Problem 5 - Thomas Calculus

Question

In Exercises \(1-8,\) let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle- 2,5\rangle .\) Find the (a) component form and \((\mathbf{b})\) magnitude (length) of the vector. $$ 2 \mathbf{u}-3 \mathbf{v} $$

Step-by-Step Solution

Answer

The vector is \( \langle 12, -19 \rangle \) with magnitude \( \sqrt{505} \approx 22.47 \).

Step 1: Calculate Component Form of 2u

We start by multiplying vector \( \mathbf{u} = \langle 3, -2 \rangle \) by 2. To do this, multiply each component of \( \mathbf{u} \) by 2: \( 2 \cdot 3 = 6 \) and \( 2 \cdot (-2) = -4 \). Thus, \( 2\mathbf{u} = \langle 6, -4 \rangle \).

Step 2: Calculate Component Form of 3v

Next, we multiply vector \( \mathbf{v} = \langle -2, 5 \rangle \) by 3. Multiply each component of \( \mathbf{v} \) by 3: \( 3 \cdot (-2) = -6 \) and \( 3 \cdot 5 = 15 \). Thus, \( 3\mathbf{v} = \langle -6, 15 \rangle \).

Step 3: Find Component Form of 2u - 3v

Subtract the components of \( 3\mathbf{v} = \langle -6, 15 \rangle \) from \( 2\mathbf{u} = \langle 6, -4 \rangle \). Calculate: \( 6 - (-6) = 12 \) and \( -4 - 15 = -19 \). Therefore, the vector \( 2\mathbf{u} - 3\mathbf{v} = \langle 12, -19 \rangle \).

Step 4: Calculate Magnitude of the Vector

Use the formula for the magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), which is \( \sqrt{a_1^2 + a_2^2} \). For \( \langle 12, -19 \rangle \), calculate: \( \sqrt{12^2 + (-19)^2} = \sqrt{144 + 361} = \sqrt{505} \approx 22.47 \).

Key Concepts

Component Form

The component form of a vector is a way of representing the vector using individual components, specifically along the x-axis and y-axis in two-dimensional space. Think of it as breaking down the vector into its horizontal and vertical parts which make calculations easier.

For instance, a vector that originates from the point \( (0,0) \) to the point \( (x,y) \) can be written as \( \langle x, y \rangle \). These components tell us how far to move along each axis to reach from the origin to the given point.

• The vector \( \mathbf{u} = \langle 3, -2 \rangle \) tells us to move 3 units right and 2 units down.
• Similarly, the vector \( \mathbf{v} = \langle -2, 5 \rangle \) means moving 2 units left and 5 units up.

To combine these vectors, we can add or subtract their components. When scalar multiplication is involved, like multiplying the vectors by numbers, it's crucial to scale each component:

• For \( 2\mathbf{u} = \langle 6, -4 \rangle \), each component of \( \mathbf{u} \) is multiplied by 2.
• For \( 3\mathbf{v} = \langle -6, 15 \rangle \), each component of \( \mathbf{v} \) is multiplied by 3.

Finally, to find \( 2\mathbf{u} - 3\mathbf{v} \), subtract the corresponding components: \( \langle 6, -4 \rangle \) minus \( \langle -6, 15 \rangle \) resulting in \( \langle 12, -19 \rangle \).

Magnitude of a Vector

The magnitude of a vector is its length, essentially measuring the distance from the vector's start point to its end point. This concept mirrors finding the length of a side in geometry and is vital in understanding the vector's size.

To calculate the magnitude of a vector in two-dimensional space, we use the Pythagorean theorem. If a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), then its magnitude, denoted as \( ||\mathbf{a}|| \), can be found using the formula:
\[||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2}\]

• For our combined vector \( \langle 12, -19 \rangle \), calculate as follows:
• Square each component: \( 12^2 = 144 \) and \( (-19)^2 = 361 \).
• Add these squares: \( 144 + 361 = 505 \).
• Finally, take the square root: \( \sqrt{505} \approx 22.47 \).

This result, approximately \( 22.47 \), tells us the actual size or length of our vector.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, which is simply a real number. It essentially changes the size of the vector but not its direction, unless the scalar is negative, which then reverses the direction. Understanding this concept is crucial for manipulating and scaling vectors in practical applications.

When you perform scalar multiplication, you multiply each component of the vector by the scalar. This operation is straightforward and is exemplified in our exercise:

• For vector \( \mathbf{u} = \langle 3, -2 \rangle \) and scalar 2, compute \( 2 \cdot \mathbf{u} = \langle 6, -4 \rangle \).
• For vector \( \mathbf{v} = \langle -2, 5 \rangle \) and scalar 3, compute \( 3 \cdot \mathbf{v} = \langle -6, 15 \rangle \).

This multiplication process is fundamental when resolving operations like \( 2\mathbf{u} - 3\mathbf{v} \). It prepares further calculations, setting the stage for addition or subtraction between scaled vectors. The beauty of scalar multiplication is its ability to proportionally extend or shrink the vector while preserving its direction characteristics.