Problem 1 - 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 $$\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]$$ $$ \vec{v}=\langle 12,-5\rangle, \vec{w}=\langle 3,4\rangle $$

Step-by-Step Solution

Answer

The calculations involve both vectors and scalars. The Parallelogram Law is verified.

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

Add the corresponding components of \( \vec{v} = \langle 12, -5 \rangle \) and \( \vec{w} = \langle 3, 4 \rangle \). \( \vec{v} + \vec{w} = \langle 12 + 3, -5 + 4 \rangle = \langle 15, -1 \rangle \).The result \( \langle 15, -1 \rangle \) is a vector.

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

Calculate \( 2\vec{v} \) first by multiplying each component of \( \vec{v} = \langle 12, -5 \rangle \) by 2:\( 2\vec{v} = \langle 24, -10 \rangle \).Then, subtract this from \( \vec{w} = \langle 3, 4 \rangle \):\( \vec{w} - 2\vec{v} = \langle 3 - 24, 4 - (-10) \rangle = \langle -21, 14 \rangle \).The result \( \langle -21, 14 \rangle \) is a vector.

Step 3: Calculate \( \|\vec{v} + \vec{w}\| \)

Use the vector obtained from Step 1: \( \vec{v} + \vec{w} = \langle 15, -1 \rangle \).Calculate its magnitude using the formula \( \|\vec{a}\| = \sqrt{a_x^2 + a_y^2} \):\[ \|\vec{v} + \vec{w}\| = \sqrt{15^2 + (-1)^2} = \sqrt{225 + 1} = \sqrt{226} \].The result \( \sqrt{226} \) is a scalar.

Step 4: Calculate \( \|\vec{v}\| + \|\vec{w}\| \)

Calculate the magnitudes of \( \vec{v} = \langle 12, -5 \rangle \) and \( \vec{w} = \langle 3, 4 \rangle \):\[ \|\vec{v}\| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \].\[ \|\vec{w}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \].Now add the two magnitudes:\( \|\vec{v}\| + \|\vec{w}\| = 13 + 5 = 18 \).The result 18 is a scalar.

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

First, use the magnitudes calculated in Step 4: \( \|\vec{v}\| = 13 \) and \( \|\vec{w}\| = 5 \).Calculate each product vector:\( \|\vec{v}\| \vec{w} = 13 \cdot \langle 3, 4 \rangle = \langle 39, 52 \rangle \).\( \|\vec{w}\| \vec{v} = 5 \cdot \langle 12, -5 \rangle = \langle 60, -25 \rangle \).Subtract the second product from the first:\( \langle 39, 52 \rangle - \langle 60, -25 \rangle = \langle 39 - 60, 52 - (-25) \rangle = \langle -21, 77 \rangle \).The result \( \langle -21, 77 \rangle \) is a vector.

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

Recall that \( \hat{v} \) is the unit vector of \( \vec{v} = \langle 12, -5 \rangle \). First calculate \( \hat{v} \):\( \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{\langle 12, -5 \rangle}{13} = \langle \frac{12}{13}, \frac{-5}{13} \rangle \).Then multiply by the magnitude \( \|\vec{w}\| = 5 \) (from Step 4):\( \|\vec{w}\| \hat{v} = 5 \cdot \langle \frac{12}{13}, \frac{-5}{13} \rangle = \langle \frac{60}{13}, \frac{-25}{13} \rangle \).The result \( \langle \frac{60}{13}, \frac{-25}{13} \rangle \) is a vector.

Step 7: Verify Parallelogram Law

We need to verify that \( \|\vec{v}\|^{2}+\|\vec{w}\|^{2} = \frac{1}{2}[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}] \).Calculate each squared magnitude:\( \|\vec{v}\|^2 = 13^2 = 169 \), \( \|\vec{w}\|^2 = 5^2 = 25 \).Thus, \( 169 + 25 = 194 \).Calculate \( \|\vec{v} + \vec{w}\|^{2} = 226 \) (from Step 3).Calculate \( \|\vec{v} - \vec{w}\|^{2} \) where \( \vec{v} - \vec{w} = \langle 9, -9 \rangle \):\( \|\vec{v} - \vec{w}\| = \sqrt{9^2 + (-9)^2} = \sqrt{162} \).Thus, \( \|\vec{v} - \vec{w}\|^2 = 162 \).Now, \( \frac{1}{2}[226 + 162] = \frac{1}{2}[388] = 194 \), which equals \( 194 \).The vectors satisfy the Parallelogram Law.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is like the length of a line in space. This length is represented via a non-negative scalar. To find the magnitude of a vector, such as \( \vec{v} = \langle a, b \rangle \), we use the formula for the Euclidean norm, \( \|\vec{v}\| = \sqrt{a^2 + b^2} \). This formula emerges from the Pythagorean theorem in two dimensions, providing a distance "straight" across from the origin to the endpoint of the vector.

• This calculation results in a scalar that represents distance or size without direction. This can be particularly useful when comparing the lengths of different vectors or understanding the size of one vector in real-world contexts.
• For example, the magnitude of the vector \( \vec{v} = \langle 12, -5 \rangle \) is computed as \( \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

Understanding this concept helps in many areas of physics and engineering, where determining vector size is crucial for calculations involving forces, velocities, and more.

Parallelogram Law

The Parallelogram Law for vectors explains a fundamental property of vector addition. When two vectors, \( \vec{v} \) and \( \vec{w} \), are positioned such that their tails coincide, the vector sum \( \vec{v} + \vec{w} \) forms the diagonal of the parallelogram. This geometric representation underscores the commutative nature of vector addition—meaning \( \vec{v} + \vec{w} = \vec{w} + \vec{v} \).

• The associated formula \( \|\vec{v}\|^2 + \|\vec{w}\|^2 = \frac{1}{2} \left[ \|\vec{v}+\vec{w}\|^2 + \|\vec{v}-\vec{w}\|^2 \right] \) is about verifying that the sum of the squares of the magnitudes of two vectors equals half the sum of the squares of the magnitudes of their sums and differences.
• This law arises naturally in physics, for example, in scenarios where forces apply concurrently but not collinearly, allowing for the effective calculation of resultant vectors.

Appreciating the Parallelogram Law supports understanding how different vector quantities interact spatially and influence one another.

Unit Vector

A unit vector is a vector of length 1, often used to specify a direction without indicating magnitude or size. It is derived by dividing any vector by its magnitude, essentially scaling it down to a length of 1 while maintaining its direction. For vector \( \vec{v} = \langle a, b \rangle \), the unit vector, noted as \( \hat{v} \), is calculated by \[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \left\langle \frac{a}{\|\vec{v}\|}, \frac{b}{\|\vec{v}\|} \right\rangle \].

• Unit vectors are crucial in defining coordinate system directions, such as \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) for the x, y, and z directions in three-dimensional space.
• In navigation and computer graphics, they help direct vectors along desired axes without scaling them—merely indicating orientation.

Knowing how to calculate and use unit vectors allows us to normalize vectors in various calculations, making them a vital tool in trigonometry and vector-based analyses.