Problem 50 - Precalculus Mathematics for Calculus

Question

A unit vector is a vector of magnitude 1 . Multiplying a vector by a scalar changes its magnitude but not its direction. (a) If a vector \(\mathbf{v}\) has magnitude \(m,\) what scalar multiple of \(\mathbf{v}\) has magnitude 1 (i.e., is a unit vector)? (b) Multiply each of the following vectors by an appropriate scalar to change them into unit vectors: $$(1,-2,2\rangle \quad(-6,8,-10\rangle \quad\langle 6,5,9\rangle$$

Step-by-Step Solution

Answer

(a) Scalar is \(\frac{1}{m}\). (b) Unit vectors are \(\left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right), \left(-\frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, -\frac{5}{\sqrt{2}}\right), \left(\frac{6}{\sqrt{142}}, \frac{5}{\sqrt{142}}, \frac{9}{\sqrt{142}}\right)\).

Step 1: Understand the Problem

We are given a vector \(\mathbf{v}\) with magnitude \(m\), and we need to find a scalar multiple of \(\mathbf{v}\) that has a magnitude of 1, thus forming a unit vector. Subsequently, we have three vectors provided, and we are tasked with transforming them into unit vectors by finding appropriate scalar multiples.

Step 2: Find the Scalar for Part (a)

To transform a vector \(\mathbf{v}\) with magnitude \(m\) into a unit vector, we need to divide \(\mathbf{v}\) by its magnitude \(m\). This means the scalar that converts \(\mathbf{v}\) into a unit vector is \(\frac{1}{m}\). The resulting unit vector is then \(\frac{1}{m}\mathbf{v}\).

Step 3: Determine Magnitude of Vectors for Part (b)

Calculate the magnitude of each vector. For a vector \(\mathbf{v}=(x,y,z)\), the magnitude is given by: \[\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\].1. \((1, -2, 2)\): \(\|\mathbf{v}_1\| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\)2. \((-6, 8, -10)\): \(\|\mathbf{v}_2\| = \sqrt{(-6)^2 + 8^2 + (-10)^2} = \sqrt{200} = \sqrt{100} \cdot \sqrt{2} = 10\sqrt{2}\)3. \((6, 5, 9)\): \(\|\mathbf{v}_3\| = \sqrt{6^2 + 5^2 + 9^2} = \sqrt{142}\)

Step 4: Apply Scalar to Vectors for Part (b)

To convert each vector into a unit vector, multiply it by the reciprocal of its magnitude calculated in the previous step:1. Vector \((1, -2, 2)\): Unit vector = \(\frac{1}{3}(1, -2, 2) = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)\)2. Vector \((-6, 8, -10)\): Unit vector = \(\frac{1}{10\sqrt{2}}(-6, 8, -10) = \left(-\frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, -\frac{5}{\sqrt{2}}\right)\)3. Vector \((6, 5, 9)\): Unit vector = \(\frac{1}{\sqrt{142}}(6, 5, 9) = \left(\frac{6}{\sqrt{142}}, \frac{5}{\sqrt{142}}, \frac{9}{\sqrt{142}}\right)\)

Key Concepts

vector magnitude

The vector magnitude is a measure of the length or size of a vector. In a three-dimensional space, if you have a vector \(\mathbf{v} = (x, y, z)\), you can find its magnitude using the formula \[\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\]. This formula comes from the Pythagorean theorem, extended into three dimensions.

To better understand this, think of the vector as an arrow pointing from the origin to the point \(x, y, z\) in space. The magnitude tells you how long this arrow is. Just like a car's speedometer tells you how fast the car is moving regardless of direction, the magnitude tells us about the size of the vector, independent of direction.

For example, if you have a vector \(\mathbf{v} = (1, -2, 2)\), the magnitude can be calculated as follows:

• Calculate \(1^2 = 1\)
• Calculate \((-2)^2 = 4\)
• Calculate \(2^2 = 4\)
• Add them up: \(1 + 4 + 4 = 9\)
• Take the square root: \(\sqrt{9} = 3\)

And there you have it, the magnitude of \(\mathbf{v} = (1, -2, 2)\) is 3, meaning the vector is 3 units long.

scalar multiplication

Scalar multiplication is the process of multiplying a vector by a scalar (a real number). This operation alters the magnitude of the vector but not its direction. It's like stretching or compressing the vector.

If you have a vector \(\mathbf{v}\) and a scalar \(k\), scalar multiplication is performed by multiplying each component of the vector by \(k\). For example, if \(\mathbf{v} = (x, y, z)\) and you multiply by \(k\), the new vector will be \(k \mathbf{v} = (kx, ky, kz)\).

When finding a unit vector, you divide the vector by its magnitude, which is equivalent to multiplying by the scalar \((1/\text{magnitude})\). This process adjusts the vector's magnitude to 1. Here’s how this applies to our exercise:

• For vector \(\mathbf{v} = (1, -2, 2)\) with magnitude 3, the unit vector is \(\frac{1}{3}\mathbf{v} = (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})\).
• It changes the length of the vector from 3 units to 1 unit.

This adjustment means that scalar multiplication helps us reshape vectors to desired lengths without affecting direction.

mathematical vectors

Mathematical vectors are essential tools in physics and engineering that represent quantities with both magnitude and direction. In mathematics, particularly vector calculus and linear algebra, vectors are objects that can be added together and multiplied by scalars.

Vectors are often denoted in components, such as \(\mathbf{v} = (x, y, z)\) in three dimensions. These components correspond to directions in the coordinate system, moving in the x, y, and z directions.

Understanding vectors involves working with:

• **Addition**: Combining two vectors by adding their corresponding components. For example, \(\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3)\).
• **Subtraction**: Finding the difference between two vectors, essentially the addition of a vector and the negative of another.
• **Magnitude**: As established, the length or size of the vector, indicating how far it stretches in space.
• **Direction**: Vectored by its components, showing both the path and slope in space it occupies.

Vectors are used to model real-world situations, such as force, velocity, and displacement, making them pivotal in many scientific and engineering calculations.