Problem 69 - Precalculus with Calculus Previews

Question

In Problems \(65-76, \mathbf{u}=\langle 1,-3,2\rangle, \mathbf{v}=\langle-1,1,1\rangle\), and \(\mathbf{w}=\langle 2,6,9\rangle .\) Find the indicated vector or scalar. $$ |\mathbf{u}+\mathbf{w}| $$

Step-by-Step Solution

Answer

The magnitude \(|\mathbf{u} + \mathbf{w}|\) is approximately 11.79.

Step 1: Add the vectors \( \mathbf{u} \) and \( \mathbf{w} \)

To find \( |\mathbf{u} + \mathbf{w}| \), we first need to determine the vector sum \( \mathbf{u} + \mathbf{w} \). Given that \( \mathbf{u} = \langle 1, -3, 2 \rangle \) and \( \mathbf{w} = \langle 2, 6, 9 \rangle \), we can add them component-wise: \( \mathbf{u} + \mathbf{w} = \langle 1+2, -3+6, 2+9 \rangle = \langle 3, 3, 11 \rangle \).

Step 2: Calculate the magnitude of the resulting vector

The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) is given by the formula \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). Applying this formula to the vector \( \langle 3, 3, 11 \rangle \), we get: \( |\langle 3, 3, 11 \rangle| = \sqrt{3^2 + 3^2 + 11^2} \).

Step 3: Compute each squared component

Calculate each squared component: \( 3^2 = 9 \), \( 3^2 = 9 \), and \( 11^2 = 121 \).

Step 4: Sum the squared components

Add the squared components together: \( 9 + 9 + 121 = 139 \).

Step 5: Take the square root of the sum

Find the square root of 139 to determine the magnitude: \(|\langle 3, 3, 11 \rangle| = \sqrt{139} \). This simplifies to \( \sqrt{139} \), which is approximately 11.79.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. It's a positive value, indicating how "long" a vector is in space. To find the magnitude of a vector, you use the formula:

• For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).

So, in the given exercise, to find the magnitude of \( \mathbf{u} + \mathbf{w} = \langle 3, 3, 11 \rangle \), you'll calculate

• \( 3^2 = 9 \)
• \( 3^2 = 9 \)
• \( 11^2 = 121 \)

Add them up to get the total under the square root. The sum is 139, and taking the square root gives you about 11.79. The magnitude here tells us how far the endpoint of the vector \( \mathbf{u} + \mathbf{w} \) is from the origin.

Vector Components

Every vector is made up of components, which are its projections along the coordinate axes. If you picture a vector as an arrow in 3D space, the components are the shadow of that vector cast on each axis.
The components break down into three values:

• \( a_1 \): the component along the x-axis.
• \( a_2 \): the component along the y-axis.
• \( a_3 \): the component along the z-axis.

When you add vectors, you combine each of these components separately. For example, if \( \mathbf{u} = \langle 1, -3, 2 \rangle \) and \( \mathbf{w} = \langle 2, 6, 9 \rangle \), the addition is done component-wise:

• \( 1 + 2 = 3 \)
• \( -3 + 6 = 3 \)
• \( 2 + 9 = 11 \)

This process results in the new vector \( \langle 3, 3, 11 \rangle \). Each component of this vector tells you how far you move along each axis.

Vector Operations

Vector operations encompass various algebraic manipulations involving vectors. The most common include vector addition, subtraction, and scalar multiplication.
**Vector Addition**
This involves adding two vectors together by summing their respective components. For instance, with vectors \( \mathbf{u} \) and \( \mathbf{w} \), the addition produces \( \mathbf{u} + \mathbf{w} = \langle 3, 3, 11 \rangle \).
**Scalar Multiplication**
This operation involves multiplying a vector by a scalar (a single number). Each component of the vector is multiplied by the scalar value. Although we didn't use scalar multiplication directly in the exercise, it's vital for scaling vectors up or down.