Problem 6 - 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 $$\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]$$ $$ \vec{v}=\left\langle\frac{3}{5}, \frac{4}{5}\right\rangle, \vec{w}=\left\langle-\frac{4}{5}, \frac{3}{5}\right\rangle $$

Step-by-Step Solution

Answer

The calculations yield vectors and scalars, confirming the Parallelogram Law.

Step 1: Calculating Vector Addition

Add the corresponding components of \(\vec{v}\) and \(\vec{w}\):\[ \vec{v} + \vec{w} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle + \left\langle -\frac{4}{5}, \frac{3}{5} \right\rangle = \left\langle \frac{3}{5} - \frac{4}{5}, \frac{4}{5} + \frac{3}{5} \right\rangle = \left\langle -\frac{1}{5}, \frac{7}{5} \right\rangle \]The result is a vector: \(\left\langle -\frac{1}{5}, \frac{7}{5} \right\rangle\).

Step 2: Calculating a Linear Combination

Compute \(\vec{w} - 2\vec{v}\):\[ \vec{w} - 2\vec{v} = \left\langle -\frac{4}{5}, \frac{3}{5} \right\rangle - 2 \times \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \left\langle -\frac{4}{5} - \frac{6}{5}, \frac{3}{5} - \frac{8}{5} \right\rangle = \left\langle -2, -1 \right\rangle \]The result is a vector: \(\left\langle -2, -1 \right\rangle\).

Step 3: Calculating Magnitude of Sum of Vectors

First, calculate \(\vec{v} + \vec{w}\) as in Step 1: \(\left\langle -\frac{1}{5}, \frac{7}{5} \right\rangle\).Find its magnitude:\[ \| \vec{v} + \vec{w} \| = \sqrt{\left(-\frac{1}{5}\right)^2 + \left(\frac{7}{5}\right)^2} = \sqrt{\frac{1}{25} + \frac{49}{25}} = \sqrt{\frac{50}{25}} = \sqrt{2} \]The result is a scalar: \(\sqrt{2}\).

Step 4: Sum of Magnitudes

Calculate the magnitudes separately:For \(\vec{v}\):\[ \| \vec{v} \| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1 \]For \(\vec{w}\):\[ \| \vec{w} \| = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{9}{25}} = \sqrt{1} = 1 \]Thus, \(\| \vec{v} \| + \| \vec{w} \| = 1 + 1 = 2\).The result is a scalar: \(2\).

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

Using step 4 results \(\|\vec{v}\| = 1\) and \(\|\vec{w}\| = 1\):\[ \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} = 1 \cdot \left\langle -\frac{4}{5}, \frac{3}{5} \right\rangle - 1 \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \left\langle -\frac{4}{5} - \frac{3}{5}, \frac{3}{5} - \frac{4}{5} \right\rangle = \left\langle -\frac{7}{5}, -\frac{1}{5} \right\rangle \]The result is a vector: \(\left\langle -\frac{7}{5}, -\frac{1}{5} \right\rangle\).

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

First, find the unit vector \(\hat{v}\):\[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle}{1} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \]Now compute \(\|\vec{w}\| \hat{v}\):\[ \|\vec{w}\| \hat{v} = 1 \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \]The result is a vector: \(\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle\).

Step 7: Verify Parallelogram Law

First calculate \(\|\vec{v} - \vec{w}\|\):\[ \vec{v} - \vec{w} = \left\langle \frac{3}{5} - \left(-\frac{4}{5}\right), \frac{4}{5} - \frac{3}{5} \right\rangle = \left\langle \frac{7}{5}, \frac{1}{5} \right\rangle \]\[ \|\vec{v} - \vec{w}\| = \sqrt{\left(\frac{7}{5}\right)^2 + \left(\frac{1}{5}\right)^2} = \sqrt{\frac{49}{25} + \frac{1}{25}} = \sqrt{2} \]Now check the formula:\[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = 1^2 + 1^2 = 2 \]\[ \frac{1}{2}[\|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2] = \frac{1}{2}[2 + 2] = 2 \]Both sides equal 2, confirming the Parallelogram Law is satisfied.

Key Concepts

Vector Addition

When you add vectors, you essentially combine their magnitudes and directions. For two vectors \( \vec{v} \) and \( \vec{w} \), vector addition involves adding their respective components separately. This is done by taking the first component of \( \vec{v} \) and adding it to the first component of \( \vec{w} \), and doing the same for the second components. This results in a new vector with these summed components.
For example, if \( \vec{v} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \) and \( \vec{w} = \left\langle -\frac{4}{5}, \frac{3}{5} \right\rangle \), their vector addition is:

• For the x-component: \( \frac{3}{5} - \frac{4}{5} = -\frac{1}{5} \)
• For the y-component: \( \frac{4}{5} + \frac{3}{5} = \frac{7}{5} \)

So, \( \vec{v} + \vec{w} = \left\langle -\frac{1}{5}, \frac{7}{5} \right\rangle \). This result is a new vector.

Vector Magnitudes

The magnitude of a vector represents its length and is a scalar. It is calculated using the Pythagorean Theorem, which is akin to finding the hypotenuse of a right triangle. For a vector \( \vec{v} = \langle a, b \rangle \), the magnitude \( \| \vec{v} \| \) is calculated as follows:
\[ \| \vec{v} \| = \sqrt{a^2 + b^2} \]
Applying this to our vectors:

• Magnitude of \( \vec{v} \):
  \[ \| \vec{v} \| = \sqrt{ \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 } = 1 \]
• Magnitude of \( \vec{w} \):
  \[ \| \vec{w} \| = \sqrt{ \left(-\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2 } = 1 \]

Remember, the magnitude is always a positive number, representing the vector's length.

Parallelogram Law

The Parallelogram Law is a beautiful way of verifying vector relationships. It states that the sum of the squares of the magnitudes of two vectors is equal to half the sum of the squares of the magnitudes of their sum and difference. If \( \vec{v} \) and \( \vec{w} \) are two vectors, this law can be expressed mathematically as:
\[ \|\vec{v}\|^{2} + \|\vec{w}\|^{2} = \frac{1}{2} \left[ \|\vec{v} + \vec{w}\|^{2} + \|\vec{v} - \vec{w}\|^{2} \right] \]
In our example:

• Compute \( \|\vec{v} - \vec{w}\| \) and \( \|\vec{v} + \vec{w}\| \)
• Verify \[ 1^2 + 1^2 = 2 \] equals \[ \frac{1}{2}[2 + 2] = 2 \]

This confirms that \( \vec{v} \) and \( \vec{w} \) follow the Parallelogram Law.

Unit Vectors

Unit vectors serve the purpose of indicating direction without considering magnitude. The magnitude of a unit vector is always 1. To find a unit vector \( \hat{v} \) from a vector \( \vec{v} \), divide \( \vec{v} \) by its magnitude:
\[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} \]
For instance, with \( \vec{v} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \) and \( \|\vec{v}\| = 1 \), the unit vector is:
\[ \hat{v} = \frac{\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle}{1} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \]
Using unit vectors helps in scaling the direction to maintain consistency across different operations.