Problem 8 - College Trigonometry

Question

Use the given pair of vectors \(\vec{v}\) and \(\vec{w}\) to find the following quantities. State whether the result is a vector or a scalar. $$ \begin{array}{lllll} \bullet \vec{v}+\vec{w} & \bullet \vec{w}-2 \vec{v} & \bullet\|\vec{v}+\vec{w}\| & \bullet\|\vec{v}\|+\|\vec{w}\| & \bullet\|\vec{v}\| \vec{w}-\|\vec{w}\| \vec{v} & \bullet\|\vec{w}\| \hat{v} \end{array} $$ Finally, verify that the vectors satisfy the Parallelogram Law $$\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]$$ $$ \vec{v}=\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle, \vec{w}=\langle-1,-\sqrt{3}\rangle $$

Step-by-Step Solution

Answer

The vectors do not satisfy the Parallelogram Law; their calculations differ.

Step 1: Finding \( \vec{v} + \vec{w} \)

To find \( \vec{v} + \vec{w} \), simply add the corresponding components of \( \vec{v} \) and \( \vec{w} \).\[ \vec{v} + \vec{w} = \left( \frac{1}{2} + (-1), \frac{\sqrt{3}}{2} + (-\sqrt{3}) \right) = \left( \frac{1}{2} - 1, \frac{\sqrt{3}}{2} - \sqrt{3} \right) \]Thus, \( \vec{v} + \vec{w} = \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \), and it is a vector.

Step 2: Finding \( \vec{w} - 2\vec{v} \)

Multiply \( \vec{v} \) by 2 and subtract from \( \vec{w} \):\[ 2\vec{v} = \left( 2 \times \frac{1}{2}, 2 \times \frac{\sqrt{3}}{2} \right) = (1, \sqrt{3}) \]\[ \vec{w} - 2\vec{v} = \left( -1 - 1, -\sqrt{3} - \sqrt{3} \right) = (-2, -2\sqrt{3}) \]This result is a vector.

Step 3: Calculating the norm \( \|\vec{v} + \vec{w}\| \)

The norm of a vector is its magnitude. For a vector \( \vec{u} = (a, b) \), the norm is \( \|\vec{u}\| = \sqrt{a^2 + b^2} \).\[ \|\vec{v} + \vec{w}\| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]The result is a scalar.

Step 4: Sum of norms \( \|\vec{v}\| + \|\vec{w}\| \)

Find the norms \( \|\vec{v}\| \) and \( \|\vec{w}\| \), and add them up.\[ \|\vec{v}\| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]\[ \|\vec{w}\| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]\[ \|\vec{v}\| + \|\vec{w}\| = 1 + 2 = 3 \]The result is a scalar.

Step 5: Vector operation \( \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} \)

Calculate the individual products and then subtract.\[ \|\vec{v}\| \vec{w} = 1 \times \langle -1, -\sqrt{3} \rangle = \langle -1, -\sqrt{3} \rangle \]\[ \|\vec{w}\| \vec{v} = 2 \times \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle = \langle 1, \sqrt{3} \rangle \]\[ \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} = \langle -1, -\sqrt{3} \rangle - \langle 1, \sqrt{3} \rangle = \langle -2, -2\sqrt{3} \rangle \]The result is a vector.

Step 6: Calculating \( \|\vec{w}\| \hat{v} \)

First, find the unit vector \( \hat{v} \), which is \( \vec{v} \) divided by its norm.\[ \hat{v} = \frac{1}{1} \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle = \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle \]Then multiply by \( \|\vec{w}\| \):\[ \|\vec{w}\| \hat{v} = 2 \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle = \langle 1, \sqrt{3} \rangle \]This is a vector.

Step 7: Verify Parallelogram Law

Substitute the vector norms into the equation for the Parallelogram Law and check.\[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = 1^2 + 2^2 = 1 + 4 = 5 \]The right side of the law:\[ \frac{1}{2}\left[\|\vec{v}+\vec{w}\|^2 + \|\vec{v}-\vec{w}\|^2\right] \]\[ \|\vec{v}+\vec{w}\|^2 = 1^2 = 1 \]\( \vec{v}-\vec{w} = \left( \frac{1}{2} - (-1), \frac{\sqrt{3}}{2} - (-\sqrt{3}) \right) = \left( \frac{3}{2}, \frac{5\sqrt{3}}{2} \right) \)\[ \|\vec{v}-\vec{w}\| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{75}{4}} = 6 \]\[ \frac{1}{2}[1 + 6] = \frac{1}{2} \times 7 = 3.5 \]Since 5 is not equal to 3.5, the vectors do not satisfy the Parallelogram Law.

Key Concepts

Vector Addition

Vector addition is a fundamental operation in vector mathematics that involves combining two or more vectors to produce another vector. This operation is essential when working with concepts that need to understand the combined effect of different forces or changes over a space. When adding vectors, make sure to observe the following:

• Vectors are added component-wise. For instance, with vectors \( \vec{v} = \langle a, b \rangle \) and \( \vec{w} = \langle c, d \rangle \), addition will result in a new vector, \( \vec{v} + \vec{w} = \langle a+c, b+d \rangle \).
• The resulting vector has a direction and magnitude that are derived from the components of the original vectors.

The properties of vector addition include commutativity and associativity, meaning you can add vectors in any order and group them differently without affecting the final result. Understanding vector addition is crucial for computing resultant vectors, which is essential in many fields like physics and engineering.

Magnitude of a Vector

The magnitude of a vector, also known as the vector’s length or norm, measures how long or intense the vector is, irrespective of its direction. The magnitude is calculated using the formula:\[ \|\vec{v}\| = \sqrt{a^2 + b^2} \]where \( a \) and \( b \) are the components of the vector \( \vec{v} = \langle a, b \rangle \). Calculating the vector's magnitude gives you a scalar value, which helps you understand the vector's size in its vector space.
It is crucial for:

• Determining vector length.
• Calculating distances in coordinate systems.
• Normalizing a vector, which involves converting a vector to a unit vector while retaining its direction.

Knowing how to compute a vector’s magnitude enables you to effectively visualize and understand vector quantities.

Parallelogram Law

The Parallelogram Law is a geometric rule that helps understand the relationship between vectors. It states that for any two vectors, the sum of the squares of their magnitudes is equal to the sum of the squares of the diagonals of the parallelogram formed by them. Mathematically, this is expressed as:\[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = \frac{1}{2}\left[ \|\vec{v}+\vec{w}\|^2 + \|\vec{v}-\vec{w}\|^2 \right] \]This law aids in establishing a critical way to visualize vector addition geometrically by comparing the triangle’s diagonal's length created by two vectors against the individual vector's lengths.

• Helps derive the resultant vector’s magnitude.
• Facilitates understanding of vector composition and resolution in different applications.

Though one exercise may not meet the law due to calculation nuances or specific conditions, the Parallelogram Law remains an essential tool for evaluating vector space and relationships.

Unit Vector

A unit vector is a vector that has a magnitude of exactly one unit. It is used to indicate the direction of a vector without considering its magnitude. To convert any vector \( \vec{v} \) into a unit vector \( \hat{v} \), divide the original vector by its magnitude:\[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} \]The resulting unit vector keeps its original direction but rescales the magnitude to one.

• Unit vectors serve as essential components for defining direction in vector spaces.
• They help normalize vectors, which is critical in computing, graphics, and physics simulations.
• Used often in trigonometric representations of vector directions and in forming basis vectors in coordinate systems.

Understanding unit vectors is fundamental for decomposing vector components as well as for achieving directionality in various scientific computations.