Lesson Explainer: Equation of a Straight Line in Space: Cartesian and Vector Forms | Nagwa Lesson Explainer: Equation of a Straight Line in Space: Cartesian and Vector Forms | Nagwa

Lesson Explainer: Equation of a Straight Line in Space: Cartesian and Vector Forms Mathematics • Third Year of Secondary School

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In this explainer, we will learn how to find the Cartesian and vector forms of the equation of a straight line in space.

When we are considering equations in space, we have coordinates in three dimensions, as (𝑥,𝑦,𝑧), rather than in two dimensions as (𝑥,𝑦).

In vector form, we consider that a line is defined by any point on the line and a direction. To find an equation representing a line, 𝐿, in three dimensions, we choose a point, 𝑃, on the line and a nonzero vector, 𝑑, parallel to the line, where 𝑟 is the position vector of 𝑃.

As 𝑑 is parallel to 𝐿, then adding any constant multiple of 𝑑 to 𝑟 will produce the position vector of a point on the line. Thus, a position vector, 𝑟, on the line can be found by the movement to point 𝑃(𝑥,𝑦,𝑧) on the line along the vector 𝑟 and then by following a constant multiple of the vector 𝑑.

We can then establish the vector form of a line as follows.

Definition: Vector Form of a Line

The position vector, 𝑟, of any general point on a line that contains the point 𝑃(𝑥,𝑦,𝑧) at position vector 𝑟 is given by 𝑟=𝑟+𝑡𝑑, where 𝑑 is a direction vector of, or along, the line and 𝑡 is any scalar multiple.

We will now see an example of how we can write a vector equation of a line, given a point on the line and a direction vector.

Example 1: Finding the Vector Equation of a Line given a Point and Its Direction Vector

Give the vector equation of the line through the point (3,7,7) with direction vector (0,5,7).

Answer

We remember that a vector can be written in the form 𝑟=𝑟+𝑡𝑑, where 𝑟 is the position vector of any general point on a line with direction vector 𝑑 containing a point at position vector 𝑟. The value 𝑡 represents a scalar multiple.

We are given a direction vector, (0,5,7).

Since the point (3,7,7) lies on the line, we can set 𝑟 to be the position vector of this point, (3,7,7).

Thus, substituting the position vector, 𝑟=(3,7,7) and the given direction vector as 𝑑=(0,5,7) into the vector equation of a line gives the solution 𝑟=(3,7,7)+𝑡(0,5,7).

We can find the direction vector 𝐴𝐵 of a straight line between two points 𝐴 and 𝐵 by subtracting the position vector of the initial point, 𝐴, from the position vector of the terminal point, 𝐵. To do this, we subtract each of the 𝑥-, 𝑦-, and 𝑧-components of the position vector of 𝐴 from the position vector of 𝐵. Let us see an example of how we can do this.

Example 2: Finding the Direction Vector of a Line given Two Points

Find the direction vector of the straight line passing through 𝐴(1,2,7) and 𝐵(4,1,3).

Answer

The direction vector of a line is a nonzero vector parallel to the line. In order to find the direction vector, 𝑑, of the straight line between the points 𝐴 and 𝐵, we notice this line must have the same direction as the vector from 𝐴 to 𝐵. We can find this vector by subtracting each of the 𝑥-, 𝑦-, and 𝑧-components of 𝑂𝐴 from those in 𝑂𝐵.

Therefore, we have 𝑑=(41,1(2),37)=(3,1,4).

We can give the solution that the direction vector of the straight line passing through 𝐴(1,2,7) and 𝐵(4,1,3) is (3,1,4).

As a side note, we were not specifically given the direction vector as 𝐴𝐵; thus, the vector 𝐵𝐴=(3,1,4) would also have been a valid direction vector. Indeed, any nonzero multiple of either of these direction vectors would be correct.

We often need to find the midpoint of a line joining two points in space. We can do this using the same process for three dimensions as we do for a line in two dimensions. Finding the mean of each of the 𝑥-, 𝑦-, and 𝑧-coordinates of both endpoints allows us to identify the coordinates of the midpoint of a three-dimensional line segment.

Recap: The Midpoint of Two Points in 3D

The midpoint, 𝑀, of any two points, (𝑥,𝑦,𝑧) and (𝑥,𝑦,𝑧), is given by 𝑀=𝑥+𝑥2,𝑦+𝑦2,𝑧+𝑧2.

We will see how this formula can be used in the next question to find the median of a triangle drawn in space.

Example 3: Finding the Vector Equation of One Median of a Triangle

The points 𝐴(8,9,2), 𝐵(0,7,6), and 𝐶(8,1,4) form a triangle. Determine, in vector form, the equation of the median drawn from 𝐶.

Answer

The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. We can consider the two-dimensional representation of the midpoint, 𝑀, on the opposite side of vertex 𝐶, that when connected to vertex 𝐶, forms the median 𝐶𝑀.

We remember that the vector equation of a line can be written in the form 𝑟=𝑟+𝑡𝑑, where 𝑟 is the position vector of any general point on a line with direction vector 𝑑 containing a point at position vector 𝑟. The value 𝑡 represents a scalar multiple.

To find the vector form of the equation of the median, we can begin by finding the midpoint, 𝑀, of 𝐴𝐵.

The midpoint, 𝑀, of two points, (𝑥,𝑦,𝑧) and (𝑥,𝑦,𝑧), is found by 𝑀=𝑥+𝑥2,𝑦+𝑦2,𝑧+𝑧2.

So, for 𝐴(8,9,2) and 𝐵(0,7,6), we have 𝑀=8+02,9+(7)2,2+62.

Evaluating this gives us 𝑀=(4,8,2).

The vector 𝐶𝑀 will be the direction vector of the median. To find 𝐶𝑀, we subtract the 𝑥-, 𝑦-, and 𝑧-components of 𝑂𝐶, (8,1,4), from 𝑂𝑀, (4,8,2), which gives 𝐶𝑀=(4(8),8(1),2(4))=(4,7,6).

We can now apply the vector form of the equation 𝑟=𝑟+𝑡𝑑, where 𝑡 is a scalar multiple, as 𝑟=𝑂𝐶+𝑡𝐶𝑀.

Substituting in 𝑂𝐶=(8,1,4) and 𝐶𝑀=(4,7,6), we obtain the solution for the vector form of the median drawn from 𝐶 as 𝑟=(8,1,4)+𝑡(4,7,6).

An alternative solution can be found by using the direction vector 𝑀𝐶 of the median. In this case, the vector form of the equation would use the position vector 𝑂𝑀 to give the equation in the form 𝑟=𝑂𝑀+𝑡𝑀𝐶, where 𝑡 is a scalar multiple.

Substituting 𝑂𝑀=(4,8,2) and 𝑀𝐶=(4,7,6) gives 𝑟=(4,8,2)+𝑡(4,7,6).

There are a number of ways of writing the equation of a line. We have the vector equation of a line written in the form 𝑟=𝑟+𝑡𝑑, where 𝑟 is the position vector of any general point on a line with direction vector 𝑑 containing a point at position vector 𝑟. The value 𝑡 represents a scalar multiple.

We can also represent vector 𝑟 as (𝑥,𝑦,𝑧) and direction vector 𝑑 as (𝑎,𝑏,𝑐). Thus, we can write the vector form alternatively as 𝑟=(𝑥,𝑦,𝑧)+𝑡(𝑎,𝑏,𝑐).

Representing 𝑟 as (𝑥,𝑦,𝑧), and simplifying the components on the right-hand side, we can write this equation as (𝑥,𝑦,𝑧)=(𝑥+𝑎𝑡,𝑦+𝑏𝑡,𝑧+𝑐𝑡).

When two vectors are equal, their components are equal. This means we have 𝑥=𝑥+𝑎𝑡𝑦=𝑦+𝑏𝑡𝑧=𝑧+𝑐𝑡.

This set of equations gives us the parametric equations of the line. To find a point on the line in this form, we could select any value of 𝑡 and substitute it into each of the equations.

We could also rearrange this set of equations to find the value of 𝑡 in each case, assuming 𝑎, 𝑏, and 𝑐 are nonzero: 𝑡=𝑥𝑥𝑎,𝑡=𝑦𝑦𝑏,𝑡=𝑧𝑧𝑐.

As the value of 𝑡 will be the same in each equation, we can set these expressions on the right-hand side equal to each other: 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐.

This is the Cartesian equation of a line.

Definition: The Cartesian Equation of a Line

The equation of a line with direction vector 𝑑=(𝑎,𝑏,𝑐), where 𝑎, 𝑏, and 𝑐 are nonzero real numbers, that passes through the point (𝑥,𝑦,𝑧) is given by 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐.

We can still use the Cartesian form of the equation of a line if one or two of the variables, 𝑎, 𝑏, or 𝑐, are equal to 0. For example, suppose that 𝑎=0 and 𝑏 and 𝑐 are nonzero. In this case, 𝑡 will not exist in the parametric equation for 𝑥, so we could only solve the parametric equations for 𝑦 and 𝑧. We set those equal to each other and note the parametric equation for 𝑥; thus, we have two equations: 𝑥=𝑥,𝑦𝑦𝑏=𝑧𝑧𝑐.

We will now see an example of how we can change a line given in vector form to one given in Cartesian form.

Example 4: Finding the Cartesian Equation of a Line given Its Vector Equation

Give the Cartesian equation of the line 𝑟=(3,2,2)+𝑡(4,2,4).

Answer

As we have the equation of a line in vector form, we can observe that we have the position vector of a point (3,2,2). We also have the direction vector (4,2,4).

We can use the fact that the equation of a line with direction vector 𝑑=(𝑎,𝑏,𝑐), where 𝑎, 𝑏, and 𝑐 are nonzero real numbers, that passes through the point (𝑥,𝑦,𝑧) is given by 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐.

We then substitute the values of (𝑥,𝑦,𝑧)=(3,2,2) and (𝑎,𝑏,𝑐)=(4,2,4) into this form.

This gives 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐𝑥(3)4=𝑦(2)2=𝑧(2)4.

Simplifying the numerators, the solution for the Cartesian equation of this line is 𝑥+34=𝑦+22=𝑧+24.

In the final example, we can see how we use two points on a line in space to find its Cartesian equation.

Example 5: Finding the Cartesian Equation of a Line given Two Points

Find the Cartesian form of the equation of the straight line passing through the points (7,3,7) and (3,10,4).

Answer

One method we can use to find the Cartesian equation of a line drawn in space is by considering it in a similar way to the vector form; that is, a line described by a point on the line and direction vector.

Given the two points 𝐴(7,3,7) and 𝐵(3,10,4), we can find the direction vector 𝐴𝐵 by subtracting the 𝑥-, 𝑦-, and 𝑧-components of the position vector of 𝐴, from those of the position vector of 𝐵.

This gives us 𝐴𝐵=(3,10,4)(7,3,7)=(3(7),10(3),4(7))=(4,7,3).

We can use the fact that the equation of a line with direction vector 𝑑=(𝑎,𝑏,𝑐), where 𝑎, 𝑏, and 𝑐 are nonzero real numbers, that passes through the point (𝑥,𝑦,𝑧) is given by 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐.

We can substitute the coordinates of the point (7,3,7) and the components of the direction vector (4,7,3) into this equation. This gives 𝑥(7)4=𝑦(3)7=𝑧(7)3.

Simplifying the numerators, the solution for the Cartesian equation of this line is 𝑥+74=𝑦+37=𝑧+73.

Key Points

  • We can find the direction vector of a line passing through the points 𝐴 and 𝐵 by finding the vector 𝐴𝐵.
  • The position vector, 𝑟, of any general point on a line that contains the point 𝑃(𝑥,𝑦,𝑧) at position vector 𝑟 is given by 𝑟=𝑟+𝑡𝑑, where 𝑑 is the direction vector of the line and 𝑡 is any scalar multiple.
  • The Cartesian form of a line with direction vector 𝑑=(𝑎,𝑏,𝑐), where 𝑎, 𝑏, and 𝑐 are nonzero real numbers, that passes through the point (𝑥,𝑦,𝑧) is given by 𝑥𝑥𝑎=𝑦𝑦𝑏=𝑧𝑧𝑐.
  • If one or two of 𝑎, 𝑏, and 𝑐 are equal to 0 in the form above (e.g., if 𝑎=0), then there are two equations: 𝑥=𝑥,𝑦𝑦𝑏=𝑧𝑧𝑐.

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