Problem 50 - Algebra and Trigonometry

Question

Unit Vectors 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 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: $$ \langle 1,-2,2\rangle \quad\langle- 6,8,-10\rangle \quad\langle 6,5,9\rangle $$

Step-by-Step Solution

Answer

(a) Multiply by \(\frac{1}{m}\). (b) Unit vectors: \(\langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \rangle\), \(\langle -\frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, -\frac{1}{\sqrt{2}} \rangle\), \(\langle \frac{6}{\sqrt{142}}, \frac{5}{\sqrt{142}}, \frac{9}{\sqrt{142}} \rangle\).

Step 1: Understanding the Problem

To change a vector \( \mathbf{v} \) into a unit vector, we need to find a scalar \( k \) such that when \( \mathbf{v} \) is multiplied by \( k \), the magnitude of the resulting vector is 1. The unit vector is calculated by dividing the vector by its magnitude.

Step 2: Solving Part (a)

Given that the vector \( \mathbf{v} \) has a magnitude \( m \), we want the magnitude of the unit vector to be 1. Therefore, we must multiply \( \mathbf{v} \) by the scalar \( \frac{1}{m} \), so \( \mathbf{u} = \frac{1}{m}\mathbf{v} \) where \( \mathbf{u} \) is the unit vector.

Step 3: Calculating Magnitude of Vectors

Calculate the magnitude of each vector using the formula: \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \).- For \( \langle 1, -2, 2 \rangle \), \( \|\mathbf{v}\| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3 \).- For \( \langle -6, 8, -10 \rangle \), \( \|\mathbf{v}\| = \sqrt{(-6)^2 + 8^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2} \).- For \( \langle 6, 5, 9 \rangle \), \( \|\mathbf{v}\| = \sqrt{6^2 + 5^2 + 9^2} = \sqrt{142} \).

Step 4: Solving Part (b)

Divide each vector by its magnitude to convert it into a unit vector:- For \( \langle 1, -2, 2 \rangle \), the unit vector is \( \frac{1}{3} \langle 1, -2, 2 \rangle = \langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \rangle \).- For \( \langle -6, 8, -10 \rangle \), the unit vector is \( \frac{1}{10\sqrt{2}} \langle -6, 8, -10 \rangle = \langle -\frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, -\frac{1}{\sqrt{2}} \rangle \).- For \( \langle 6, 5, 9 \rangle \), the unit vector is \( \frac{1}{\sqrt{142}} \langle 6, 5, 9 \rangle = \langle \frac{6}{\sqrt{142}}, \frac{5}{\sqrt{142}}, \frac{9}{\sqrt{142}} \rangle \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. Think of it as the distance the vector travels from its starting point to its endpoint. For a vector represented in component form, such as \( \mathbf{v} = \langle x, y, z \rangle \), its magnitude is calculated using the formula: \[ \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \].
This formula finds the 'straight line' distance using the Pythagorean theorem in three dimensions. Consider this with the vector \( \langle 1, -2, 2 \rangle \). Each component (1, -2, and 2) is squared and summed, then we take the square root of that total: \( \|\mathbf{v}\| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3 \).
Magnitude is crucial when normalizing a vector, meaning you adjust its length while keeping its direction the same.

Scalar Multiplication

Scalar multiplication involves taking a vector and multiplying each of its components by a scalar (a real number). This operation alters the vector's magnitude but maintains its direction. For instance, if you multiply a vector \( \mathbf{v} \) by a scalar \( k \), the resulting vector is \( k\mathbf{v} = \langle kx, ky, kz \rangle \).
The key result here is how the scaling factor \( k \) influences the magnitude of the vector. For example, if \( |k| > 1 \), the vector gets longer, while if \( 0 < |k| < 1 \), it gets shorter.
In exercises involving unit vectors, we often use scalar multiplication to adjust a vector's magnitude to exactly 1. For example, to make the vector \( \langle 6, 5, 9 \rangle \) into a unit vector, we multiply by the reciprocal of its current magnitude, \( \frac{1}{\sqrt{142}} \). This operation ensures that the resulting unit vector maintains the direction of \( \mathbf{v} \) but reduces its magnitude precisely to 1.

Vector Normalization

Vector normalization is the process of converting any given vector into a unit vector, which has a magnitude of exactly 1. This is done by dividing each component of the vector by the magnitude of the vector.
Normalization is essential in various applications, such as simplifying vector equations or in computer graphics for defining direction without altering length. The formula for creating a unit vector \( \mathbf{u} \) from a vector \( \mathbf{v} \) with magnitude \( m \) is: \[ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{1}{m}\mathbf{v} \].
Following this method ensures the direction remains unchanged while the vector size becomes 1. For instance, transforming \( \langle 1, -2, 2 \rangle \) into a unit vector involves dividing it by its magnitude: \( \langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \rangle \). This results in a vector pointing in the same direction as \( \mathbf{v} \) but with a smaller, standardized length.