Problem 11 - 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. $$ \mathbf{a}=\frac{1}{2} \mathbf{i}-\mathbf{j}+\frac{2}{3} \mathbf{k}, \quad \mathbf{b}=6 \mathbf{i}-12 \mathbf{j}-6 \mathbf{k} $$

Step-by-Step Solution

Answer

A perpendicular vector is \(2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}\), and the unit vector is \(\frac{2}{\sqrt{38}}\mathbf{i} + \frac{5}{\sqrt{38}}\mathbf{j} + \frac{3}{\sqrt{38}}\mathbf{k}\).

Step 1: Compute the Cross Product

To find a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), compute the cross product \( \mathbf{a} \times \mathbf{b} \). Use the determinant formula for the cross product:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{1}{2} & -1 & \frac{2}{3} \6 & -12 & -6\end{vmatrix}\]Calculate the determinant to find:\[\mathbf{a} \times \mathbf{b} = \mathbf{i} \left( (-1)(-6) - (-12) \left( \frac{2}{3} \right) \right) - \mathbf{j} \left( \frac{1}{2}(-6) - 6 \left( \frac{2}{3} \right) \right) + \mathbf{k} \left( \frac{1}{2}(-12) - 6(-1) \right)\]Simplifying gives:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(2) - \mathbf{j}(5) + \mathbf{k}(3) \Rightarrow \mathbf{a} \times \mathbf{b} = 2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}\]

Step 2: Calculate the Magnitude of the Cross Product

Find the magnitude of the vector \( \mathbf{a} \times \mathbf{b} = 2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \) to determine the length of the vector:\[|\mathbf{a} \times \mathbf{b}| = \sqrt{2^2 + 5^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}\]

Step 3: Find the Unit Vector

To find a unit vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), divide the cross product vector by its magnitude:\[\mathbf{u} = \frac{1}{\sqrt{38}}(2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = \frac{2}{\sqrt{38}}\mathbf{i} + \frac{5}{\sqrt{38}}\mathbf{j} + \frac{3}{\sqrt{38}}\mathbf{k}\]

Step 4: Conclusion

The vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is \( 2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \). The unit vector perpendicular to both is \( \frac{2}{\sqrt{38}}\mathbf{i} + \frac{5}{\sqrt{38}}\mathbf{j} + \frac{3}{\sqrt{38}}\mathbf{k} \).

Key Concepts

Unit Vector

A unit vector is essentially a vector that has a magnitude or length of exactly 1. This concept is crucial because unit vectors are often used to indicate direction without scaling the magnitude. To create a unit vector, you take any vector and divide each of its components by the vector's magnitude. This process normalizes the vector, meaning it retains its direction but has a standardized length.
The formula for converting a vector \( \mathbf{v} \) to a unit vector \( \mathbf{u} \) is:

• Calculate the magnitude \( |\mathbf{v}| \) of the vector.
• Divide each component of the vector by this magnitude: \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \).

By normalizing, you can easily illustrate directions within a multi-dimensional space like 2D and 3D, which is particularly useful in physics and engineering.

Vector Magnitude

Vector magnitude, also referred to as the length or size of a vector, is a measure of how long the vector is. It provides a scalar quantity that describes the extent of the vector, distinct from its directional component. To calculate the magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) in three-dimensional space, use the formula: \[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]This formula allows you to determine the hypotenuse of the imaginary triangle that the vector forms with respect to the coordinate axes.

Understanding the magnitude is key in physics and engineering where it often represents quantities like force, velocity, or displacement. The relation between components and magnitude helps in designing solutions and visualizing vectors in practical scenarios.

Perpendicular Vectors

Perpendicular vectors, also called orthogonal vectors, are vectors that meet at a right angle. This characteristic is critical in applications ranging from physics to geometry, where orthogonality can imply independence of directions or separation of influences. By definition, two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular if their dot product is zero: \[ \mathbf{a} \cdot \mathbf{b} = 0 \] However, when you need a vector that is perpendicular to two given vectors, the cross product comes into play. The cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) yields a new vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
This characteristic is particularly useful in 3D space for tasks like finding normals to surfaces or solving mechanical equilibrium problems. Knowing how to find perpendicular vectors, whether through dot products or cross products, is a foundational skill in many scientific and engineering disciplines.