Problem 5 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The component form is \( \langle 12, -19 \rangle \); magnitude is \( \sqrt{505} \)."

Step 1: Multiply Vector u by 2

To find the component form of the vector expression \( 2\mathbf{u} \), start by multiplying each component of \( \mathbf{u} = \langle 3, -2 \rangle \) by 2: \[ 2\mathbf{u} = 2 \times \langle 3, -2 \rangle = \langle 2\times 3, 2\times (-2) \rangle = \langle 6, -4 \rangle.\]

Step 2: Multiply Vector v by 3

Next, we find \( 3\mathbf{v} \). Multiply each component of \( \mathbf{v} = \langle -2, 5 \rangle \) by 3: \[ 3\mathbf{v} = 3 \times \langle -2, 5 \rangle = \langle 3\times (-2), 3\times 5 \rangle = \langle -6, 15 \rangle.\]

Step 3: Subtract Vectors

Now, find the vector expression \( 2\mathbf{u} - 3\mathbf{v} \) by subtracting the components of \( 3\mathbf{v} \) from \( 2\mathbf{u} \): \[ 2\mathbf{u} - 3\mathbf{v} = \langle 6, -4 \rangle - \langle -6, 15 \rangle.\] Performing the subtraction, we get: \[ = \langle 6 - (-6), -4 - 15 \rangle = \langle 12, -19 \rangle.\] Thus, the component form is \( \langle 12, -19 \rangle.\)

Step 4: Calculate Magnitude of the Resulting Vector

To find the magnitude (length), use the formula for the magnitude of a vector \( \langle a, b \rangle \): \[ \| \langle a, b \rangle \| = \sqrt{a^2 + b^2}.\] For the vector \( \langle 12, -19 \rangle \), we have: \[ \| \langle 12, -19 \rangle \| = \sqrt{12^2 + (-19)^2} = \sqrt{144 + 361} = \sqrt{505}.\] Therefore, the magnitude is \( \sqrt{505} \).

Key Concepts

Vector Magnitude

To find the magnitude of a vector, you always depend on a simple and straightforward formula that involves the Pythagorean theorem. The magnitude, or length, of a vector \( \langle a, b \rangle \), can be imagined like finding the hypotenuse of a right triangle. It is calculated using the formula: \[ \| \langle a, b \rangle \| = \sqrt{a^2 + b^2} \].This operation helps you understand how long the vector actually is, regardless of its direction. For instance, when you have a vector like \( \langle 12, -19 \rangle \), imagine it as a diagonal within a rectangle extending 12 units in the x-direction and 19 units in the opposite y-direction. Using the formula, first, square each component:

• \( 12^2 = 144 \)
• \( (-19)^2 = 361 \)

Add them together resulting in 505, and then take the square root to find \( \sqrt{505} \). This value is how we measure the length of the vector in space, which tells us how far the point is from its vector origin.

Vector Subtraction

Vector subtraction takes an interesting role in vector arithmetic, where you subtract one vector from another to find the resulting vector. If you have two vectors, \( \mathbf{A} = \langle a_1, b_1 \rangle \) and \( \mathbf{B} = \langle a_2, b_2 \rangle \), subtracting them is simply about dealing with each component separately:\[ \mathbf{A} - \mathbf{B} = \langle a_1 - a_2, b_1 - b_2 \rangle \].In the given exercise, after doing scalar multiplications, we subtracted vector \( 3\mathbf{v} = \langle -6, 15 \rangle \) from vector \( 2\mathbf{u} = \langle 6, -4 \rangle \). So, here's the breakdown:

• Start by subtracting the x-components: \( 6 - (-6) = 12 \)
• Then, subtract the y-components: \( -4 - 15 = -19 \)

This results in the new vector, \( \langle 12, -19 \rangle \). That's it! The subtraction is straightforward, and it provides the components of the new vector "2\mathbf{u} - 3\mathbf{v}".

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number), to scale the vector's size without changing its direction. This concept is key when looking at how vectors are manipulated in different applications, such as physics or engineering.For a vector \( \mathbf{v} = \langle x, y \rangle \) and a scalar \( c \), scalar multiplication is simply calculated by: \[ c \mathbf{v} = \langle c \times x, c \times y \rangle \].Let's look at our task: You have the vector \( \mathbf{u} = \langle 3, -2 \rangle \). When you multiply it by 2, you multiply each component separately:

• The X-component becomes \( 2 \times 3 = 6 \).
• The Y-component becomes \( 2 \times (-2) = -4 \).

The result of this is the vector \( 2\mathbf{u} = \langle 6, -4 \rangle \). Similarly, apply it to \( \mathbf{v} = \langle -2, 5 \rangle \):

• 3 multiplied by the X-component gives \( 3 \times (-2) = -6 \).
• 3 multiplied by the Y-component gives \( 3 \times 5 = 15 \).

Hence, \( 3\mathbf{v} = \langle -6, 15 \rangle \). Scalar multiplication effectively rescales the vector and when combined with other operations, helps in constructing different vectors for modeling real-world phenomena.