Question Video: Determining the Relationship between Two Lines Mathematics

Video Transcript

Consider the two lines 𝐿 sub one: 𝐫 is equal to the vector three, four, one plus 𝑡 times the vector one, negative one, one and 𝐿 sub two: 𝐫 is equal to the vector two, zero, five plus 𝑡 times the vector negative two, negative three, zero. Choose the correct statement about them. (A) 𝐿 sub one and 𝐿 sub two are coincident lines. (B) 𝐿 sub one and 𝐿 sub two are parallel but not coincident. (C) 𝐿 sub one and 𝐿 sub two are skew lines. (D) 𝐿 sub one and 𝐿 sub two intersect at a point.

In this question, we’re given the vector equations of two different lines, 𝐿 sub one and 𝐿 sub two. We need to choose which of four given options is the correct statement about the two given lines. And if we look at the four given options, we can actually see that the four options are just the four possible types of pairs of lines in three-dimensional space. The two lines can be coincident lines, which means they’re exactly the same line. They can be parallel but not coincident, which means they have the same direction; however, they’re not the same line. They can be skew lines, which means the two lines are not parallel and they don’t intersect. Or alternatively, the two lines can intersect at a single point.

We can go about checking these statements in any order. However, usually, it’s easiest to check whether the lines are coincident and parallel first. This is because if we have two coincident lines or parallel lines which are not coincident, then they must have the same direction. We can recall in terms of the direction vectors, this just means that direction vectors are nonzero scalar multiples of each other. And we’re given the vector forms of the equations of both lines, so we can just read off the direction vectors of each line. Therefore, for the direction vectors to be scalar multiples of each other, we need the vector one, negative one, one to be equal to 𝑘 multiplied by the vector negative two, negative three, zero for some scalar value of 𝑘 not equal to zero.

Let’s try and solve this equation for 𝑘. We’ll start by multiplying all of the components of our vector by our scalar 𝑘. This gives us the vector one, negative one, one needs to be equal to the vector negative two 𝑘, negative three 𝑘, zero. And for two vectors to be equal, all of their corresponding components must be equal. However, if we compare the third component of these vectors, we see that these cannot be equal. Therefore, the direction vectors of the two lines are not parallel. So, the lines are not coincident and they’re not parallel. So, we can eliminate these two options. And this only leaves two possibilities, either the lines are skew lines or they intersect at a point.

And since it’s quite difficult to check if two lines are skew lines, let’s determine if the two lines intersect at a point. To do this, we recall in the vector equation of the line, the vector 𝐫 represents the position vector of any point on the line. So, if the two lines intersect, then the position vector of this point must be equal in both equations. So, we can find a point of intersection by setting the vector equations equal to each other. Well, we need to remember we need to relabel the scalar values. We’ll call these 𝑡 sub one and 𝑡 sub two, respectively. This gives us the equation the vector three, four, one plus 𝑡 sub one times the vector one, negative one, one needs to be equal to the vector two, zero, five plus 𝑡 sub two multiplied by the vector negative two, negative three, zero. And we want to solve this vector equation for 𝑡 sub one and 𝑡 sub two.

We’ll do this by first evaluating the scalar multiplication on both sides of the equation. Then, we’ll also evaluate the vector addition. This then gives us that the vector three plus 𝑡 sub one, four minus 𝑡 sub one, one plus 𝑡 sub one is equal to the vector two minus two 𝑡 sub two, negative three 𝑡 sub two, five. And remember, for two vectors of the same dimension to be equal, all of their corresponding components must be equal. So, this gives us three equations in terms of 𝑡 sub one and 𝑡 sub two. We can solve this in a few different ways. However, the easiest way is to note that if we equate the third component of each vector, we only get an equation in one variable.

We get that one plus 𝑡 sub one needs to be equal to five. And we can solve this for 𝑡 sub one by subtracting one from both sides of the equation. This doesn’t yet confirm the two lines intersect. We also need the other two components of the vectors to be equal. So, let’s now equate the first component of each vector. We get that three plus 𝑡 sub one needs to be equal to two minus two 𝑡 sub two. Of course, we’ve already shown that 𝑡 sub one is equal to four. So, we can substitute this into our equation. This gives us seven is equal to two minus two 𝑡 sub two. And we can solve this equation for 𝑡 sub two. We subtract two from both sides of the equation and divide through by negative two. We get 𝑡 sub two is equal to negative five over two.

And we might be tempted to stop here. But remember, for these two vectors to be equal, all of their components need to be equal, and this is where we can note a problem. The second component of our first vector is four minus 𝑡 sub one. We’ve shown 𝑡 sub one needs to be equal to four, so this gives us four minus four, which is equal to zero. However, the second component in our second vector is negative three 𝑡 sub two. If we set 𝑡 sub two equal to negative five over two, then the second component is negative three multiplied by negative five over two. And we can see that this is not equal to the second component in the other vector. In other words, we’ve shown there’s no values of 𝑡 sub one and 𝑡 sub two where the position vector of the points on both lines are equal.

Therefore, we’ve shown that there’s no point of intersection between 𝐿 sub one and 𝐿 sub two. And now, we can recall that skew lines are lines which are not parallel; however, they don’t intersect. And since we’ve just shown that the lines are not parallel and that they don’t intersect, we can conclude the answer is option (C). 𝐿 sub one and 𝐿 sub two are skew lines.