Problem 18 - Algebra and Trigonometry

Question

Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \mathbf{u}=\langle- 2,5\rangle, \quad \mathbf{v}=\langle 2,-8\rangle $$

Step-by-Step Solution

Answer

\(2\mathbf{u}=\langle -4, 10 \rangle, -3\mathbf{v}=\langle -6, 24 \rangle, \mathbf{u}+\mathbf{v}=\langle 0, -3 \rangle, 3\mathbf{u}-4\mathbf{v}=\langle -14, 47 \rangle.\)

Step 1: Scalar Multiplication of u

To calculate \(2 \mathbf{u}\), multiply each component of \(\mathbf{u}\) by 2. For \(\mathbf{u} = \langle -2, 5 \rangle\), this gives: \[ 2 \mathbf{u} = \langle 2(-2), 2(5) \rangle = \langle -4, 10 \rangle. \]

Step 2: Scalar Multiplication of v

To calculate \(-3 \mathbf{v}\), multiply each component of \(\mathbf{v}\) by -3. For \(\mathbf{v} = \langle 2, -8 \rangle\), this results in: \[ -3 \mathbf{v} = \langle -3(2), -3(-8) \rangle = \langle -6, 24 \rangle. \]

Step 3: Vector Addition of u and v

To find \(\mathbf{u} + \mathbf{v}\), add the corresponding components of \(\mathbf{u}\) and \(\mathbf{v}\). This yields: \[ \mathbf{u} + \mathbf{v} = \langle -2 + 2, 5 + (-8) \rangle = \langle 0, -3 \rangle. \]

Step 4: Combined Vector Operation

To find \(3 \mathbf{u} - 4 \mathbf{v}\), first multiply \(\mathbf{u}\) by 3 and \(\mathbf{v}\) by 4, then subtract the results. Calculate: \[ 3 \mathbf{u} = \langle 3(-2), 3(5) \rangle = \langle -6, 15 \rangle, \] and \[ 4 \mathbf{v} = \langle 4(2), 4(-8) \rangle = \langle 8, -32 \rangle. \] Thus, \[ 3 \mathbf{u} - 4 \mathbf{v} = \langle -6 - 8, 15 - (-32) \rangle = \langle -14, 47 \rangle. \]

Key Concepts

Scalar Multiplication

Scalar multiplication is the process of multiplying a vector by a scalar (a real number). In doing so, you enlarge or shrink the vector. Each component of the vector is multiplied by the scalar.

Here's the magic at play. Suppose you have a vector \( \mathbf{u} = \langle x, y \rangle \) and you want to multiply it by some scalar \( a \). The result of this operation is a new vector \( a \mathbf{u} = \langle ax, ay \rangle \). This transformation maintains the direction of the vector \( \mathbf{u} \), but alters its magnitude.

For example, given the vector \( \mathbf{u} = \langle -2, 5 \rangle \), if we perform scalar multiplication with the scalar 2, we calculate as follows:

• Multiply each component by 2: \( 2 \times -2 = -4 \) and \( 2 \times 5 = 10 \).
• The result is \( 2 \mathbf{u} = \langle -4, 10 \rangle \).

Another example with a different vector \( \mathbf{v} = \langle 2, -8 \rangle \) and scalar \(-3\), results in:

• \(-3 \times 2 = -6 \)
• \(-3 \times -8 = 24 \)
• So, \(-3 \mathbf{v} = \langle -6, 24 \rangle \).

Vector Addition

Vector addition is all about combining vectors to get a resultant vector. If you have two vectors \( \mathbf{u} = \langle x_1, y_1 \rangle \) and \( \mathbf{v} = \langle x_2, y_2 \rangle \), their sum \( \mathbf{u} + \mathbf{v} \) is found by adding their corresponding components.

This operation adjusts both the direction and magnitude by summing up each component pair:

• \( x_1 + x_2 \)
• \( y_1 + y_2 \)

The result is a new vector \( \mathbf{u} + \mathbf{v} = \langle x_1 + x_2, y_1 + y_2 \rangle \). In the specific case of finding \( \mathbf{u} + \mathbf{v} \) for \( \mathbf{u} = \langle -2, 5 \rangle \) and \( \mathbf{v} = \langle 2, -8 \rangle \), we have:

• Adding the first components: \( -2 + 2 = 0 \)
• Adding the second components: \( 5 - 8 = -3 \)
• Thus, \( \mathbf{u} + \mathbf{v} = \langle 0, -3 \rangle \).

This operation is relatively straightforward and shows how vectors relate and combine forces, often used in physics to represent combined forces or velocities.

Vector Subtraction

Vector subtraction is similar to vector addition but involves subtracting the components of one vector from another. If you have vectors \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), the result of subtraction \( \mathbf{a} - \mathbf{b} \) is achieved by subtracting the corresponding components.

This process creates a new vector:

• Subtract the first components: \( x_1 - x_2 \)
• Subtract the second components: \( y_1 - y_2 \)

Expressed as \( \mathbf{a} - \mathbf{b} = \langle x_1 - x_2, y_1 - y_2 \rangle \).

For a more complex operation like \( 3 \mathbf{u} - 4 \mathbf{v} \), first perform scalar multiplication on each vector, then subtract:

• Calculate \( 3 \mathbf{u} = \langle -6, 15 \rangle \) and \( 4 \mathbf{v} = \langle 8, -32 \rangle \).
• Subtract the corresponding components: \( -6 - 8 = -14 \) for the x-component, and \( 15 - (-32) = 47 \) for the y-component.
• The result is \( 3 \mathbf{u} - 4 \mathbf{v} = \langle -14, 47 \rangle \).

This illustrates how vector subtraction can adjust directions or magnitudes, often useful in determining relative distances or differences in motion.