a unit vector in the same direction. The standard approach to find a unit vector involves dividing each component of the vector by its magnitude. The magnitude of vector \(\langle 2, -2, 2 \rangle\) is calculated using the formula for magnitude: \[||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2}\] For \(\langle 2, -2, 2 \rangle\), the magnitude \(||\mathbf{v}||\) is calculated as:

• \(||\mathbf{v}|| = \sqrt{2^2 + (-2)^2 + 2^2}\)
• \(||\mathbf{v}|| = \sqrt{4 + 4 + 4}\)
• \(||\mathbf{v}|| = \sqrt{12} = 2\sqrt{3}\)

To find the unit vector \(\hat{v}\) in the same direction: \[\hat{v} = \left\langle \frac{2}{2\sqrt{3}}, \frac{-2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}} \right\rangle = \left\langle \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle\] The resulting unit vector is what we've transformed from the original cross product result, maintaining its direction but standardizing its length.

Perpendicular Vectors

Perpendicular vectors, often referred to as orthogonal vectors, have a significant role in geometry and physics. Two vectors are perpendicular if their dot product is zero. However, when vectors are three-dimensional, finding a vector perpendicular to two other vectors requires a different approach, which involves the cross product.In the exercise, finding a vector perpendicular to both \(x\) and \(y\) automatically happens through the computed cross product \(x \times y = \langle 2, -2, 2 \rangle\). The property of the cross product dictates that the resulting vector is always perpendicular to the original pair of vectors used in its calculation. The geometric principle behind perpendicularity is valuable for:

• Determining normal vectors to surfaces,
• Calculating torque in physics,
• Finding normal directions in 3D space modeling.

In our example, the vector \(\langle 2, -2, 2 \rangle\), derived from the cross product of vectors \(x\) and \(y\), is orthogonal to both, beautifully illustrating the principle of perpendicular vectors in vector calculus.