Problem 47 - Advanced Engineering Mathematics

Question

\(\mathbf{a}=\langle 1,-3,2\rangle, \mathbf{b}=\langle-1,1,1\rangle\), and \(\mathbf{c}=\langle 2,6,9\rangle .\) Find the indicated vector or scalar. \(\left\|\frac{\mathbf{a}}{\|\mathbf{a}\|}\right\|+5\left\|\frac{\mathbf{b}}{\|\mathbf{b}\|}\right\|\)

Step-by-Step Solution

Answer

The value is 6.

Step 1: Find the Magnitude of Vector \( \mathbf{a} \)

The magnitude of vector \( \mathbf{a} = \langle 1, -3, 2 \rangle \) is given by the formula: \( \|\mathbf{a}\| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \).

Step 2: Compute the Magnitude of \( \frac{\mathbf{a}}{\|\mathbf{a}\|} \)

A unit vector has a magnitude of 1, therefore \( \left\| \frac{\mathbf{a}}{\|\mathbf{a}\|} \right\| = 1 \).

Step 3: Find the Magnitude of Vector \( \mathbf{b} \)

The magnitude of vector \( \mathbf{b} = \langle -1, 1, 1 \rangle \) is calculated as: \( \|\mathbf{b}\| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \).

Step 4: Compute the Magnitude of \( \frac{\mathbf{b}}{\|\mathbf{b}\|} \)

Since \( \frac{\mathbf{b}}{\|\mathbf{b}\|} \) is a unit vector, \( \left\| \frac{\mathbf{b}}{\|\mathbf{b}\|} \right\| = 1 \).

Step 5: Calculate the Expression

Substitute the magnitudes into the expression: \( \left\| \frac{\mathbf{a}}{\|\mathbf{a}\|} \right\| + 5 \left\| \frac{\mathbf{b}}{\|\mathbf{b}\|} \right\| = 1 + 5 \times 1 = 6 \).

Key Concepts

Vector Magnitude

In vector calculus, the magnitude of a vector represents the length or size of the vector. To find the magnitude of a vector in three-dimensional space, we use the formula: \[ \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \]where \( x, y, \) and \( z \) are the components of the vector.

• For vector \( \mathbf{a} = \langle 1, -3, 2 \rangle \), the magnitude is calculated as \( \|\mathbf{a}\| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{14} \).
• For vector \( \mathbf{b} = \langle -1, 1, 1 \rangle \), we use the formula to determine \( \|\mathbf{b}\| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \).

Understanding vector magnitude is crucial as it helps us determine the length irrespective of its direction. The magnitude is always a non-negative number.

Unit Vectors

A unit vector is a vector with a magnitude of exactly one. Unit vectors are often used to represent direction without considering magnitude. To convert any vector to a unit vector, divide each component of the vector by its magnitude:\[ \text{Unit vector of } \mathbf{v} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \]For example:

• For vector \( \mathbf{a} \), the unit vector is \( \frac{\mathbf{a}}{\|\mathbf{a}\|} = \frac{\langle 1, -3, 2 \rangle}{\sqrt{14}} \).
• For vector \( \mathbf{b} \), the conversion gives the unit vector as \( \frac{\mathbf{b}}{\|\mathbf{b}\|} = \frac{\langle -1, 1, 1 \rangle}{\sqrt{3}} \).

It's important to remember that the magnitude of any unit vector is always 1. This unique property is what makes them critical in simplifying vector operations and analyses.

Scalar Multiplication

Scalar multiplication in vector calculus involves multiplying a vector by a real number or scalar. This operation results in a new vector that is either stretched or compressed but maintains the same direction as the original vector (unless the scalar is negative, which reverses the direction).Consider the expression involving scalar multiplication:\[ 5 \times \left\| \frac{\mathbf{b}}{\|\mathbf{b}\|} \right\| \]Here, \(5\) is the scalar, and the unit vector \( \frac{\mathbf{b}}{\|\mathbf{b}\|} \) remains unchanged in magnitude since we are multiplying it by 1. The essence of scalar multiplication highlights that multiplying by 1 or using a unit vector does not alter the vector's essential properties aside from its magnitude.Make sure to:

• Understand that scalar multiplication adheres to distributive laws: \( c(u+v) = cu + cv \) and \( c(du) = (cd)u \).
• Acknowledge that a zero scalar reduces any vector to a zero vector \( \mathbf{0} \).

By recognizing these principles, scalar multiplication helps us manipulate vectors efficiently in various applications.