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.
