Problem 7 - 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 $$\|\vec{v}\|^{2}+\|\vec{w}\|^{2}=\frac{1}{2}\left[\|\vec{v}+\vec{w}\|^{2}+\|\vec{v}-\vec{w}\|^{2}\right]$$ $$ \vec{v}=\left\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right\rangle, \vec{w}=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle $$

Step-by-Step Solution

Answer

The quantities are 0 (scalar), \( \vec{v} - \vec{w} \) (vector), and 2 (scalar), \( \left\langle -\sqrt{2}, \sqrt{2} \right\rangle \) (vector), \( \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \) (vector), and they satisfy the Parallelogram Law.

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

Add the corresponding components of \( \vec{v} \) and \( \vec{w} \): \[ \vec{v} + \vec{w} = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle + \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle = \left\langle 0, 0 \right\rangle \]The result is a vector.

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

Compute by multiplying \( \vec{v} \) by 2 and subtract coordinates:\[ \vec{w} - 2\vec{v} = \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle - \left\langle \sqrt{2}, -\sqrt{2} \right\rangle = \left\langle -\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2} \right\rangle \]The result is a vector.

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

Since \( \vec{v} + \vec{w} = \left\langle 0, 0 \right\rangle \):\[ \|\vec{v} + \vec{w}\| = \sqrt{0^2 + 0^2} = 0 \]The result is a scalar.

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

Calculate the magnitudes separately:\[ \|\vec{v}\| = \sqrt{ \left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2 } = 1 \]\[ \|\vec{w}\| = \sqrt{ \left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 } = 1 \]\[ \|\vec{v}\| + \|\vec{w}\| = 2 \]The result is a scalar.

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

Use the magnitudes calculated previously:\[ \|\vec{v}\| \vec{w} = \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \]\[ \|\vec{w}\| \vec{v} = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \]\[ \|\vec{v}\| \vec{w} - \|\vec{w}\| \vec{v} = \left\langle -\sqrt{2}, \sqrt{2} \right\rangle \]The result is a vector.

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

Since \( \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} \), we find:\[ \|\vec{w}\| \hat{v} = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \]The result is a vector.

Step 7: Verify the Parallelogram Law

Calculate both sides of the equation:Left side: \( \|\vec{v}\|^2 + \|\vec{w}\|^2 = 1^2 + 1^2 = 2 \)Right side:- \( \|\vec{v} + \vec{w}\|^2 = 0^2 = 0 \)- \( \|\vec{v} - \vec{w}\| = 2 \)- \( \|\vec{v} - \vec{w}\|^2 = 2^2 = 4 \)Thus, \( \frac{1}{2}(0 + 4) = 2 \).Both sides are equal, so the Parallelogram Law holds.

Key Concepts

Vector Addition

Vector addition is the process of combining two or more vectors to form a new vector. Let's consider two vectors, \( \vec{v} \) and \( \vec{w} \). Each vector has components along the x and y axes. To add these vectors, you simply add their corresponding components:

• Add the x-components together.
• Add the y-components together.

For instance, if the vectors are \( \vec{v} = \langle a, b \rangle \) and \( \vec{w} = \langle c, d \rangle \), then their vector sum is \( \vec{v} + \vec{w} = \langle a + c, b + d \rangle \). This resultant vector represents the diagonal of the parallelogram that the two vectors form when placed tail-to-tail. It's important to note that the result of vector addition is always another vector.
Vector addition obeys the commutative property, meaning \( \vec{v} + \vec{w} = \vec{w} + \vec{v} \). Try visualizing this by sketching them on a graph to see how their components line up together.

Vector Subtraction

Vector subtraction is pretty similar to vector addition, but instead of adding the corresponding components, you subtract them. This process gives the vector difference, which is still a directional quantity. Suppose you have two vectors \( \vec{v} \) and \( \vec{w} \) given by \( \langle a, b \rangle \) and \( \langle c, d \rangle \) respectively, then:

• Subtract the x-components: \( a - c \).
• Subtract the y-components: \( b - d \).

Thus, the vector difference becomes \( \vec{v} - \vec{w} = \langle a - c, b - d \rangle \). This difference vector can be viewed as the result of reversing one vector's direction and then performing vector addition. Vector subtraction can be used to find vectors pointing from one endpoint to another, which is often useful in navigation or physics problems dealing with displacement or force fields.

Magnitude of a Vector

The magnitude of a vector provides a measure of its length and is a scalar quantity—meaning it has no direction, only size. For a vector \( \vec{v} = \langle x, y \rangle \), the magnitude is calculated using the Pythagorean theorem:

• First, square each component of the vector.
• Then, sum these squared values.
• Finally, take the square root of this sum.

Mathematically, it is denoted by \( \|\vec{v}\| \) and can be expressed as \( \|\vec{v}\| = \sqrt{x^2 + y^2} \). The magnitude tells us only how long the vector is, without any reference to its direction. Understanding this concept is fundamental for applications like physics, where knowing how strong a force is, regardless of the direction it acts in, can determine the outcomes of many calculations.

Parallelogram Law

The Parallelogram Law is a fascinating principle in vector mathematics that provides a geometric approach to the addition of vectors. This law states that the sum of the squares of the magnitudes of two vectors equals half the sum of the squares of the magnitudes of the sum and the difference of the two vectors. Mathematically, it's represented as: \[ \|\vec{v}\|^2 + \|\vec{w}\|^2 = \frac{1}{2} \left[ \|\vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2 \right] \] You can visualize this by placing two vectors tail-to-tail in a plane, forming a parallelogram. The diagonals of the parallelogram represent the sum (\( \vec{v} + \vec{w} \)) and the difference (\( \vec{v} - \vec{w} \)) of the vectors. The Parallelogram Law nicely ties together concepts of magnitude and addition, requiring you to explore both geometric intuition and algebraic manipulation.