Problem 32 - Algebra and Trigonometry

Question

\(31-36\) 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

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

Step 1: Calculate \(2 \mathbf{u}\)

To find \(2 \mathbf{u}\), multiply each component of the vector \(\mathbf{u} = \langle -2, 5 \rangle\) by 2:\[\begin{align*} 2 \mathbf{u} &= 2 \times \langle -2, 5 \rangle \&= \langle 2 \times (-2), 2 \times 5 \rangle \&= \langle -4, 10 \rangle. \end{align*}\]

Step 2: Calculate \(-3 \mathbf{v}\)

To find \(-3 \mathbf{v}\), multiply each component of vector \(\mathbf{v} = \langle 2, -8 \rangle\) by -3:\[\begin{align*} -3 \mathbf{v} &= -3 \times \langle 2, -8 \rangle \&= \langle -3 \times 2, -3 \times (-8) \rangle \&= \langle -6, 24 \rangle. \end{align*}\]

Step 3: Calculate \(\mathbf{u} + \mathbf{v}\)

To find \(\mathbf{u} + \mathbf{v}\), add corresponding components of \(\mathbf{u} = \langle -2, 5 \rangle\) and \(\mathbf{v} = \langle 2, -8 \rangle\):\[\begin{align*} \mathbf{u} + \mathbf{v} &= \langle -2 + 2, 5 + (-8) \rangle \&= \langle 0, -3 \rangle. \end{align*}\]

Step 4: Calculate \(3 \mathbf{u} - 4 \mathbf{v}\)

To find \(3 \mathbf{u} - 4 \mathbf{v}\), calculate \(3 \mathbf{u}\) and \(4 \mathbf{v}\) first, then subtract \(4 \mathbf{v}\) from \(3 \mathbf{u}\). First, calculate \(3 \mathbf{u}\):\[3 \mathbf{u} = 3 \times \langle -2, 5 \rangle = \langle -6, 15 \rangle.\] Next, calculate \(4 \mathbf{v}\):\[4 \mathbf{v} = 4 \times \langle 2, -8 \rangle = \langle 8, -32 \rangle.\] Now, perform the subtraction:\[\begin{align*} 3 \mathbf{u} - 4 \mathbf{v} &= \langle -6, 15 \rangle - \langle 8, -32 \rangle \&= \langle -6 - 8, 15 + 32 \rangle \&= \langle -14, 47 \rangle. \end{align*}\]

Key Concepts

Scalar Multiplication

Scalar multiplication is an essential operation in vector mathematics that allows you to change the length (or magnitude) of a vector without altering its direction. In this operation, each component of the vector is multiplied by the scalar (a real number). For example, if you have a vector \( \mathbf{a} = \langle x, y \rangle \) and you want to multiply it by a scalar \( k \), the new vector would be \( k \mathbf{a} = \langle kx, ky \rangle \).

This concept can be better understood with an example using the vector \( \mathbf{u} = \langle -2, 5 \rangle \). When we multiply it by the scalar 2, we perform the following operation:

• (-2) \( \times 2 = -4 \)
• (5) \( \times 2 = 10 \)

This results in the vector \( 2\mathbf{u} = \langle -4, 10 \rangle \). Such manipulation only affects the vector's length, not its direction.

Remember that if the scalar is negative, the vector not only changes in magnitude but also flips direction.

Vector Addition

Vector addition involves combining two or more vectors to form a resultant vector. To add two vectors, you simply add their corresponding components. If you have vectors \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), their sum \( \mathbf{a} + \mathbf{b} = \langle x_1 + x_2, y_1 + y_2 \rangle \).

In our example, let's add the vectors \( \mathbf{u} = \langle -2, 5 \rangle \) and \( \mathbf{v} = \langle 2, -8 \rangle \):

• Add the first components: \(-2 + 2 = 0\)
• Add the second components: \(5 + (-8) = -3\)

The resulting vector from this operation is \( \mathbf{u} + \mathbf{v} = \langle 0, -3 \rangle \). This vector represents the combined effect of the two original vectors.

Vector addition is commutative, meaning that \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \). It is pictorially represented as placing the tail of one vector at the head of another.

Vector Subtraction

Vector subtraction is similar to vector addition, but instead of adding corresponding components, you subtract them. The rule is: if you have vectors \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), the subtraction \( \mathbf{a} - \mathbf{b} = \langle x_1 - x_2, y_1 - y_2 \rangle \).

In the worked exercise, where we find \( 3\mathbf{u} - 4\mathbf{v} \), the process is as follows: First, compute \( 3\mathbf{u} = \langle -6, 15 \rangle \) and \( 4\mathbf{v} = \langle 8, -32 \rangle \). Now, perform the subtraction:

• Subtract the first components: \(-6 - 8 = -14\)
• Subtract the second components: \(15 - (-32) = 47\)

The resulting vector is \( 3\mathbf{u} - 4\mathbf{v} = \langle -14, 47 \rangle \).

This subtraction effectively gives you the direction and magnitude pointing from the end of the second vector to the end of the first one when arranged tail-to-tail. Vector subtraction can also be viewed as adding the negative of a vector, which involves flipping the signs of all components of the vector being subtracted.