Question

a. Find a unit vector in the direction from \(\left[\begin{array}{r}3 \\ -1 \\\ 4\end{array}\right]\) to \(\left[\begin{array}{l}1 \\ 3 \\\ 5\end{array}\right]\) b. If \(\mathbf{u} \neq \mathbf{0}\), for which values of \(a\) is \(a \mathbf{u}\) a unit vector?

Short answer

The unit vector is \( \begin{bmatrix} \frac{-2}{\sqrt{21}} \\ \frac{4}{\sqrt{21}} \\ \frac{1}{\sqrt{21}} \end{bmatrix} \). \( a = \pm 1 \).

Step-by-step solution

Find the Direction Vector

To find the direction vector from \( \begin{bmatrix} 3 \ -1 \ 4 \end{bmatrix} \) to \( \begin{bmatrix} 1 \ 3 \ 5 \end{bmatrix} \), subtract the initial point from the terminal point. The direction vector \( \mathbf{v} \) is \( \begin{bmatrix} 1-3 \ 3-(-1) \ 5-4 \end{bmatrix} = \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} \).

Calculate the Magnitude of the Direction Vector

Calculate the magnitude of the direction vector \( \mathbf{v} = \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} \). The magnitude \| \mathbf{v} \| is given by \( \sqrt{(-2)^2 + 4^2 + 1^2} = \sqrt{4 + 16 + 1} = \sqrt{21} \).

Determine the Unit Vector

A unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is \( \frac{\mathbf{v}}{\| \mathbf{v} \|} = \frac{1}{\sqrt{21}} \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{-2}{\sqrt{21}} \ \frac{4}{\sqrt{21}} \ \frac{1}{\sqrt{21}} \end{bmatrix} \).

Analyze the Unit Vector Condition for a Scalar Multiple

If \( \mathbf{u} eq \mathbf{0} \), the vector \( a\mathbf{u} \) is a unit vector if \( \| a\mathbf{u} \| = 1 \). This means \( |a| \times \| \mathbf{u} \| = 1 \). Given that \( \| \mathbf{u} \| = 1 \), we deduce that \( |a| = 1 \).

Solution for Scalar Multiple Condition

Thus, the scalar \( a \) can have values such that \( a = 1 \) or \( a = -1 \) to ensure that \( a\mathbf{u} \) remains a unit vector.

Key concepts

Direction Vectors

Understanding direction vectors is essential when working with vectors in geometry and physics. A direction vector is a vector that represents the direction and length from one point to another. In this exercise, we find the direction vector from

• \[ \begin{bmatrix} 3 \ -1 \ 4 \end{bmatrix} \]
• to \[ \begin{bmatrix} 1 \ 3 \ 5 \end{bmatrix} \]

We do this by subtracting the coordinates of the initial point from the terminal point:

• Direction vector \( \mathbf{v} \) is \[ \begin{bmatrix} 1-3 \ 3-(-1) \ 5-4 \end{bmatrix} = \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} \]

This vector points from the first point to the second point, showing both the direction and distance between them.
Direction vectors are fundamental when you need to navigate from one place to another in space.

Magnitude Calculation

The magnitude of a vector, often called its length or norm, is crucial for understanding how long the vector is, irrespective of direction. Calculating magnitude involves finding the square root of the sum of the squares of the vector's components.
For our direction vector

• \( \mathbf{v} = \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} \)

we calculate the magnitude using the formula:

• \[ \| \mathbf{v} \| = \sqrt{(-2)^2 + 4^2 + 1^2} \]
• \[ = \sqrt{4 + 16 + 1} = \sqrt{21} \]

This tells us the vector's "length" in a geometrical sense.
Knowing how to find the magnitude is helpful when normalizing vectors or comparing the lengths of different vectors.

Scalar Multiples

Scalar multiples of a vector are created by multiplying the vector by a scalar (a single number). This operation stretches or shrinks the vector by the scalar's value.
In our case, we want to ensure a scalar multiple is a unit vector, meaning its magnitude should be 1. If a vector

• \( \mathbf{u} eq \mathbf{0} \)

produces an equation where

• \[ \| a \mathbf{u} \| = 1 \]

then

• \( |a| \times \| \mathbf{u} \| = 1 \)

Since

• \( \| \mathbf{u} \| = 1 \),

we realize

• \( |a| = 1 \).
• Hence, \( a = 1 \) or \( a = -1 \).

Scalar multiples are a simple yet powerful way to manipulate vectors for various applications such as scaling in graphical environments or physics calculations.