# Lesson Video: Equation of a Straight Line in Space: Cartesian and Vector Forms Mathematics

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

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### Video Transcript

In this video, we’ll see how we can find the equation of a straight line in space. So that means we’re looking at coordinates in three dimensions rather than just in two dimensions. We’ll see how we can write this equation in Cartesian form, which is sometimes called the general form. And we’ll also see how we can write it in vector form. Let’s begin by having a look at the vector form.

The vector form of a line can be described as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫, 𝐫 sub zero, and 𝐯 are all vectors. 𝐫 is the position vector of any general point on the line. 𝐫 sub zero is the position vector of a given point on the line. 𝐯 is the direction vector of or along the line. And 𝑡 is a scalar multiple. Vector form can be used in both two dimensions and three dimensions. The difference is that in three dimensions, all of our vectors will have 𝑥-, 𝑦-, and 𝑧-components. When we’re describing a line in vector form, remember that we’re thinking about how we navigate from the origin to a specific point. And then we’re moving along that line in scalar multiples of the vector 𝐯.

We’ll now have a look at a question where we need to write the vector equation of a line given a point on the line and a direction vector.

Give the vector equation of the line through the point three, seven, negative seven with direction vector zero, negative five, seven.

We should remember that when we need to write an equation in vector form, it will be in the form 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫 is the position vector of a general point on the line, 𝐫 sub zero is a position vector of a given point on the line, and 𝐯 is the direction vector. 𝑡 is a scalar multiple. If we look at the information that we’re given in the question, we can see that we have a direction vector. And we’ve got a point on the line which can be written as a position vector. As we navigate from the origin to the point three, seven, negative seven, then we can write this as the position vector three, seven, negative seven.

We can then simply plug in these two vectors into the vector form. 𝐫 equals the position vector three, seven, negative seven plus 𝑡 times the direction vector zero, negative five, seven. And so that’s the answer for the vector equation of the line.

In the following question, we’ll see how we can calculate a direction vector given two points.

Find the direction vector of the straight line passing through 𝐴 one, negative two, seven and 𝐵 four, negative one, three.

In this question, we’re given the position vectors of two points in space, 𝐴 and 𝐵, and we’re asked to find the direction vector. When we want to find a direction vector 𝐀𝐁, 𝐴 is the starting point and 𝐵 is the terminal point, we subtract the starting point from the terminal point. In order to find the direction vector, we can subtract each of the 𝑥-, 𝑦-, and 𝑧-components in 𝐴 from those in 𝐵. To begin then, we’ll have four subtract one, giving us three. Then we’ll have negative one subtract negative two, which is equivalent to negative one plus two, which is one. And finally, we’ll have three subtract seven, giving us negative four. And so we’ve got the answer for the direction vector of 𝐝 as three, one, negative four.

In this question, however, we didn’t necessarily need to find the direction vector 𝐀𝐁. We could also have found the direction vector 𝐁𝐀. In this case, we would’ve got the vector inverse of 𝐝 equals negative three, negative one, four, which would also have been a valid answer.

In the next question, we’ll see a slightly more complex example where we need to find the vector equation of a median of a triangle drawn in three-dimensional space.

The points 𝐴 negative eight, negative nine, negative two; 𝐵 zero, negative seven, six; and 𝐶 negative eight, negative one, negative four form a triangle. Determine in vector form the equation of the median drawn from 𝐶.

In this question, we have three points 𝐴, 𝐵, and 𝐶, which are given in three-dimensional space. These three points we’re told form a triangle. We’re told that there is a median drawn from 𝐶, and so it would be useful to recall that a median is a line segment joining a vertex to the midpoint of the opposite side. For example, if we drew this two-dimensional triangle 𝐴𝐵𝐶, the median from 𝐶 would look like this.

Perhaps the best way to begin this question is to see if we can find the point that is the midpoint of 𝐴𝐵. Let’s define this with the letter 𝑀. The formula to find the midpoint of two points in space is very similar to that which we might use for two coordinates in two-dimensional space. To find the midpoint 𝑀 of 𝑥 one, 𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two, we have that 𝑀 is equal to 𝑥 one plus 𝑥 two over two, 𝑦 one plus 𝑦 two over two, 𝑧 one plus 𝑧 two over two. When we fill our values into this formula, we need to make sure we’re using the values for 𝐴 and 𝐵 as, after all, we need to find the midpoint of 𝐴𝐵.

Note that when we’re plugging in our values, it doesn’t matter which point we use with our 𝑥 one, 𝑦 one, 𝑧 one values or the 𝑥 two, 𝑦 two, 𝑧 two values. So we have that the midpoint 𝑀 is equal to negative eight plus zero over two, negative nine plus negative seven over two, and negative two plus six over two. Simplifying this, we have that 𝑀 is equal to negative four, negative eight, two. We can now clear some space so we can begin to think about the vector form of the equation of this median. The vector form of an equation can be written in the form 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫 is the position vector of a general point on the line, 𝐫 sub zero is the position vector of a given point on the line, and 𝐯 is the direction vector. 𝑡 is a scalar multiple.

Let’s think about what would happen if we model these three points in three-dimensional space. We’d have the triangle 𝐴𝐵𝐶 and the median, which would be the line segment of 𝐶𝑀. So when it comes to writing the median in vector form, the position vector can be the point 𝐶. But we still need to work out the direction vector of 𝐂𝐌. To find the vector 𝐂𝐌, we subtract the starting point 𝐶 from the terminal point 𝑀. So we have negative four subtract negative eight, negative eight subtract negative one, and two subtract negative four. Simplifying this, we have that vector 𝐂𝐌 is equal to four, negative seven, six.

Now we have all the information that we need to plug in to the vector form of the line. 𝐫 sub zero will be the position vector representing point 𝐶. Vector 𝐯 will be represented by the vector 𝐂𝐌. Therefore, the answer for the equation of the median from 𝐶 is 𝐫 equals negative eight, negative one, negative four plus 𝑡 four, negative seven, six.

So far in this video, we’ve looked at equations of lines in vector form. Now we’ll think about how we change a line given in vector form to one in Cartesian form. You may be confused by the terminology of Cartesian form, but a line in two dimensions in Cartesian form can be written in the form 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient and 𝑏 is the 𝑦-intercept. But of course, it’s different in three-dimensional space as we need an equation that describes the 𝑥-, 𝑦-, and 𝑧-variables.

So, to find the equation of a line in Cartesian form, we can say that given the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 which passes through the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one, then it’s given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers. So we can see how this information of an equation with a point and a direction, i.e., one in vector form, can be changed into one in Cartesian form. We’ll now have a look at two questions where we could apply this formula.

Give the Cartesian equation of the line 𝐫 equals negative three, negative two, negative two plus 𝑡 four, two, four.

In this question, we’re given this equation in vector form. Negative three, negative two, negative two is the position vector of a given point, and four, two, four is the direction vector. In order to change the equation in vector form into an equation in Cartesian form, there is a formula we can apply. The equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers.

We now need to take the direction vector four, two, four to have the values of 𝑙, 𝑚, and 𝑛, respectively. We can do the same and designate the coordinate 𝑥 sub one, 𝑦 sub one, 𝑧 sub one with the values negative three, negative two, negative two. Plugging these values into the formula, we have 𝑥 minus negative three over four equals 𝑦 minus negative two over two equals 𝑧 minus negative two over four. Simplifying the numerators, we have 𝑥 plus three over four equals 𝑦 plus two over two equals 𝑧 plus two over four. And that’s the answer for the Cartesian equation of the given line.

Let’s have a look at one final question.

Find the Cartesian form of the equation of the straight line passing through the points negative seven, negative three, negative seven and negative three, negative 10, negative four.

In this question, although we’re asked for the Cartesian form of the equation, it might be useful to think about what this line would be like in a vector form. If we considered our two points as 𝐴 and 𝐵 and we wanted to find the direction vector of 𝐀𝐁, then we would subtract all the points on our starting point 𝐴 from those on 𝐵. So we’d have 𝐀𝐁 is equal to negative three subtract negative seven, which is equivalent to negative three plus seven, giving us four. Negative 10 subtract negative three is equivalent to negative 10 plus three, which is negative seven. And then we’d have negative four subtract negative seven, which gives us three.

Now that we have a direction vector and a point on a line, we can find the Cartesian form of the equation of the line joining these two points. We should remember that the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛. Note that 𝑙, 𝑚, and 𝑛 are nonzero real numbers.

We can now use the direction vector of 𝐀𝐁 for the values of 𝑙, 𝑚, and 𝑛 and the point negative seven, negative three, negative seven for the values 𝑥 sub one, 𝑦 sub one, and 𝑧 sub one. Filling these values into the formula, we have 𝑥 subtract negative seven over four equals 𝑦 subtract negative three over negative seven equals 𝑧 subtract negative seven over three. Simplifying the numerators gives us the answer in Cartesian form 𝑥 plus seven over four equals 𝑦 plus three over negative seven equals 𝑧 plus seven over three.

We can now summarize the key points of this video. Firstly, we saw the equation of a line given in vector form as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯. 𝐫 is the position vector of a general point on the line, 𝐫 sub zero is the position vector of a given point on the line, and 𝐯 is the direction vector. 𝑡 is a scalar multiple. To find a direction vector 𝐀𝐁, we subtract the starting point 𝐴 from the terminal point 𝐵.

Finally, we saw that the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers. This final formula is very useful for changing the equation of a line given in vector form to an equation given in Cartesian form.