Question

(a) Write down the vector \(\overrightarrow{\mathrm{AB}}\) joining the points A and \(\mathrm{B}\) with coordinates \((3,2,7)\) and \((-1,2,3)\) respectively. (b) Find the equation of the straight line through \(\mathrm{A}\) and \(\mathrm{B}\).

Short answer

Question: Determine the vector joining points A(3, 2, 7) and B(-1, 2, 3), and write the equation of the straight line passing through these points. Answer: The vector joining points A and B is $\begin{pmatrix} -4 \\ 0 \\ -4 \end{pmatrix}$. The equation of the straight line passing through points A and B is given by: $x = 3 - 4t,$ $y = 2,$ $z = 7 - 4t$.

Step-by-step solution

Find the coordinates of the vector AB

The coordinates of the vector joining points A(3, 2, 7) and B(-1, 2, 3) can be calculated using the following formula: $$\overrightarrow{AB} = \begin{pmatrix} B_x - A_x \\ B_y - A_y \\ B_z - A_z\end{pmatrix}$$ Let's calculate the coordinates of vector AB:

Calculate the direction vector of the straight line AB

The direction vector of the straight line AB is the same as the coordinates of vector AB. In this case, the direction vector is: $$\textbf{d} = \begin{pmatrix} -4 \\ 0 \\ -4 \end{pmatrix}$$

Write the equation of the straight line passing A and B using the direction vector and a point on the line

Now, we can write the equation of the straight line AB using the point A and the direction vector. Recall that the equation of a line in 3D space is given by the following parametric form: $$\textbf{r}(t) = \textbf{a} + t\textbf{d}$$ Where \(\textbf{r}(t)\) represents a point on the line, \(\textbf{a}\) is the position vector of a point on the line, and \(\textbf{d}\) is the direction vector. In our case, the position vector of point A is: $$\textbf{a} = \begin{pmatrix} 3 \\ 2 \\ 7 \end{pmatrix}$$ And the direction vector is: $$\textbf{d} = \begin{pmatrix} -4 \\ 0 \\ -4 \end{pmatrix}$$ Therefore, the equation of the straight line AB is: $$\textbf{r}(t) = \begin{pmatrix} 3 \\ 2 \\ 7 \end{pmatrix} + t\begin{pmatrix} -4 \\ 0 \\ -4 \end{pmatrix}$$ Which can be written in component form as: $$x = 3 - 4t$$ $$y = 2$$ $$z = 7 - 4t$$ So the equation of the straight line AB is given by: $$\boxed{x = 3 - 4t,~~ y = 2,~~ z = 7 - 4t}$$

Key concepts

Vector Coordinates

Understanding the vector coordinates is essential to grasping the basics of vector algebra, especially when we move into three-dimensional (3D) spaces. In 3D space, every point can be represented as a position vector with coordinates that define its location relative to the origin. In the context of the exercise, vector \(\overrightarrow{\mathrm{AB}}\) signifies a vector that starts from point A, located at \( (3,2,7) \) and ends at point B, with coordinates \( (-1,2,3) \).

To find these coordinates, we simply subtract the coordinates of A from B resulting in \(\overrightarrow{\mathrm{AB}} = \begin{pmatrix} -1 - 3 \ 2 - 2 \ 3 - 7 \end{pmatrix} = \begin{pmatrix} -4 \ 0 \ -4 \end{pmatrix}\). This subtracted vector reflects the change in position from A to B and thus effectively defines the \(\overrightarrow{\mathrm{AB}}\)'s direction and magnitude in space.

Direction Vector

A direction vector is a vector that gives direction to a line or plane, serving as a sort of 'arrow' showing which way the line or plane 'points'. In the exercise, the direction vector is particularly straightforward because it aligns with \(\overrightarrow{\mathrm{AB}}\), the vector from point A to B.

Specifically, the direction vector \(\textbf{d}\) is \(\begin{pmatrix} -4 \ 0 \ -4 \end{pmatrix}\), indicating that the line passes through points A and B and moves 4 units in the negative x-direction and 4 units in the negative z-direction for each unit of length along the line. It's paramount to ensure this vector does not have all zero coordinates, or it could not serve as a direction, since it would then indicate no movement.

Parametric Equation

The parametric equation of a line provides a means to calculate any point along a line in 3D space given a single scalar parameter, typically denoted by \(t\). It's often represented as \(\textbf{r}(t) = \textbf{a} + t\textbf{d}\), where \(\textbf{r}(t)\) describes a vector to any point on the line, \(\textbf{a}\) is a known point's position vector (like point A in our exercise), and \(\textbf{d}\) is the direction vector.

In this exercise, the parametric equation for the line through points A and B is given by \(\textbf{r}(t) = \begin{pmatrix} 3 \ 2 \ 7 \end{pmatrix} + t\begin{pmatrix} -4 \ 0 \ -4 \end{pmatrix}\), outlining the relationship between the line's points and the parameter \(t\). As \(t\) changes, we can find every point along this line by plugging the value into the equation and solving for the coordinates.

3D Space Geometry

In 3D space geometry, we analyze shapes and objects within a three-dimensional context, using x, y, and z coordinates to describe positions and movements along the length, width, and depth of the space. This is more complicated than 2D geometry because we add an extra dimension which gives depth to our calculations and visualizations.

Our exercise is an example of solving problems within this 3D space, showing that even a simple task like connecting two points with a line involves understanding how the coordinates interact across three axes. The final equation of the line through A and B, presented in the component form \(x = 3 - 4t\), \(y = 2\), \(z = 7 - 4t\), portrays the line as a collection of points whose x and z coordinates vary with \(t\) while the y-coordinate remains constant, illustrating a 3D object's properties within this geometric framework.