Lesson Video: Equation of a Straight Line in Space: Parametric Form | Nagwa Lesson Video: Equation of a Straight Line in Space: Parametric Form | Nagwa

Lesson Video: Equation of a Straight Line in Space: Parametric Form Mathematics

In this video, we will learn how to find the parametric equations of straight lines in space.

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Video Transcript

In this video, we will learn how to find the parametric equations of straight lines in space. We will begin by recalling the different ways of writing the equation of a straight line in the π₯π¦-plane.

The form π¦ equals ππ₯ plus π is known as slope-intercept form, where π is the slope or gradient of the line and π is the π¦-intercept. Writing the equation of a line in slope-point form, we have π¦ minus π¦ sub zero is equal to π multiplied by π₯ minus π₯ sub zero. Once again, π is the slope of the line and the point with coordinates π₯ sub zero, π¦ sub zero lies on the line. Finally, we have the general form written ππ₯ plus ππ¦ plus π equals zero. This form is usually found by rearranging one of the first two. The values of π, π, and π are constants, where π is nonnegative and only one of π and π can be zero.

We can use this information to find the vector equation of a straight line in two dimensions. The position vector π« of any point on a line containing the point π with coordinates π₯ sub zero, π¦ sub zero with position vector π« sub zero is given by π« is equal to π« sub zero plus π‘ multiplied by π, where π is the direction vector of the line and π‘ is any scalar value. This equation can be written in terms of its components, where the vector π« has components π₯ and π¦. The position vector π« sub zero is equal to π₯ sub zero, π¦ sub zero, and we will let the direction vector π have components π and π.

We can then write this equation in parametric form. Looking at the first components, we have π₯ is equal to π₯ sub zero plus π‘ multiplied by π. Equating the second components, we have π¦ is equal to π¦ sub zero plus π‘ multiplied by π. These two equations combined give us the equation of a straight line in two dimensions written in parametric form. We will now consider how this can be extended to three dimensions.

The parametric equations of a line in space are a nonunique set of three equations of the form π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point on the line. The vector π₯, π¦, π§ is a direction vector of the line. And π‘ is a real number that varies from negative β to β. We will now look at an example where we need to find the parametric equations of a line given a point that lies on the line and its direction vector.

Give the parametric equation of the line on point two, negative four, four with direction vector one, negative one, five.

We begin by recalling that the parametric equations of a line are as follows. π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point that lies on the line. And the vector π₯, π¦, π§ is a direction vector of the line. We also have the parameter π‘, which is a scalar quantity. In this question, we are told that the point two, negative four, four lies on the line. So these will be the values of π₯ sub zero, π¦ sub zero, and π§ sub zero, respectively. We are also given a direction vector one, negative one, five, which will be the values of π₯, π¦, and π§.

Substituting the values of π₯ sub zero and π₯, we have π₯ is equal to two plus π‘. Next, we have π¦ is equal to negative four minus π‘. Finally, substituting the values four and five for π§ sub zero and π§, we have π§ is equal to four plus five π‘. The parametric equation of the line on point two, negative four, four with direction vector one, negative one, five is as shown.

It is important to note that this set of equations is not unique. For example, we could multiply each component of the direction vector by two, giving us two, negative two, 10. We could then use this as the direction vector together with the point two, negative four, four, giving us a set of parametric equations π₯ is equal to two plus two π‘, π¦ is equal to negative four minus two π‘, and π§ is equal to four plus 10π‘. This would also be a valid solution to this question. Alternatively, we could find another point that lies on the line by choosing a value to substitute for π‘. We could then use this point together with a direction vector to find a different set of parametric equations that satisfy this question.

In our next question, weβll find the parametric equation of a line that passes through two points in space.

Write the equation of the straight line πΏ passing through the points π sub one β which is equal to four, one, five β and π sub two β which is equal to negative two, one, three β in parametric form. Is it option (A) π₯ is equal to negative two minus six π‘, π¦ is equal to one, and π§ is equal to three plus two π‘? Option (B) π₯ is equal to four minus six π‘, π¦ is equal to one, and π§ is equal to five minus two π‘. Is it option (C) π₯ is equal to four plus two π‘, π¦ is equal to one plus π‘, and π§ is equal to five plus three π‘? Option (D) π₯ is equal to negative two plus six π‘, π¦ is equal to one plus π‘, and π§ is equal to three plus two π‘. Or is it option (E) π₯ is equal to six plus four π‘, π¦ is equal to one, and π§ is equal to five plus two π‘? For all five options, we are told that π‘ is greater than negative β and less than β.

We begin by recalling that the parametric equations of a straight line are written in the form π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point that lies on the line and the vector π₯, π¦, π§ is a direction vector of the line. We also know that π‘ is a scalar quantity that lies between negative β and β. We are given two points that lie on the line: four, one, five and negative two, one, three. So we can substitute either of these for π₯ sub zero, π¦ sub zero, and π§ sub zero.

In options (A) and (D), the coordinates of π sub two have been used, whereas in options (B) and (C), the coordinates of π sub one have been used. π₯ sub zero equals four, π¦ sub zero equals one, and π§ sub zero equals five. Whilst the values of π₯ sub zero, π¦ sub zero, and π§ sub zero in option (E) do not match those coordinates in π sub one or π sub two, this is not enough to say that the point six, one, five does not lie on the line. This is because we can use any point that lies on the straight line and not just one of the two that are given.

We therefore need to focus on calculating a direction vector from the information given. One way of doing this would be to work out the vector π sub one, π sub two. We can do this by subtracting the vector four, one, five from the vector negative two, one, three. Subtracting the corresponding components gives us negative six, zero, negative two. This is one possible direction vector of the straight line πΏ. Since we are given a limited number of options in this question, we can compare this direction vector with the direction vectors in options (A) to (E).

In option (A), the direction vector is equal to negative six, zero, two. In option (B), we have negative six, zero, negative two. In options (C), (D), and (E), the direction vectors are two, one, three; six, one, two; and four, zero, two, respectively. The only option that has a direction vector equal to negative six, zero, negative two is option (B).

As already mentioned, this also passes through the point π sub one. We can therefore conclude that from the options given, the equation of the straight line πΏ is π₯ equals four minus six π‘, π¦ is equal to one, and π§ is equal to five minus two π‘. Option (B) is correct. It is important to note here that the parametric equations of straight lines are not unique. However, as none of our other four options have direction vectors that are equal to or parallel to the direction vector negative six, zero, negative two, then these are not correct.

Before looking at one final example, we will consider how we can write the parametric form of the equation of a straight line from its Cartesian form.

We begin by recalling that the Cartesian form of the equation of a line can be written π₯ minus π₯ sub zero over π₯ is equal to π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub zero over π§. Once again, the point π₯ sub zero, π¦ sub zero, π§ sub zero lies on the line and the vector π₯, π¦, π§ is a direction vector. These values of π₯, π¦, and π§ must be nonzero real numbers. This is closely related to the set of parametric equations as they simply give three values of π‘.

Firstly, by considering π‘ is equal to π₯ minus π₯ sub zero over π₯, we can multiply both sides of this equation by π₯ and then add π₯ sub zero to both sides, giving us π₯ is equal to π₯ sub zero plus π‘ multiplied by π₯. In the same way, we have π¦ is equal to π¦ sub zero plus π‘ multiplied by π¦ and π§ is equal to π§ sub zero plus π‘ multiplied by π§. These are the three parametric equations that we have already seen in this video.

At this point, it is also worth noting what happens when one component of the direction vector equals zero, for example, if the direction vector of the line is equal to π₯, π¦, zero. This means that the third parametric equation becomes π§ is equal to π§ sub zero. In this case, the line is perpendicular to the π§-axis and is in a plane that is parallel to the π₯π¦-plane. Letβs now consider what happens when two components of the direction vector equal zero, for example, π₯, zero, zero. If both π¦ and π§ are equal to zero, then π¦ is equal to π¦ sub zero and π§ is equal to π§ sub zero. This means that the direction vector is undimensional. And if this direction vector is parallel to the π₯-axis, then π₯ equals π‘.

We will now look at an example where we need to convert Cartesian equations into parametric equations.

Find the parametric equations of the straight line three π₯ minus seven over negative nine is equal to eight π¦ minus three over four, which is equal to negative eight minus six π§ over negative nine.

We begin by recalling that the standard form of a Cartesian equation of a straight line is π₯ minus π₯ sub zero over π₯ is equal to π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub zero over π§. The three parts of the equation in this question have been slightly rearranged from the general form. However, this does not matter, and we can simply set each of the equations equal to our parameter, in this case π‘. Firstly, we have π‘ is equal to three π₯ minus seven over negative nine. We need to rearrange this so that it is in the form π₯ is equal to π₯ sub zero plus π‘ multiplied by π₯. This means that we need to make π₯ the subject.

Multiplying through by negative nine and then adding seven to both sides gives us seven minus nine π‘ is equal to three π₯. We can then divide through by three, giving us π₯ is equal to seven-thirds minus three π‘. We will then repeat this process by rearranging π‘ is equal to eight π¦ minus three over four so that it is in the form π¦ is equal to π¦ sub zero plus π‘π¦. Multiplying through by four and adding three to both sides gives us three plus four π‘ is equal to eight π¦. And then dividing through by eight, we have π¦ is equal to three-eighths plus a half π‘. Finally, we need to rearrange the equation π‘ is equal to negative eight minus six π§ over negative nine such that π§ is the subject. This gives us π§ is equal to negative four-thirds plus three over two π‘.

We now have the three parametric equations in their required form. The straight line three π₯ minus seven over negative nine, which is equal to eight π¦ minus three over four, which is equal to negative eight minus six π§ over negative nine, written in parametric form, is π₯ equals seven-thirds minus three π‘, π¦ equals three-eighths plus a half π‘, and π§ equals negative four-thirds plus three over two π‘.

We will now summarize the key points from this video. The parametric equations of a line are a nonunique set of equations of the form π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point on the line. π₯, π¦, π§ is a direction vector of the line. And π‘ is a real number that varies from negative β to β. Where π₯, π¦, and π§ are all nonzero real numbers, the parametric equations can be derived from the Cartesian equations by writing π‘ is equal to π₯ minus π₯ sub zero over π₯, which is equal to π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub zero over π§. We can then rearrange each of the three equations so theyβre written in parametric form.

When one component of the direction vector is zero, it means that the corresponding coordinates of all the points lying on the line are constant. For example, if the direction vector is equal to π₯, π¦, zero, we have π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero. When two components of the direction vector are zero, the line is parallel to an axis, meaning that only one coordinate varies, while the other two are fixed. For example, the parametric equations for a line parallel to the π₯-axis, where the direction vector is π₯, zero, zero, are π₯ equals π‘, π¦ is equal to π¦ sub zero, and π§ is equal to π§ sub zero.