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

Step-by-Step Solution

Answer

\(\vec{v} + \vec{w} = \langle 0, 3 \rangle\), another vector. The Parallelogram Law is verified.

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

To add the vectors \( \vec{v} \) and \( \vec{w} \), add their corresponding components. \[ \vec{v} + \vec{w} = \langle 2, -1 \rangle + \langle -2, 4 \rangle = \langle 2 + (-2), -1 + 4 \rangle = \langle 0, 3 \rangle \] This result is a vector.

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

First, scale \( \vec{v} \) by 2 and then subtract it from \( \vec{w} \): \[ 2\vec{v} = 2\langle 2, -1 \rangle = \langle 4, -2 \rangle \] \[ \vec{w} - 2\vec{v} = \langle -2, 4 \rangle - \langle 4, -2 \rangle = \langle -2 - 4, 4 + 2 \rangle = \langle -6, 6 \rangle \] This result is a vector.

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

The magnitude of the vector \( \vec{v} + \vec{w} = \langle 0, 3 \rangle \) is calculated as follows: \[ \|\vec{v} + \vec{w}\| = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \] This result is a scalar.

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

First, find the magnitudes of the individual vectors \( \vec{v} \) and \( \vec{w} \): \[ \|\vec{v}\| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \] \[ \|\vec{w}\| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \] Add them together: \[ \|\vec{v}\| + \|\vec{w}\| = \sqrt{5} + 2\sqrt{5} = 3\sqrt{5} \] This result is a scalar.

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

First, use the magnitudes from Step 4, then compute: \[ \|\vec{v}\| \vec{w} = \sqrt{5} \langle -2, 4 \rangle = \langle -2\sqrt{5}, 4\sqrt{5} \rangle \] \[ \|\vec{w}\| \vec{v} = 2\sqrt{5} \langle 2, -1 \rangle = \langle 4\sqrt{5}, -2\sqrt{5} \rangle \] Then subtract: \[ \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} = \langle -2\sqrt{5} - 4\sqrt{5}, 4\sqrt{5} + 2\sqrt{5} \rangle = \langle -6\sqrt{5}, 6\sqrt{5} \rangle \] This result is a vector.

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

First, find the unit vector \( \hat{v} \) by dividing \( \vec{v} \) by its magnitude: \[ \hat{v} = \frac{1}{\sqrt{5}} \langle 2, -1 \rangle = \langle \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \rangle \] Then, multiply by \( \|\vec{w}\| \): \[ \|\vec{w}\| \hat{v} = 2\sqrt{5} \langle \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \rangle = \langle 4, -2 \rangle \] This result is a vector.

Step 7: Verify Parallelogram Law

First, calculate all terms: \[ \|\vec{v}\|^2 = 5, \quad \|\vec{w}\|^2 = 20 \] \[ \|\vec{v} - \vec{w}\|^2 = \|\langle 4, -5 \rangle\|^2 = 4^2 + (-5)^2 = 41 \] \[ \|\vec{v} + \vec{w}\|^2 = \|\langle 0, 3 \rangle\|^2 = 0^2 + 3^2 = 9 \] Now verify the equation: \[ 5 + 20 = \frac{1}{2}(9 + 41) \] \[ 25 = \frac{1}{2}(50) \] \[ 25 = 25 \] Thus, the Parallelogram Law holds.

Key Concepts

Magnitude of a Vector

When dealing with vectors, one of the foundational concepts is understanding their magnitude. The magnitude of a vector is a measure of its length and is denoted as \( \|\vec{v}\| \). It's the equivalent of finding the length of a line segment between the vector's origin and its endpoint in a coordinate plane. To find the magnitude, we use the Pythagorean theorem, applying it to the vector's coordinates.

For a vector \( \vec{v} = \langle a, b \rangle \), its magnitude is computed as:

• \( \|\vec{v}\| = \sqrt{a^2 + b^2} \)

This involves squaring each component, summing them, and then taking the square root of the result. The magnitude is always a scalar quantity, representing just the size, not the direction. In our example with \( \vec{v} = \langle 2, -1 \rangle \), the magnitude is \( \sqrt{5} \). This number tells us how long the vector is without pointing us toward any direction.

Vector Addition

Adding vectors involves combining their individual components. When you add two vectors together, you're essentially finding a new vector that represents the combined effect of both original vectors. The process is straightforward: add the corresponding components of each vector.

Given two vectors, \( \vec{v} = \langle a, b \rangle \) and \( \vec{w} = \langle c, d \rangle \), the sum \( \vec{v} + \vec{w} \) is computed component-wise:

• \( \vec{v} + \vec{w} = \langle a + c, b + d \rangle \)

It's crucial to note that the result of vector addition is another vector. In our original example, by adding \( \vec{v} = \langle 2, -1 \rangle \) and \( \vec{w} = \langle -2, 4 \rangle \), we obtain the vector \( \langle 0, 3 \rangle \). This new vector represents the total effect of both vectors when acting together.

Parallelogram Law

The Parallelogram Law is a geometric representation of vector addition. It states that if you place two vectors \( \vec{v} \) and \( \vec{w} \) to have the same initial point, they form two sides of a parallelogram. The diagonals of this parallelogram represent the sums and differences of the two vectors.

Mathematically, the law can be verified using the expression:

• \( \|\vec{v}\|^2 + \|\vec{w}\|^2 = \frac{1}{2}\left[\|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right] \)

This equation shows that the sum of the squares of the magnitudes of the vectors is equal to half the sum of the squares of the magnitudes of the diagonals. In our case, we confirmed it held true with vectors identified as \( \vec{v} \) and \( \vec{w} \). This property highlights the interconnection between vector geometry and algebra.

Vector Subtraction

Subtracting vectors is slightly different from adding them, but it follows a similar logic. The idea is to find a vector that represents the change from one vector to another. To perform vector subtraction, subtract the corresponding components of the two vectors.

For vectors \( \vec{v} = \langle a, b \rangle \) and \( \vec{w} = \langle c, d \rangle \), the difference \( \vec{v} - \vec{w} \) can be calculated as:

• \( \vec{v} - \vec{w} = \langle a - c, b - d \rangle \)

The result is again a vector, indicating how much you move from one vector's position to the other. In our exercise, when we computed \( \vec{w} - 2\vec{v} \), we essentially scaled \( \vec{v} \) by \( 2 \) before subtracting it from \( \vec{w} \), resulting in the vector \( \langle -6, 6 \rangle \). This operation helps in understanding the relative positioning of vectors.