Problem 20 - Precalculus

Question

Assume that the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c},\) and \(\mathbf{d}\) are defined as follows: $$\mathbf{a}=\langle 2,3\rangle \quad \mathbf{b}=\langle 5,4\rangle \quad \mathbf{c}=\langle 6,-1\rangle \quad \mathbf{d}=\langle-2,0\rangle$$ Compute each of the indicated quantities. $$.1 /|\mathbf{d}|$$

Step-by-Step Solution

Answer

The value is \( \frac{1}{2} \).

Step 1: Understand the Vector

Vector \( \mathbf{d} \) is given as \( \langle -2, 0 \rangle \). This means that the vector has components \( x = -2 \) and \( y = 0 \).

Step 2: Use the Magnitude Formula

The magnitude (or length) of a vector \( \mathbf{v} = \langle x, y \rangle \) is computed with the formula: \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). Here, apply this to vector \( \mathbf{d} = \langle -2, 0 \rangle \).

Step 3: Substitute and Compute

Substitute the components of \( \mathbf{d} \) into the formula: \( |\mathbf{d}| = \sqrt{(-2)^2 + (0)^2} \). Compute the squared components and the square root.

Step 4: Calculate the Output

Compute \((-2)^2 = 4\) and \((0)^2 = 0\). Thus, \(|\mathbf{d}| = \sqrt{4 + 0} = \sqrt{4}\). This simplifies to \(|\mathbf{d}| = 2\).

Step 5: Apply Division

The problem asks for \( \frac{1}{|\mathbf{d}|} \). Since \(|\mathbf{d}| = 2\), we have \( \frac{1}{2} \).

Key Concepts

Vector Components

Every vector can be represented by its components. For vectors in a two-dimensional space, like in this exercise, components are often labeled as \( x \) and \( y \).

For example, the vector \( \mathbf{d} = \langle -2, 0 \rangle \) has components:

• \( x = -2 \)
• \( y = 0 \)

These components tell us that the vector has a horizontal movement of \(-2\) units. Meanwhile, it has no vertical movement. These vector components define the direction and strength in each dimension.

Understanding vector components is crucial for resolving vectors into their respective basic dimensions.

Magnitude Formula

The magnitude of a vector is a measure of its length or size. It tells us how "strong" or "long" the vector is without considering its direction.

To find the magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \), we use the formula:\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]This formula is derived from the Pythagorean theorem. It calculates the distance from the origin to the point \( \langle x, y \rangle \) in the coordinate plane.

In our example, for the vector \( \mathbf{d} = \langle -2, 0 \rangle \), its magnitude would be calculated by:

• \( |\mathbf{d}| = \sqrt{(-2)^2 + (0)^2} \)
• \( = \sqrt{4 + 0} = \sqrt{4} \)
• \( = 2 \)

So, the magnitude of \( \mathbf{d} \) is 2. This tells us that the vector extends 2 units from the origin, purely in the negative x-direction since there's no y-component.

Division of Vectors

Dividing by a vector magnitude provides a normalized form of that vector's information. However, in this exercise, division by the magnitude gives a scalar value.

It involves calculating \( \frac{1}{|\mathbf{d}|} \). From our earlier steps, we know:

• \( |\mathbf{d}| = 2 \)
• Hence, \( \frac{1}{|\mathbf{d}|} = \frac{1}{2} \)

This process shows how scaling down a vector magnitude works. Often used in physics and engineering, division by a vector magnitude provides unit vectors or scalar factors

which are key in direction-focused calculations.

Step-by-Step Solution

Following a step-by-step procedure can greatly improve understanding and solve complex problems systematically.

Let's go through our solution:

Step 1: Identifying components of \( \mathbf{d} = \langle -2, 0 \rangle \). Find \( x = -2 \), \( y = 0 \).

Step 2: Apply the magnitude formula. \( |\mathbf{d}| = \sqrt{x^2 + y^2} \).

Step 3: Substitute values into the formula and compute. \( \sqrt{(-2)^2 + (0)^2} = 2 \).

Step 4: Solve for \( |\mathbf{d}| \), resulting in 2.

Step 5: Perform graphical division, get \( \frac{1}{|\mathbf{d}|} = \frac{1}{2} \).

Using this structured approach ensures accuracy and helps in understanding each mathematical move. It is a logical pathway from problem to solution.