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

Step-by-Step Solution

Answer

The results are: vector, vector, scalar, scalar, vector, vector. The law is verified.

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

The sum of two vectors \( \vec{v} = \langle -\sqrt{3}, 1 \rangle \) and \( \vec{w} = \langle 2\sqrt{3}, 2 \rangle \) is computed component-wise: \( \vec{v} + \vec{w} = \langle -\sqrt{3} + 2\sqrt{3}, 1 + 2 \rangle = \langle \sqrt{3}, 3 \rangle \). This result is a vector.

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

Multiply vector \( \vec{v} \) by 2: \( 2\vec{v} = \langle -2\sqrt{3}, 2 \rangle \). Then subtract from \( \vec{w} \): \( \vec{w} - 2\vec{v} = \langle 2\sqrt{3}, 2 \rangle - \langle -2\sqrt{3}, 2 \rangle = \langle 4\sqrt{3}, 0 \rangle \). This result is a vector.

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

First, find \( \|\vec{v}+\vec{w}\| = \|\langle \sqrt{3}, 3 \rangle\| \). Compute the magnitude: \( \sqrt{(\sqrt{3})^2 + (3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \). This result is a scalar.

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

Compute \( \|\vec{v}\| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2 \) and \( \|\vec{w}\| = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12+4} = 4 \). Summing these magnitudes gives \( 2 + 4 = 6 \). This result is a scalar.

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

Substitute magnitudes: \( \|\vec{v}\| = 2 \) and \( \|\vec{w}\| = 4 \). Compute: \( 2 \langle 2\sqrt{3}, 2 \rangle - 4 \langle -\sqrt{3}, 1 \rangle = \langle 4\sqrt{3}, 4 \rangle + \langle 4\sqrt{3}, 4 \rangle = \langle 8\sqrt{3}, 8 \rangle \). This result is a vector.

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

Unit vector \( \hat{v} = \frac{1}{2}\langle -\sqrt{3}, 1 \rangle \). Then \( \|\vec{w}\| \hat{v} = 4 \times \frac{1}{2} \langle -\sqrt{3}, 1 \rangle = 2 \langle -\sqrt{3}, 1 \rangle = \langle -2\sqrt{3}, 2 \rangle \). This result is a vector.

Step 7: Verify Parallelogram Law

Calculate each part: \( \|\vec{v}\|^2 = 4 \), \( \|\vec{w}\|^2 = 16 \), \( \|\vec{v} + \vec{w}\|^2 = 12 \), and \( \|\vec{v} - \vec{w}\| = \|\langle -3\sqrt{3}, -1 \rangle\| = \sqrt{27+1} = \sqrt{28} = 2\sqrt{7} \). Hence: \( \|\vec{v} - \vec{w}\|^2 = 28 \). Verify: \( 4 + 16 = 20 = \frac{1}{2}(12 + 28) = 20 \), confirming the law.

Key Concepts

Understanding Vector Addition

Vectors are mathematical objects that have both magnitude and direction. To add vectors, perform component-wise addition, meaning you add the corresponding components of the vectors separately. Let's consider vectors \(\vec{v} = \langle -\sqrt{3}, 1 \rangle\) and \(\vec{w} = \langle 2\sqrt{3}, 2 \rangle\). When adding these vectors, sum the \(x\)-components and \(y\)-components separately:

• The \(x\)-component: \(-\sqrt{3} + 2\sqrt{3} = \sqrt{3}\).
• The \(y\)-component: \(1 + 2 = 3\).

This gives the resulting vector as \(\langle \sqrt{3}, 3 \rangle\). This result is indeed a vector, as it retains both direction and magnitude.

Grasping Vector Subtraction

Vector subtraction follows a similar component-wise approach as vector addition, but here, we subtract each corresponding component. If you have a vector \(\vec{w} = \langle 2\sqrt{3}, 2 \rangle\) and you want to subtract double of the vector \(\vec{v} = \langle -\sqrt{3}, 1 \rangle\), you first multiply \(\vec{v}\) by 2, resulting in \(2\vec{v} = \langle -2\sqrt{3}, 2 \rangle\). Next, subtract these calculated components:

• The \(x\)-component: \(2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}\).
• The \(y\)-component: \(2 - 2 = 0\).

Thus, the resulting vector from this subtraction is \(\langle 4\sqrt{3}, 0 \rangle\). Again, this result is a vector, complete with both direction and length.

Finding Vector Magnitude

The magnitude of a vector, often called its length, is a measure of how long the vector is. It doesn't account for direction, only size and is found using the Pythagorean theorem. For any vector \(\langle a, b \rangle\), its magnitude \(\| \langle a, b \rangle \|\) is calculated by: \[ \sqrt{a^2 + b^2} \]. For example, for the vector \(\vec{v} = \langle -\sqrt{3}, 1 \rangle\):

• \(\| \vec{v} \| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\).

This same approach applies to other vectors as well. Therefore, the magnitude is a scalar, providing the vector's size but not its direction.

The Parallelogram Law Explained

The Parallelogram Law is a fundamental concept that describes the relationship of magnitudes in a parallelogram formed by two vectors. In simple terms, this law states: "The sum of the squares of the magnitudes of two vectors is equal to one half the sum of the squares of the magnitudes of their sum and their difference." This is mathematically expressed as: \[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = \frac{1}{2}\left[\|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right] \] It's a useful check for problems involving vector operations. Consider vectors \(\vec{v}\) and \(\vec{w}\) forming a parallelogram:

• Calculate \(\|\vec{v}\|^2 + \|\vec{w}\|^2\) and ensure it matches \(\frac{1}{2}\left[\|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right]\).

For our vectors, verification of this law showed both sides of the equation equal 20, demonstrating the principle's accuracy in the scenario. This clarifies that vector operations within a plane behave consistently with geometric principles.