Problem 32 - Precalculus Mathematics for Calculus

Question

Find \(2 u,-3 v, u+v,\) and \(3 u-4 v\) 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

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

Step 1: Calculate 2u

To find \(2\mathbf{u}\), multiply each component of \(\mathbf{u}\) by 2. \(\mathbf{u} = \langle -2, 5 \rangle\), so \(2\mathbf{u} = \langle 2 \times -2, 2 \times 5 \rangle = \langle -4, 10 \rangle\).

Step 2: Calculate -3v

To determine \(-3\mathbf{v}\), multiply each component of \(\mathbf{v}\) by -3. \(\mathbf{v} = \langle 2, -8 \rangle\), so \(-3\mathbf{v} = \langle -3 \times 2, -3 \times -8 \rangle = \langle -6, 24 \rangle\).

Step 3: Calculate u+v

For \(\mathbf{u} + \mathbf{v}\), add the corresponding components of \(\mathbf{u}\) and \(\mathbf{v}\). \(\mathbf{u} = \langle -2, 5 \rangle\) and \(\mathbf{v} = \langle 2, -8 \rangle\), so \(\mathbf{u} + \mathbf{v} = \langle -2 + 2, 5 + (-8) \rangle = \langle 0, -3 \rangle\).

Step 4: Calculate 3u-4v

To find \(3\mathbf{u} - 4\mathbf{v}\), first calculate \(3\mathbf{u}\) and \(4\mathbf{v}\), and then subtract the two results. \(\mathbf{u} = \langle -2, 5 \rangle\), so \(3\mathbf{u} = \langle 3 \times -2, 3 \times 5 \rangle = \langle -6, 15 \rangle\). For \(4\mathbf{v}\), \(\mathbf{v} = \langle 2, -8 \rangle\), so \(4\mathbf{v} = \langle 4 \times 2, 4 \times -8 \rangle = \langle 8, -32 \rangle\). Now, subtract: \(3\mathbf{u} - 4\mathbf{v} = \langle -6 - 8, 15 - (-32) \rangle = \langle -14, 47 \rangle\).

Key Concepts

Vector Multiplication

Vector multiplication comes in different forms, like the dot product and cross product. However, for basic vector arithmetic involving scalars (i.e., numbers), vector multiplication refers to multiplying the vector by a scalar. If you have a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), multiplying this vector by a scalar \( k \) involves multiplying each component of the vector by \( k \) resulting in \( k \mathbf{a} = \langle k \times a_1, k \times a_2 \rangle \).

This kind of multiplication is straightforward:

• For the vector \( \mathbf{u} = \langle -2, 5 \rangle \) and scalar 2, the result is \(2 \mathbf{u} = \langle 2 \times -2, 2 \times 5 \rangle = \langle -4, 10 \rangle \).
• For the vector \( \mathbf{v} = \langle 2, -8 \rangle \) and scalar \(-3\), the result is \(-3 \mathbf{v} = \langle -3 \times 2, -3 \times -8 \rangle = \langle -6, 24 \rangle \).

Each component gets individually multiplied, changing the length and possibly the direction of the vector but not its position in the space.

Vector Addition

In vector addition, the rule is simple: you add the corresponding components of the vectors. Suppose you have two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \). Their sum is \( \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle \).

This rule works with vectors of the same size only, ensuring that each element pairs correctly:

• For \( \mathbf{u} = \langle -2, 5 \rangle \) and \( \mathbf{v} = \langle 2, -8 \rangle \), their addition is: \( \mathbf{u} + \mathbf{v} = \langle -2 + 2, 5 + (-8) \rangle = \langle 0, -3 \rangle \).

This operation affects the endpoint of the resulting vector, not its components separately. Visualizing them on a graph, vector addition equates to placing one vector's start at another’s tip, then drawing from the free start to the free end.

Vector Subtraction

Vector subtraction is quite similar to vector addition but instead involves subtracting the corresponding components from each other. For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), subtraction is performed as \( \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle \).

Think of subtraction as adding the first vector to the negative of the second:

• For a problem where \(3\mathbf{u} - 4\mathbf{v} \) must be found:
  • First find \(3\mathbf{u} = \langle 3 \times -2, 3 \times 5 \rangle = \langle -6, 15 \rangle \).
  • Then calculate \(4\mathbf{v} = \langle 4 \times 2, 4 \times -8 \rangle = \langle 8, -32 \rangle \).
  • Finally, subtract: \(3\mathbf{u} - 4\mathbf{v} = \langle -6 - 8, 15 - (-32) \rangle = \langle -14, 47 \rangle \).

Visualize it as moving in the opposite direction of the second vector from the end of the first vector.

Scalar Multiplication

Scalar multiplication of vectors is an operation where a vector is multiplied by a number, or scalar. This is distinct from vector multiplication because it exclusively involves a scalar and adjusts the magnitude of the vector while maintaining or inverting its direction depending on the scalar's sign. If the scalar is negative, the vector reverses direction.

Consider the same vectors \( \mathbf{u} = \langle -2, 5 \rangle \) and \( \mathbf{v} = \langle 2, -8 \rangle \). By applying scalar multiplication:

• For \(2\mathbf{u}, \) each component of \( \mathbf{u} \) is doubled, leading to \(2 \mathbf{u} = \langle -4, 10 \rangle \).
• In the case of \(-3\mathbf{v},\) the vector \(\mathbf{v}\) components are tripled and inverted, so that \(-3 \mathbf{v} = \langle -6, 24 \rangle \).

This multiplication expands or contracts the vector along its original path, or flips it when the scalar is negative.