Question

In each case, find \(\overrightarrow{P Q}\) and \(\|\overrightarrow{P Q}\| .\) a. \(P(1,-1,3), Q(3,1,0)\) b. \(P(2,0,1), Q(1,-1,6)\) c. \(P(1,0,1), Q(1,0,-3)\) d. \(P(1,-1,2), Q(1,-1,2)\) e. \(P(1,0,-3), Q(-1,0,3)\) f. \(P(3,-1,6), Q(1,1,4)\)

Short answer

a) \( \overrightarrow{P Q} = (2, 2, -3), \|\overrightarrow{P Q}\| = \sqrt{17} \); b) \( (-1, -1, 5), \sqrt{27} \); c) \( (0, 0, -4), 4 \); d) \( (0, 0, 0), 0 \); e) \( (-2, 0, 6), \sqrt{40} \); f) \( (-2, 2, -2), \sqrt{12} \).

Step-by-step solution

Understanding Vectors

To find the vector \( \overrightarrow{P Q} \), we need to calculate the difference between the coordinates of points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\). The vector is given by:\[ \overrightarrow{P Q} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \]We then find the magnitude \( \|\overrightarrow{P Q}\| \) using the formula:\[ \|\overrightarrow{P Q}\| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Calculate \( \overrightarrow{P Q} \) for each part

a. Using \(P(1,-1,3)\) and \(Q(3,1,0)\), \( \overrightarrow{P Q} = (3-1, 1 - (-1), 0-3) = (2, 2, -3)\).b. Using \(P(2,0,1)\) and \(Q(1,-1,6)\), \( \overrightarrow{P Q} = (1-2, -1-0, 6-1) = (-1, -1, 5)\).c. Using \(P(1,0,1)\) and \(Q(1,0,-3)\), \( \overrightarrow{P Q} = (1-1, 0-0, -3-1) = (0, 0, -4)\). d. Using \(P(1,-1,2)\) and \(Q(1,-1,2)\), \( \overrightarrow{P Q} = (1-1, -1-(-1), 2-2) = (0, 0, 0)\).e. Using \(P(1,0,-3)\) and \(Q(-1,0,3)\), \( \overrightarrow{P Q} = (-1-1, 0-0, 3-(-3)) = (-2, 0, 6)\).f. Using \(P(3,-1,6)\) and \(Q(1,1,4)\), \( \overrightarrow{P Q} = (1-3, 1-(-1), 4-6) = (-2, 2, -2)\).

Calculate \( \|\overrightarrow{P Q}\| \) for each part

a. \[ \|\overrightarrow{P Q}\| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \] b. \[ \|\overrightarrow{P Q}\| = \sqrt{(-1)^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} \] c. \[ \|\overrightarrow{P Q}\| = \sqrt{0^2 + 0^2 + (-4)^2} = \sqrt{16} = 4 \] d. Since all components are zero, \( \|\overrightarrow{P Q}\| = 0 \). e. \[ \|\overrightarrow{P Q}\| = \sqrt{(-2)^2 + 0^2 + 6^2} = \sqrt{4 + 0 + 36} = \sqrt{40} \] f. \[ \|\overrightarrow{P Q}\| = \sqrt{(-2)^2 + 2^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} \]

Key concepts

Vectors in 3D Space

In vector mathematics, a vector is a mathematical entity defined by both a direction and a magnitude, classically drawn as an arrow. In 3D space, vectors have three components, corresponding to the x, y, and z dimensions. These components are coordinates in a three-dimensional space. A vector is often represented as \( \overrightarrow{PQ} = (x, y, z) \), where \(x, y,\) and \(z\) are the differences between the corresponding components of two points P and Q.
The vector between two points, P (\(x_1, y_1, z_1\)) and Q (\(x_2, y_2, z_2\)), is calculated by taking the difference of their components: \( \overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \). This gives us a new vector that points from P to Q in three-dimensional space.
Vectors in 3D are essential for many applications such as computer graphics, physics simulations, and engineering, where they help describe motion and changes in velocity or forces acting in three-dimensional space.

Magnitude of a Vector

The magnitude of a vector, often denoted \( \|\overrightarrow{PQ}\| \), describes its length or size, irrespective of its direction. To find the magnitude of a vector in 3D space, you can use the distance formula, which is an extension of the Pythagorean theorem.
Given a vector \( \overrightarrow{PQ} = (a, b, c) \), its magnitude is calculated using the formula: \[ \|\overrightarrow{PQ}\| = \sqrt{a^2 + b^2 + c^2} \] This formula provides the 'straight-line' distance between the starting point and the end-point of the vector, equivalent to the vector's length in 3D space.
The magnitude helps in quantifying the vector, allowing engineers and mathematicians to handle forces and mechanical systems precisely. It is also crucial in unitizing vectors, where a vector of one unit length, known as a unit vector, is derived by dividing the vector by its magnitude.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the field of mathematics where we use coordinate points to solve geometrical problems. It combines algebra with geometry, allowing us to represent geometric figures analytically through equations and formulas. In 3D space, coordinate geometry becomes even more powerful.
In the context of vectors, coordinate geometry allows for calculations such as vector addition, subtraction, and scalar multiplication through the manipulation of points and lines in Cartesian coordinates.

• The vector \( \overrightarrow{PQ} \) represents the line segment between points P and Q.
• The coordinates help in describing vector directions, calculating distances, or finding midpoints efficiently.

Coordinate geometry provides a practical approach to tackle complex geometrical problems using simple algebraic manipulations, essential in fields like computer graphics and physics, where spatial reasoning is necessary.

Vector Subtraction

Vector subtraction is a process used to find the vector between two distinct points, P and Q. In other words, it calculates the change or difference as you move from one point to the other. This is an operation often used when assessing displacements or differences.
The operation is performed component-wise, enabling you to find new coordinates for the resulting vector. Given two points, P (\(x_1, y_1, z_1\)) and Q (\(x_2, y_2, z_2\)), vector subtraction is performed as follows:

• The difference in the x-coordinates: \(x_2 - x_1 \)
• The difference in the y-coordinates: \(y_2 - y_1 \)
• The difference in the z-coordinates: \(z_2 - z_1 \)

This yields a new vector \( \overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \).
By calculating the differences in each component, vector subtraction allows us to determine the precise vector pointing from P to Q. This is a foundational concept in physics, where determining the displacement of objects is crucial for understanding motion.