Question

Vector A extends from the origin to $(1,2,3)$, and vector $\mathbf{B}$ extends from the origin to $(2,3,-2)$. Find $(a)$ the unit vector in the direction of $(\mathbf{A}-\mathbf{B})$; (b) the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of $\mathbf{A}$ and $\mathbf{B}$.

Short answer

In short: (a) The unit vector in the direction of $(\mathbf{A}-\mathbf{B})$ is ${\hat{\mathbf{u}}}_{(\mathbf{A}-\mathbf{B})}=\left(\begin{array}{c}-1/\sqrt{27}\\ -1/\sqrt{27}\\ 5/\sqrt{27}\end{array}\right)$. (b) The unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of $\mathbf{A}$ and $\mathbf{B}$ is ${\hat{\mathbf{u}}}_M=\left(\begin{array}{c}1.5/\sqrt{7.25}\\ 2.5/\sqrt{7.25}\\ 0.5/\sqrt{7.25}\end{array}\right)$.

Step-by-step solution

Calculate the difference vector

Subtract the components of vector $\mathbf{B}$ from the components of vector $\mathbf{A}$ to find the difference vector $(\mathbf{A}-\mathbf{B})$. $(\mathbf{A}-\mathbf{B})=\left(\begin{array}{c}1-2\\ 2-3\\ 3-(-2)\end{array}\right)=\left(\begin{array}{c}-1\\ -1\\ 5\end{array}\right)$

Find the magnitude of the difference vector

Calculate the magnitude of $(\mathbf{A}-\mathbf{B})$ using the Pythagorean theorem: $\Vert (\mathbf{A}-\mathbf{B})\Vert =\sqrt{(-1{)}^2+(-1{)}^2+5^2}=\sqrt{27}$

Calculate the unit vector in the direction of $(\mathbf{A}-\mathbf{B})$

Divide each component of $(\mathbf{A}-\mathbf{B})$ by its magnitude to find the unit vector: ${\hat{\mathbf{u}}}_{(\mathbf{A}-\mathbf{B})}=\frac{(\mathbf{A}-\mathbf{B})}{\Vert (\mathbf{A}-\mathbf{B})\Vert }=\left(\begin{array}{c}-1/\sqrt{27}\\ -1/\sqrt{27}\\ 5/\sqrt{27}\end{array}\right)$

Calculate the midpoint of the line segment

Add the coordinates of vector $\mathbf{A}$ and vector $\mathbf{B}$, and divide the sum by 2 to find the midpoint: $\mathbf{M}=\frac{\mathbf{A}+\mathbf{B}}{2}=\left(\begin{array}{c}(1+2)/2\\ (2+3)/2\\ (3-2)/2\end{array}\right)=\left(\begin{array}{c}1.5\\ 2.5\\ 0.5\end{array}\right)$

Find the magnitude of the midpoint vector

Calculate the magnitude of $\mathbf{M}$ using the Pythagorean theorem: $\Vert \mathbf{M}\Vert =\sqrt{(1.5{)}^2+(2.5{)}^2+(0.5{)}^2}=\sqrt{7.25}$

Calculate the unit vector in the direction of the midpoint vector

Divide each component of $\mathbf{M}$ by its magnitude to find the unit vector in the direction of the desired line: ${\hat{\mathbf{u}}}_M=\frac{\mathbf{M}}{\Vert \mathbf{M}\Vert }=\left(\begin{array}{c}1.5/\sqrt{7.25}\\ 2.5/\sqrt{7.25}\\ 0.5/\sqrt{7.25}\end{array}\right)$ To summarize: (a) The unit vector in the direction of $(\mathbf{A}-\mathbf{B})$ is ${\hat{\mathbf{u}}}_{(\mathbf{A}-\mathbf{B})}=\left(\begin{array}{c}-1/\sqrt{27}\\ -1/\sqrt{27}\\ 5/\sqrt{27}\end{array}\right)$ (b) The unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of $\mathbf{A}$ and $\mathbf{B}$ is ${\hat{\mathbf{u}}}_M=\left(\begin{array}{c}1.5/\sqrt{7.25}\\ 2.5/\sqrt{7.25}\\ 0.5/\sqrt{7.25}\end{array}\right)$

Key concepts

Unit Vector

Understanding unit vectors is crucial when dealing with vectors in physics and mathematics. A unit vector is a vector with a magnitude of 1. It essentially points in the same direction as the original vector but is scaled down to a length of exactly one unit, which makes it very useful for indicating direction without being affected by the vector's magnitude.

For example, to find the unit vector in the direction of a difference vector like $\mathbf{A}-\mathbf{B}$, the process involves two main steps. First, we calculate the difference vector which gives us the direction, and then we normalize this difference vector by dividing it by its magnitude. This results in a unit vector.

The formula to calculate a unit vector $\hat{\mathbf{u}}$ is given by:
$$\hat{\mathbf{u}}=\frac{\mathbf{v}}{||\mathbf{v}||}$$
where $\mathbf{v}$ is the original vector, and $||\mathbf{v}||$ is its magnitude. This concept plays a pivotal role in vector operations, including vector subtraction as shown in the example problem.

Vector Magnitude

The magnitude of a vector, often referred to as the vector's length, is a measure of how long the vector is. The magnitude of a vector $\mathbf{v}$ in three-dimensional space with components $x$ , $y$ , and $z$ is found using the three-dimensional version of Pythagoras' theorem:
$$||\mathbf{v}||=\sqrt{x^2+y^2+z^2}$$
The magnitude is always a non-negative number and provides a quantitative measure of the vector's size. To subtract vectors, like in our exercise, we subtract their respective components. After subtraction, when we speak about finding the unit vector, we first need to calculate the magnitude of the resultant difference vector. It is this magnitude which we use to normalize the vector, converting it into a unit vector as shown in the error-free steps of the exercise solution.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, most famously used to relate the sides of a right-angled triangle. In terms of vector operations, it aids in finding the magnitude of vectors, especially when moving beyond two dimensions. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

When calculating the magnitude of a vector in three dimensions, we are effectively using a three-dimensional extension of the Pythagorean theorem:
$$c=\sqrt{a^2+b^2+c^2}$$
where $a$ , $b$ , and $c$ are the vector components. This concept is applied both to find the magnitude of the difference vector $\mathbf{A}-\mathbf{B}$ and the midpoint vector $\mathbf{M}$ in the exercise. A thorough understanding of Pythagorean theorem is thus essential in vector operations, including vector subtraction and normalization.