Problem 4 - 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 $$\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]$$ $$ \vec{v}=\langle 10,4\rangle, \vec{w}=\langle-2,5\rangle $$

Step-by-Step Solution

Answer

The results are: 1. \(\langle 8, 9 \rangle\) (Vector) 2. \(\langle -22, -3 \rangle\) (Vector) 3. \(\sqrt{145}\) (Scalar) 4. \(\sqrt{116} + \sqrt{29}\) (Scalar) 5. Vector 6. Vector The Parallelogram Law is verified.

Step 1: Adding the vectors

To find \(\vec{v} + \vec{w}\), we add the corresponding components of the vectors \(\vec{v}\) and \(\vec{w}\): \[ \vec{v} + \vec{w} = \langle 10, 4 \rangle + \langle -2, 5 \rangle = \langle 10 + (-2), 4 + 5 \rangle = \langle 8, 9 \rangle \]. This result is a vector.

Step 2: Subtracting the vectors with scaling

For \(\vec{w} - 2\vec{v}\), first scale \(\vec{v}\) by 2: \(2\vec{v} = 2 \times \langle 10, 4 \rangle = \langle 20, 8 \rangle\). Now, subtract it from \(\vec{w}\): \(\vec{w} - 2\vec{v} = \langle -2, 5 \rangle - \langle 20, 8 \rangle = \langle -2 - 20, 5 - 8 \rangle = \langle -22, -3 \rangle\). This result is a vector.

Step 3: Magnitude of the sum of vectors

Calculate \(\|\vec{v} + \vec{w}\|\) using the result from Step 1: \[ \|\vec{v} + \vec{w}\| = \sqrt{8^2 + 9^2} = \sqrt{64 + 81} = \sqrt{145} \]. This result is a scalar.

Step 4: Sum of magnitudes

Find the magnitudes \(\|\vec{v}\|\) and \(\|\vec{w}\|\):- \(\|\vec{v}\| = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116}\)- \(\|\vec{w}\| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}\)Then, calculate \(\|\vec{v}\| + \|\vec{w}\| = \sqrt{116} + \sqrt{29}\). This result is a scalar.

Step 5: Scaling and subtracting vectors

First, compute \(\|\vec{v}\| \vec{w}\):\[ \|\vec{v}\| \vec{w} = \sqrt{116} \langle -2, 5 \rangle = \langle -2\sqrt{116}, 5\sqrt{116} \rangle \]Then \(\|\vec{w}\| \vec{v}\):\[ \|\vec{w}\| \vec{v} = \sqrt{29} \langle 10, 4 \rangle = \langle 10\sqrt{29}, 4\sqrt{29} \rangle \]Finally, subtract:\[ \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} = \langle -2\sqrt{116}, 5\sqrt{116} \rangle - \langle 10\sqrt{29}, 4\sqrt{29} \rangle \]\[ = \langle -2\sqrt{116} - 10\sqrt{29}, 5\sqrt{116} - 4\sqrt{29} \rangle \]. This result is a vector.

Step 6: Scaling a unit vector

Calculate the unit vector \(\hat{v}\) by dividing \(\vec{v}\) by its magnitude \(\|\vec{v}\|\): \[ \hat{v} = \frac{1}{\sqrt{116}} \langle 10, 4 \rangle = \langle \frac{10}{\sqrt{116}}, \frac{4}{\sqrt{116}} \rangle \].Now scale by \(\|\vec{w}\|\):\(\|\vec{w}\| \hat{v} = \sqrt{29} \langle \frac{10}{\sqrt{116}}, \frac{4}{\sqrt{116}} \rangle = \langle \frac{10\sqrt{29}}{\sqrt{116}}, \frac{4\sqrt{29}}{\sqrt{116}} \rangle \). This result is a vector.

Step 7: Verifying the Parallelogram Law

First compute \(\|\vec{v} - \vec{w}\|\):\[\vec{v} - \vec{w} = \langle 10, 4 \rangle - \langle -2, 5 \rangle = \langle 12, -1 \rangle \]\(\|\vec{v} - \vec{w}\| = \sqrt{12^2 + (-1)^2} = \sqrt{144 + 1} = \sqrt{145}\).Now verify: \[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = 116 + 29 = 145 \]The right side is:\[ \frac{1}{2} \left(\|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right) = \frac{1}{2} (145 + 145) = 145 \]Both sides are equal, verifying the law.

Key Concepts

Vector Addition

Vector addition is a fundamental operation that allows us to combine two vectors to create a new vector. Think of it like adding two arrows on a graph, where each arrow represents a vector. The end point of the new vector represents the sum of the two original vectors. In this exercise, the vectors \( \vec{v} = \langle 10, 4 \rangle \) and \( \vec{w} = \langle -2, 5 \rangle \) are added by combining their corresponding components:

• Add the x-components: \( 10 + (-2) = 8 \)
• Add the y-components: \( 4 + 5 = 9 \)

The resulting vector is \( \langle 8, 9 \rangle \). This illustrates that vector addition results in another vector. The new vector essentially represents a direction and magnitude derived from both \( \vec{v} \) and \( \vec{w} \). Remember, components are simply projections on the axes, and adding them directly gives the components of the resultant vector.

Vector Subtraction

Vector subtraction works similarly to vector addition, but think of it as reversing one of the vectors before combining them. In this exercise, we are given \( \vec{w} - 2\vec{v} \). This means we scale \( \vec{v} \) by 2 and then subtract it from \( \vec{w} \).

• Scale \( \vec{v} \) by 2: \( 2\vec{v} = \langle 20, 8 \rangle \)
• Subtract from \( \vec{w} \): \( \vec{w} - 2\vec{v} = \langle -2, 5 \rangle - \langle 20, 8 \rangle = \langle -22, -3 \rangle \)

The result is \( \langle -22, -3 \rangle \), a vector. In essence, subtraction here gives us a vector that represents the directional and magnitude difference between \( \vec{w} \) and the scaled \( \vec{v} \). It's as if \( \vec{w} \) moves a certain way, and scaling \( \vec{v} \) will shift in the opposite direction.

Magnitude of Vectors

The magnitude of a vector is like its length or size. It is crucial because it quantifies how long a vector is without considering direction. You compute it using the Pythagorean Theorem:

• For \( \vec{v} \), \( \|\vec{v}\| = \sqrt{10^2 + 4^2} = \sqrt{116} \)
• For \( \vec{w} \), \( \|\vec{w}\| = \sqrt{(-2)^2 + 5^2} = \sqrt{29} \)

The magnitude is always a positive number and is a scalar. Calculating the magnitude gives a clearer understanding of a vector's contribution by focusing on its absolute capacity without direction. When considering the combination of vectors, like \( \|\vec{v} + \vec{w}\| \), you apply the same concept to the resulting vector \( \langle 8, 9 \rangle \) to find its magnitude.

Parallelogram Law

The Parallelogram Law is a visible conceptual method for understanding the addition of two vectors. It states that for any two vectors \( \vec{v} \) and \( \vec{w} \), the sum of the squares of their magnitudes equals one half of the sum of the squares of the lengths of the diagonals of the parallelogram they form.Mathematically: \[\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]\]When vectors are arranged tail-to-tail, they form a parallelogram. The formula essentially expresses how the vector sums and differences relate geometrically. In the original exercise using the specific vectors provided, this law holds true. Both sides of the equation are equal to 145, confirming your understanding. The Parallelogram Law is a practical representation, helping visualize how vector actions translate into geometric forms.