Problem 1 - 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. $$3 \mathbf{u}$$

Step-by-Step Solution

Answer

The component form is \(\langle 9, -6\rangle\) and the magnitude is \(\sqrt{117}\).

Step 1: Understanding Scalar Multiplication

Scalar multiplication involves multiplying each component of a vector by the given scalar. Here, we need to multiply each component of the vector \( \mathbf{u} = \langle 3, -2 \rangle \) by 3.

Step 2: Perform Scalar Multiplication

Multiply the components of \( \mathbf{u} \) by 3: \[ 3 \cdot \langle 3, -2 \rangle = \langle 3 \times 3, 3 \times (-2) \rangle = \langle 9, -6 \rangle. \] Thus, the component form of \( 3 \mathbf{u} \) is \( \langle 9, -6 \rangle \).

Step 3: Calculate the Magnitude

The magnitude (length) of a vector \( \mathbf{a} = \langle x, y \rangle \) is found using the formula \[ \| \mathbf{a} \| = \sqrt{x^2 + y^2}. \] For \( 3\mathbf{u} = \langle 9, -6 \rangle \), calculate the magnitude as follows:\[ \| 3\mathbf{u} \| = \sqrt{9^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117}. \]

Key Concepts

Scalar Multiplication

Scalar multiplication is a simple yet foundational concept in vector calculus. It refers to the multiplication of a vector by a scalar (a real number).
This operation changes the magnitude of the vector but not its direction, unless the scalar is negative, in which case the direction is reversed as well.

• To perform scalar multiplication, multiply the scalar by each component of the vector separately.
• For example, multiplying vector \( \mathbf{u} = \langle 3, -2 \rangle \) by 3 involves multiplying both 3 and -2 by 3.

This results in the vector \( \langle 9, -6 \rangle \). This new vector is said to be scaled by the factor of 3. The operation doesn't alter the direction qualitatively, although in physical terms, the vector is now 3 times its original 'size', pointing in the same or opposite direction, depending on the scalar's sign.

Magnitude of a Vector

The magnitude of a vector, often thought of as the vector's length, provides a measure of how long the vector is in geometric terms.
This is essential when looking to understand the size of a vector regardless of its direction.

• For a two-dimensional vector \( \mathbf{a} = \langle x, y \rangle \), the magnitude is calculated using the formula: \( \| \mathbf{a} \| = \sqrt{x^2 + y^2} \).
• In the context of the vector \( 3\mathbf{u} = \langle 9, -6 \rangle \), its magnitude is calculated as \( \| 3\mathbf{u} \| = \sqrt{9^2 + (-6)^2} = \sqrt{117} \).

Calculating the magnitude is akin to finding the hypotenuse of a right triangle where the vector's components serve as the two legs.
The magnitude is always a non-negative value, representing the overall displacement or length along the vector's path.

Component Form

The component form of a vector expresses the vector in terms of its individual parts, mainly its horizontal and vertical components in a 2D space.
This representation is essential as it allows for easy computations and visualizations.

• A vector \( \mathbf{v} = \langle a, b \rangle \) can be broken down into its component parts: \( a \) is the horizontal component and \( b \) is the vertical component.
• For example, the vector \( 3\mathbf{u} \) in component form is \( \langle 9, -6 \rangle \), indicating 9 units to the right and 6 units downward.

This breakdown helps in executing operations like addition, subtraction, and scalar multiplication with vectors seamlessly.
Breaking a vector into its components clarifies the vector's direction and allows for the direct application of trigonometric and algebraic methods to further analyze its properties.