Lesson Video: Parallel, Perpendicular, and Intersecting Lines in Space Mathematics

In this video, we will learn how to find the equation of a straight line that is parallel or perpendicular to another one in space and find the intersection point between two lines.

16:29

Video Transcript

In this video, we’ll learn how to find the equation of a straight line that is parallel or perpendicular to another one in space and find the intersection point between two lines.

We might recall that we can characterize a line in three dimensions in vector form, using two pieces of information. We need a point on the line and a nonzero vector that’s parallel to that line. Now this is called the direction vector. And vectors are parallel when they’re scalar multiples of one another. Specifically the vector equation of a line in three dimensions is given by 𝐫 of 𝑑 equals π‘₯ nought, 𝑦 nought, 𝑧 nought plus 𝑑 times π‘Ž, 𝑏, 𝑐. Where π‘₯ nought, 𝑦 nought, 𝑧 nought is the position vector of the point that belongs to the line and π‘Ž, 𝑏, 𝑐 is the nonzero vector that it’s parallel to.

Now naturally, the components of the direction vector and the coordinates of the point that the line passes through explicitly appear in the equation of the line in a variety of forms. So let’s recall the other two. The Cartesian form is π‘₯ minus π‘₯ nought over π‘Ž equals 𝑦 minus 𝑦 nought over 𝑏 equals 𝑧 minus 𝑧 nought over 𝑐 for nonzero values of π‘Ž, 𝑏, and 𝑐. And the parametric form is π‘₯ equals π‘₯ nought plus π‘Žπ‘‘, 𝑦 equals 𝑦 nought plus 𝑏𝑑, and 𝑧 equals 𝑧 nought plus 𝑐𝑑.

For instance, suppose we know that a line passes through the point negative one, negative five, four and it’s parallel to the vector negative three, five, one. We can define π‘₯ nought to be equal to negative one, 𝑦 nought to be equal to negative five, and 𝑧 nought to be equal to four. And of course, it’s parallel to this nonzero vector. So π‘Ž is negative three, 𝑏 is five, and 𝑐 is one. This means that the vector equation of the line is 𝐫 of 𝑑 equals negative one, negative five, four plus 𝑑 times negative three, five, one.

So, given all of this information, how do we determine whether two lines given in vector form are parallel? Well, it follows that it doesn’t really matter the point through which the lines pass, but if their direction vectors are parallel, if they’re traveling in the same direction, then the lines themselves must also be parallel. And of course, whilst we’ve represented this in two dimensions, this statement, of course, holds for vectors in three and in fact higher dimensions. We say that two lines are parallel then if their direction vectors are parallel. So we’ll begin by demonstrating how to find the vector form of the equation of a line when we know it’s parallel to a given line.

Find the vector form of the equation of the straight line passing through the point 𝐴 two, five, five and parallel to the straight line passing through the two points 𝐡 negative three, negative two, negative six and 𝐢 five, zero, negative nine.

We’re told to give the vector form of the equation of the line. So let’s begin by reminding ourselves what that looks like. Given a straight line which passes through the point π‘₯ nought, 𝑦 nought, 𝑧 nought and which is parallel to the vector π‘Ž, 𝑏, 𝑐, its equation is 𝐫 equals the vector π‘₯ nought, 𝑦 nought, 𝑧 nought plus 𝑑 times the vector π‘Ž, 𝑏, 𝑐. And we call this vector π‘Ž, 𝑏, 𝑐 the position vector of the line. We quite clearly have the point through which our line passes through. That’s 𝐴 two, five, five. So according to the definition of the vector equation of a line, we can define π‘₯ nought to be equal to two, 𝑦 nought to be equal to five, and 𝑧 nought to be equal to five.

But how do we find the direction vector of our line? We know it’s parallel to the line that passes through points 𝐡 and 𝐢. So let’s find the vector πš©π‚. And of course, the vector πš©π‚ is given by the vector πŽπ‚ minus the vector 𝐎𝚩. Since the coordinates themselves tell us the position of each point with respect to the origin, then the vector πŽπ‚ is simply the vector five, zero, negative nine and the vector 𝐎𝚩 is negative three, negative two, negative six. And so, we find the difference between these two vectors by simply subtracting their individual components. That means to find the first component we subtract negative three from five and we get eight.

Similarly, our second component is zero minus negative two, which is two. Then our third component is negative nine minus negative six, which is negative three. And so the vector πš©π‚ is the vector eight, two, negative three. Now, of course, if two vectors are parallel, that tells us that one vector is a scalar multiple of the other. And so the direction vector of our line must be some scalar multiple of the vector eight, two, negative three. Now, of course, that scalar multiple can be one. So we’re going to define the direction vector of our line, let’s call that 𝐝, to be equal to eight, two, negative three.

And so, of course, all that remains is to substitute what we know about our line into the general formula for the vector equation of a straight line. And so the vector equation is 𝐫 equals two, five, five plus 𝑑 times eight, two, negative three.

In our first example, we demonstrated how to work with parallel lines. But if two lines in two dimensions are not parallel nor are they coincident β€” in other words, lie on top of one another β€” it follows that they must intersect at a point. This is not, however, the case in three dimensions. In this case, they can actually also be skewed. That is, they don’t intersect, nor are they parallel to one another. Now of course, if we’re given that the lines intersect, then we should be able to find the point of intersection of two lines given to us in three dimensions. Let’s see what that looks like.

Find the Cartesian form of the equation of the straight line passing through the origin and the intersection point of the two straight lines 𝐿 one, which is 𝐫 equals one, one, negative two plus 𝑑 one, four, three, and 𝐿 two, which is π‘₯ equals three and 𝑦 minus five over negative four equals 𝑧 minus three over negative one.

Remember, if we know that a straight line passes through the point π‘₯ nought, 𝑦 nought, 𝑧 nought and it’s parallel to the vector π‘Ž, 𝑏, 𝑐. Then its Cartesian form is π‘₯ minus π‘₯ nought over π‘Ž equals 𝑦 minus 𝑦 nought over 𝑏 equals 𝑧 minus 𝑧 nought over 𝑐. So it should be quite clear that to find the equation of our line, we’re going to need to find a point through which it passes and a vector which is parallel to it. In fact, we’re told it passes through the origin, which has coordinates zero, zero, zero. So we can define π‘₯ nought to be equal to zero, 𝑦 nought to be equal to zero, and 𝑧 nought to be equal to zero.

Now since we know that our vector passes through the origin and the intersection point of our lines 𝐿 sub one and 𝐿 sub two, it follows that we’re going to need to find the value of that intersection point. Once we have that, we can use information about that point and the origin to find its direction vector.

Let’s begin by writing each equation in parametric form, since that’s generally easier to work with. To achieve this, we distribute the 𝑑 over the second vector and then we add the individual components. So π‘₯ is one plus one times 𝑑. That’s one plus 𝑑. 𝑦 is one plus four 𝑑. And 𝑧 is given by negative two plus three 𝑑. For the second line, we’re already told that π‘₯ is equal to three, but we’re going to need to do some work with the second half of this equation. We’ll set each part equal to 𝑑 since this is equivalent. And then we will rearrange to make 𝑦 and 𝑧 the subject, respectively.

Multiplying by negative four and then adding five to both sides of our first equation, we find 𝑦 is five minus four 𝑑. Similarly, we find that 𝑧 is equal to three minus 𝑑. And so we have both of our equations represented in parametric form. So we know that our lines intersect at some point π‘₯, 𝑦, 𝑧. So this point must be given by some parameter value 𝑑 β€” let’s let that be 𝑑 one for our first line β€” and some other parameter value; let’s let that be 𝑑 sub two for our second line. Then we’ll substitute each value into the parametric form of the equations of our line and then set them equal to one another.

For our π‘₯-coordinate, that’s one plus 𝑑 sub one is simply equal to three, which we can solve to find that 𝑑 sub one is equal to two. Then, for our 𝑦-coordinate, one plus four 𝑑 sub one is equal to five minus four 𝑑 sub two. We then substitute 𝑑 sub one equals two into this equation. And we get one plus eight equals five minus four 𝑑 sub two. That simplifies to four 𝑑 sub two equals negative four. So 𝑑 sub two must be equal to negative one. And it should become a quite apparent that we don’t actually need to perform any further calculations here, although we could substitute this into the equivalent equation for 𝑧 just to check that it works.

If we substitute 𝑑 sub one equals two and 𝑑 sub two equals negative one, we get four equals four, which indicates to us that we’ve probably done our calculations correctly. And so we choose one of these parameters, and we substitute it into the respective equation to find the point of intersection. Let’s use 𝑑 sub one equals two and substitute it into the formulae for 𝐿 sub one. We get one plus two for the π‘₯-coordinate, one plus four times two to the 𝑦-coordinate, and negative two plus three times two for our 𝑧-coordinate. And so the point of intersection is three, nine, four.

Now, if we let this point of intersection be equal to point 𝑃, then we know that the direction vector, our line, is the vector that joins 𝑂 to 𝑃, the origin, to point 𝑃. And that’s simply three, nine, four. And so, if we go back right to the beginning and recall that formula that we had for the Cartesian equation of the line, we can define π‘Ž to be equal to three, 𝑏 to be equal to nine, and 𝑐 to be equal to four. Let’s substitute all of that into the equation of a straight line. And we find that the equation is π‘₯ over three equals 𝑦 over nine, which equals 𝑧 over four.

So far, we’ve covered the cases of lines being parallel, not parallel, coincident, or skew. If two lines intersect, however, there is another final possibility. That is, they might be perpendicular or orthogonal to one another. In other words, they meet at an angle of 90 degrees. Let’s define this formally and say that two lines are perpendicular if they intersect at a point and their direction vectors are perpendicular. Here, it’s also useful to remind ourselves that two vectors 𝐚 and 𝐛 are perpendicular if and only if their scalar or dot product is equal to zero. So here, 𝐚 dot 𝐛 equals zero. In our next example, we’ll demonstrate an application of these definitions.

Consider the two lines π‘₯ equals four plus two 𝑑, 𝑦 equals six plus 𝑑, 𝑧 equals two minus two 𝑑, and 𝐫 equals six, seven, zero plus 𝑑 times five, four, seven. Determine whether they are parallel or perpendicular.

Now we can determine whether two lines are parallel or perpendicular by considering their direction vectors. And so what we’re going to do is begin by rewriting the equation of our first line in vector form so we can then extract its direction vector. And of course, that’s of the form π‘₯ nought, 𝑦 nought, 𝑧 nought plus 𝑑 times π‘Ž, 𝑏, 𝑐, where π‘Ž, 𝑏, 𝑐 is the direction vector and π‘₯ nought, 𝑦 nought, 𝑧 nought is the point which the line passes through. To achieve this, we begin by writing each expression as the component of our vector. Then we separate the components as shown. So we get four, six, two plus two 𝑑, 𝑑, negative two 𝑑. And then we take out that constant factor of 𝑑. So the vector form of our first line is four, six, two plus 𝑑 times two, one, negative two.

So let’s write down then the direction vector of each of our lines. Let’s call our first line 𝐿 sub one. Its direction vector is two, one, negative two. Then the direction vector of our second line, we’ll call that 𝐿 sub two, is five, four, seven. And then we recall that if two vectors are parallel, specifically direction vectors, then one is a scalar multiple of the other. So for our lines to be parallel, we can say that the direction vector two, one, negative two can be written as some scalar multiple π‘˜ of the vector five, four, seven.

Now, if we look at the second component one and four, we might deduce that for this to be true, π‘˜ would have to be equal to one-quarter, since one-quarter of four is equal to one. But one-quarter of five is not equal to two, and a quarter of seven is not equal to negative two. And so this is not true. These vectors cannot be written as scalar multiples of one another, and so the lines cannot be parallel. We might deduce then that they’re perpendicular. But we’re going to check. For two vectors to be perpendicular, of course, their dot product or scalar products have to be equal to zero. They also have to intersect at a point, but we’ll deal with that in a moment.

Let’s just double-check that their dot product is indeed equal to zero. Well, their dot product or their scalar product is two times five plus one times four plus negative two times seven, which is indeed equal to zero. So this element so far is good. We do, of course, need to double-check that they intersect at a point. Now if the two lines intersect at a point, there must be some parameter values 𝑑 sub one in our first equation and 𝑑 sub two in our second that make the π‘₯-coordinates, 𝑦-coordinates, and 𝑧 coordinates, respectively, equal.

For our π‘₯-coordinates, let’s say that’s four plus two 𝑑 sub one is equal to six plus five 𝑑 sub two. For our 𝑦-coordinate, it’s six plus 𝑑 sub one equals seven plus four 𝑑 sub two. And then we have a corresponding equation for our 𝑧-coordinate. Solving any two of these equations simultaneously, we obtain 𝑑 sub one equals one and 𝑑 sub two equals zero. Of course, we can check these by then substituting into the third equation, the third one that we hadn’t used. Substituting 𝑑 sub one equals one into the equation of our first line and we find that this point of intersection is four plus two, six plus one, two minus two, which is six, seven, zero.

Let’s double-check. We get the same values when we substitute into 𝐿 sub two. As expected, we do indeed get the value of six, seven, zero. And of course, if we had substituted 𝑑 sub one equals one and 𝑑 sub two equal zero into that third equation that we didn’t use, we would have expected this result. So since the direction vectors of the two lines are perpendicular and they intersect at a point, we can say that the lines themselves must also be perpendicular.

Let’s recap some of the key concepts from this lesson. In this video, we learned that two lines are parallel if their direction vectors are parallel. We also saw that they’re perpendicular if they intersect at a point and their direction vectors themselves are perpendicular. If two parallel lines intersect at a point, then they must overlap entirely. And in this case, we call them coincident. And in three dimensions, we have skew lines. Now these are not parallel, but they also don’t intersect at any point.

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