Problem 6 - 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. $$-2 \mathbf{u}+5 \mathbf{v}$$

Step-by-Step Solution

Answer

Component form is \(\langle -16, 29 \rangle\); magnitude is \(\sqrt{1097}\).

Step 1: Find the Scalar Multiple of Vector u

Given vector \( \mathbf{u} = \langle 3, -2 \rangle \). We want to find \(-2 \mathbf{u}\).\[-2 \mathbf{u} = -2 \times \langle 3, -2 \rangle = \langle -2 \times 3, -2 \times (-2) \rangle = \langle -6, 4 \rangle\]

Step 2: Find the Scalar Multiple of Vector v

Given vector \( \mathbf{v} = \langle -2, 5 \rangle \). We want to find \(5 \mathbf{v}\).\[5 \mathbf{v} = 5 \times \langle -2, 5 \rangle = \langle 5 \times (-2), 5 \times 5 \rangle = \langle -10, 25 \rangle\]

Step 3: Add the Resulting Vectors

Now sum the resulting vectors from Step 1 and Step 2 to get the component form of the vector.\[-2 \mathbf{u} + 5 \mathbf{v} = \langle -6, 4 \rangle + \langle -10, 25 \rangle = \langle -6 + (-10), 4 + 25 \rangle = \langle -16, 29 \rangle\]

Step 4: Calculate the Magnitude of the Resulting Vector

Use the formula for the magnitude of a vector \( \langle a, b \rangle \):\[|| \langle a, b \rangle || = \sqrt{a^2 + b^2}\]Substitute for \( a = -16 \) and \( b = 29 \):\[|| \langle -16, 29 \rangle || = \sqrt{(-16)^2 + 29^2} = \sqrt{256 + 841} = \sqrt{1097}\]

Key Concepts

Vector Operations

Vector operations are crucial for understanding movements and forces in two or more dimensions. Typically, operations on vectors include addition, subtraction, and scalar multiplication.

In vector addition or subtraction, you add or subtract corresponding components. For example, adding vectors \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \) results in the vector \( \langle x_1 + x_2, y_1 + y_2 \rangle \). You calculate each component separately to determine the resultant vector.

Subtracting vectors involves a similar process, except you subtract the corresponding components instead of adding them.

• Vector addition: Sum the corresponding components.
• Vector subtraction: Subtract the corresponding components.

Understanding vector operations allows you to determine resultant vectors that could represent anything from movement directions to forces acting on a point.

Scalar Multiplication

Scalar multiplication is a simple yet powerful operation involving vectors. It refers to multiplying a vector by a single number (scalar), changing the vector's magnitude without altering its direction.

To perform scalar multiplication, you multiply each component of the vector by the scalar. For instance, given vector \( \mathbf{v} = \langle x, y \rangle \) and scalar \( c \), the result of scalar multiplication \( c \mathbf{v} \) is \( \langle cx, cy \rangle \). This operation elongates or shortens the vector based on the scalar value's magnitude.

If the scalar is negative, it not only alters the magnitude but also reverses the vector's direction. This change is evident in exercises where vectors are multiplied by negative scalars.

• Positive scalar: Magnitude changes, direction remains same.
• Negative scalar: Both magnitude and direction change.

Executing scalar multiplication appropriately is vital in vector calculations and transformations.

Magnitude of a Vector

The magnitude of a vector quantifies its length in geometrical terms. It's a measure of how long the vector reaches from one point to another in space.

The formula to calculate the magnitude for a two-dimensional vector \( \langle a, b \rangle \) is \( ||\langle a, b \rangle|| = \sqrt{a^2 + b^2} \). Using the Pythagorean theorem, this formula derives from treating the components as legs of a right triangle, with the magnitude as the hypotenuse.

Applying this in the context of the previous steps provided, if you calculate the magnitude for the vector \( \langle -16, 29 \rangle \), you find this vector's exact measure of length or distance in space.

• Magnitude formula: Direct application of the Pythagorean theorem.
• Useful for: Determining vector length or distance.

Understanding the magnitude allows for better insight into the scalar length of vectors, aiding in comprehending vector size without concern for direction.