Problem 10 - Algebra and Trigonometry

Question

Two vectors a and b are given. (a) Find a vector perpendicular to both a and b. (b) Find a unit vector perpendicular to both a and b. $$ \mathbf{a}=\langle 2,5,3\rangle, \quad \mathbf{b}=\langle 3,-2,-1\rangle $$

Step-by-Step Solution

Answer

(a) \(\langle 1, 11, -19 \rangle\), (b) \(\left\langle \frac{1}{\sqrt{483}}, \frac{11}{\sqrt{483}}, \frac{-19}{\sqrt{483}} \right\rangle\).

Step 1: Understanding Perpendicular Vectors

To find a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\), we need to use the cross product. The cross product of two vectors yields a vector that is perpendicular to both.

Step 2: Compute the Cross Product

The cross product \(\mathbf{a} \times \mathbf{b}\) is calculated as the determinant:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 5 & 3 \ 3 & -2 & -1 \end{vmatrix}\]Expanding this, we find:\[= \mathbf{i}(5 \cdot (-1) - 3 \cdot (-2)) - \mathbf{j}(2 \cdot (-1) - 3 \cdot 3) + \mathbf{k}(2 \cdot (-2) - 5 \cdot 3)\]\[= \mathbf{i}(-5 + 6) - \mathbf{j}(-2 - 9) + \mathbf{k}(-4 - 15)\]\[= \mathbf{i} + 11\mathbf{j} - 19\mathbf{k}\]Thus, \(\mathbf{a} \times \mathbf{b} = \langle 1, 11, -19 \rangle\).

Step 3: Find a Perpendicular Vector

From Step 2, the vector \(\langle 1, 11, -19 \rangle\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

Step 4: Find the Magnitude of the Perpendicular Vector

To find a unit vector, compute the magnitude of the perpendicular vector \(\langle 1, 11, -19 \rangle\):\[\| \langle 1, 11, -19 \rangle \| = \sqrt{1^2 + 11^2 + (-19)^2}\]\[= \sqrt{1 + 121 + 361} = \sqrt{483}\]

Step 5: Compute the Unit Vector

Divide each component of \(\langle 1, 11, -19 \rangle\) by its magnitude to find the unit vector:\[\text{Unit vector} = \left\langle \frac{1}{\sqrt{483}}, \frac{11}{\sqrt{483}}, \frac{-19}{\sqrt{483}} \right\rangle\]

Step 6: Verify the Solution

Finally, ensure each component of the unit vector is simplified if possible. The unit vector \(\left\langle \frac{1}{\sqrt{483}}, \frac{11}{\sqrt{483}}, \frac{-19}{\sqrt{483}} \right\rangle\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) with a magnitude of 1.

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of exactly 1. It is used to indicate direction, and is derived from any given vector to display the same direction as its original vector, but without influence of scale. Creating a unit vector involves:

• Calculating the magnitude of the original vector.
• Dividing each component of the vector by this magnitude.

In the context of the given exercise, once we find the perpendicular vector by using the cross product, we calculate its magnitude. For the vector \( \langle 1, 11, -19 \rangle \), the magnitude was found to be \( \sqrt{483} \). To convert it to a unit vector, we divide each of its components by \( \sqrt{483} \). Thus, the unit vector is \( \left\langle \frac{1}{\sqrt{483}}, \frac{11}{\sqrt{483}}, \frac{-19}{\sqrt{483}} \right\rangle \), maintaining its perpendicular nature.

Three-Dimensional Vectors

Vectors that are represented in three dimensions have three components and are commonly expressed as \( \langle x, y, z \rangle \). In our problem, both \( \mathbf{a} \) and \( \mathbf{b} \) are three-dimensional vectors, meaning they exist in three-dimensional space and have components along the x, y, and z axes. Such vectors can represent physical quantities like force, velocity, or position in space, where each component corresponds to the contribution along each axis. The ability to mathematically manipulate these vectors, such as finding a vector perpendicular to both via the cross product, is crucial for many applications, like in physics and engineering for implementing force calculations in equilibrium conditions.

Magnitude of a Vector

The magnitude of a vector represents its length in space. For a vector \( \mathbf{v} = \langle x, y, z \rangle \), the magnitude is calculated using the formula:\[\| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2}\]
In the provided exercise, we calculated the magnitude of the vector \( \langle 1, 11, -19 \rangle \) by plugging in its components into the magnitude formula:\[\| \langle 1, 11, -19 \rangle \| = \sqrt{1^2 + 11^2 + (-19)^2} = \sqrt{483}\]
Finding the vector's magnitude is essential because it allows us to then shift to the unit vector by scaling the original vector components down to achieve a length of 1. This process is pivotal in vector normalization, a key in creating unit vectors used to depict direction precisely without the effect of size.