### Video Transcript

In this video, we will learn how to
recognize parallel and perpendicular vectors in two dimensions. Remember that if we’re thinking
about lines, then two parallel lines will be two lines that never cross. Two perpendicular lines would be
two lines which cross at right angles. So let’s begin by reminding
ourselves about vectors and thinking about how we might recognize parallel
vectors.

Let’s start by looking at this
vector 𝐮. We could write this vector in terms
of its horizontal and vertical components as the vector two, one. We can now introduce some other
vectors, let’s say this vector 𝐚, which we can write in terms of the horizontal and
vertical components as four, two. We can even draw another vector,
vector 𝐛, which looks similar to vector 𝐮, only this time it’s going in the
opposite direction. We could even draw in one final
vector, vector 𝐜, written as negative six, negative three.

What do we notice about the four
vectors that we’ve drawn? Well, they’re all parallel. All of these vectors are scalar
multiples of each other. For example, if we multiplied
vector 𝐮 by two, we would get vector 𝐚. Even if we think about how we would
go from vector 𝐮 to vector 𝐛, we could multiply vector 𝐮 by negative one to get
vector 𝐛. We can write a formal definition of
parallel vectors in the following way. We can say that for any two vectors
𝐮 and 𝐯, they are parallel if 𝐮 is equal to 𝑘 times 𝐯 for some scalar 𝑘, where
𝑘 is not equal to zero. So, if we look at our diagram on
vectors 𝐛 and 𝐜, we could write that vector 𝐛 is equal to 𝑘 times vector 𝐜. We could even work out that 𝑘 in
this case would be one-third.

However, if we didn’t have a
diagram and we didn’t have any information about the lengths of 𝐛 or 𝐜, then we
would know that they must be parallel because they’re scalar multiples of each
other. We’re now going to look at some
questions. And in the first one, we’ll
identify if two given vectors are parallel or not.

True or false: vectors 𝐀 equals
two, one and 𝐁 equals six, three are parallel. Option (A) true or option (B)
false.

In this question, we’re given
information about two vectors 𝐀 and 𝐁, and we’re asked to work out if these two
vectors are parallel. We can recall that two vectors are
parallel if they are scalar multiples of each other. If vectors 𝐀 and 𝐁 are parallel,
then we could write that vector 𝐀 is equal to 𝑘 times vector 𝐁 for some scalar
quantity 𝑘, where 𝑘 is not equal to zero. Given the information about vectors
𝐀 and 𝐁, then we can fill these into this equation and check if it’s valid. You might notice that vector 𝐁 is
a nice multiple of vector 𝐀. But it’s worth checking the 𝑥- and
𝑦-components separately.

Evaluating the 𝑥-components, we’d
have two is equal to six 𝑘. Dividing both sides by six would
give us two-sixths is equal to 𝑘. Simplifying would give us a third
equals 𝑘 or 𝑘 is equal to one-third. In order for these vectors to be
parallel, then when we evaluate the 𝑦-components, we need to get the same value of
𝑘. So, let’s check. When we evaluate the 𝑦-components,
we get that one is equal to three 𝑘. Dividing both sides by three, we’d
get that one-third is equal to 𝑘 which, of course, means that 𝑘 is equal to
one-third. We can then say that vector 𝐀 does
equal 𝑘 times vector 𝐁 because we can write that vector 𝐀 is equal to one-third
times vector 𝐁. We could therefore give the answer
true since the statement “vectors 𝐀 and 𝐁 are parallel” is true.

One alternative way to answer this
question would be to represent these vectors on a diagram. Vector 𝐀 is given as two, one, and
vector 𝐁 is given as six, three. We can see visually that vectors 𝐀
and 𝐁 are parallel, confirming the answer that the statement is true.

We’ll now have a look at how we
identify perpendicular vectors. We say that two vectors 𝐮 equals
𝑥 one, 𝑦 one and 𝐯 equals 𝑥 two, 𝑦 two are perpendicular if the dot product of
𝐮 and 𝐯 is equal to zero. We recall that we can find the dot
product of vectors 𝐮 and 𝐯 by finding the product of 𝑥 one and 𝑥 two and adding
it to the product of 𝑦 one and 𝑦 two. So, let’s see how we can use this
definition to help us to identify perpendicular vectors.

Which of the following vector pairs
are perpendicular? Option (A) vector two, zero and
vector three, negative six. Option (B) vector one, four and
vector two, eight. Option (C) vector zero, seven and
vector zero, nine. Or option (D) vector three, zero
and vector zero, six.

Let’s begin by recalling how we
identify if vectors are perpendicular. If we have two vectors 𝐮 and 𝐯
which are perpendicular, then the dot product of vector 𝐮 and 𝐯 is equal to
zero. We can remind ourselves of how to
find the dot product by saying that if vector 𝐮 is given by 𝑥 one, 𝑦 one and
vector 𝐯 is given by 𝑥 two, 𝑦 two, then the dot product 𝐮𝐯 is equal to 𝑥 one
times 𝑥 two plus 𝑦 one times 𝑦 two. So, in each of these options, (A)
to (D), we’ll work out this dot product, and if it’s equal to zero, then the vector
pair will be perpendicular. So, let’s start with the vector
pair given in option (A). And we can identify 𝑥 one, 𝑦 one,
𝑥 two, and 𝑦 two values, although it doesn’t matter which vector we choose for
each of the 𝑥 one and 𝑦 one values.

To find the dot product then, we’ll
have two times three plus zero times negative six. Evaluating this, two times three
gives us six. And be careful because, of course,
zero times negative six is zero. Six plus zero simplifies to
six. So, did we calculate the dot
product equal to zero? No, we did not. Therefore, the vector pair given in
option (A) are not perpendicular. We can follow the same process then
for the vectors given in option (B). When we calculate the dot product
here, we have one times two plus four times eight. One times two is two, and four
times eight is 32. Adding these together gives us the
value of 34. As this dot product is not equal to
zero, then the vectors given in option (B) are not perpendicular.

Applying the same method for the
vectors given in option (C), we’re multiplying zero by zero and adding it to seven
times nine, which gives us 63. 63 is not equal to zero, so the
vectors in option (C) are not perpendicular. Finally, in option (D), each of the
products of 𝑥 one, 𝑥 two and 𝑦 one, 𝑦 two will give us zero. So, when we add these together, we
get zero. As we have found a dot product of
vectors which is equal to zero, then these two vectors given in option (D) are
perpendicular. Therefore, we can give the answer
that vector three, zero and vector zero, six are perpendicular.

We can confirm this by drawing
these two vectors. The vector three, zero could be
represented by a line going three units to the right and zero units up. Vector zero, six could be
represented by a line which goes zero units horizontally and six units upwards. The first vector is a horizontal
line, and the second vector is a vertical line, indicating that these two vectors
are indeed perpendicular and so confirming that the vector pair which is
perpendicular are those given in option (D).

We’ll now have a look at a few more
questions involving parallel and perpendicular vectors. In the next question, we’ll need to
find a missing value in a pair of parallel vectors.

If vector 𝐀 is equal to ℎ, ℎ plus
two and vector 𝐁 is equal to three ℎ, four ℎ minus one, then one of the values of ℎ
that makes vector 𝐀 parallel to vector 𝐁 is what. Option (A) seven, option (B) five,
option (C) negative five, or option (D) negative seven.

In this question, we’re given two
vectors 𝐀 and 𝐁. Their components include this
unknown value of ℎ. We need to find one of the values
of ℎ such that vector 𝐀 and 𝐁 will be parallel. So, let’s begin by recalling what
it means to have two parallel vectors. In this case, as we have vector 𝐀
and 𝐁, we would say that they’re parallel if we can write that vector 𝐀 is equal
to 𝑘 times vector 𝐁 for some scalar quantity 𝑘, where 𝑘 is not equal to
zero. So, let’s set our two vectors as
parallel vectors by saying that vector 𝐀 is equal to 𝑘 times vector 𝐁. We can then fill in the values of
the vectors that we’re given. Vector 𝐀 is ℎ, ℎ plus two, and
vector 𝐁 is three ℎ, four ℎ minus one.

On the right-hand side, we can
multiply 𝑘 by each of the 𝑥- and 𝑦-components. We can then evaluate the
𝑥-components. So we would have ℎ is equal to
three 𝑘ℎ. Dividing both sides by ℎ, we have
that one is equal to three 𝑘. Dividing three by three, we have
one-third is equal to 𝑘 or 𝑘 is equal to one-third. Next, let’s consider the
𝑦-components. We can write that ℎ plus two is
equal to 𝑘 times four ℎ minus one. We have already established that 𝑘
is equal to one-third, so let’s substitute this into the equation. We then have that ℎ plus two is
equal to one-third times four ℎ minus one. Now let’s expand the parentheses on
the right-hand side. One-third multiplied by four ℎ will
give us four-thirds ℎ, and one-third multiplied by negative one gives us negative
one-third.

We can then subtract ℎ from both
sides, remembering that if we have four-thirds ℎ and we take away ℎ, that will leave
us with one-third ℎ. We can then add one-third to both
sides, remembering that two and one-third is equivalent to seven-thirds. Finally, we can multiply by three
on both sides of this equation, which leaves us with seven is equal to one ℎ or
simply seven is equal to ℎ. Therefore, one of the values of ℎ
that makes 𝐀 parallel to 𝐁 is seven. So, the answer is that given in
option (A) seven. We can check this by plugging the
value of seven into vectors 𝐀 and 𝐁. When ℎ is equal to seven, vector 𝐀
will be seven, and seven plus two is nine. For vector 𝐁, when ℎ is equal to
seven, three times seven is 21 and four times seven is 28; subtract one will give us
27.

So, are vectors 𝐀 and 𝐁 scalar
multiples? Yes, they are because we could
write that vector 𝐀 is equal to one-third of vector 𝐁. So, we’ve confirmed that when ℎ is
seven, vectors 𝐀 and 𝐁 are parallel. But let’s check if any of the
options given in (B), (C), or (D) for the value of ℎ will also make the vectors 𝐀
and 𝐁 parallel. This time, we’ll take the value
given in the answer option for the value of ℎ and then check if the vector 𝐀 is
parallel to vector 𝐁.

So, for option (B), we’re checking
the value of ℎ is equal to five. And vector 𝐀 will therefore be
five, and five plus two is seven. And vector 𝐁 will be three times
five, which is 15, and four times five is 20 minus one is 19. So, is there a scalar value 𝑘 for
which the vector five, seven is equal to 𝑘 times the vector 15, 19? Well, no, there’s not. If we look at the 𝑥-components,
we’ll have five is equal to 15𝑘. That would mean that 𝑘 is equal to
one-third. However, 𝑘 is equal to one-third
would not fit the equation seven is equal to 𝑘 times 19. So, that means when ℎ is equal to
five, these two vectors five, seven and 15, 19 are not parallel. So, we can eliminate option
(B).

Let’s check option (C). This time we’re checking the value
of ℎ is equal to negative five. So, vector 𝐀 will be negative
five, negative three, and vector 𝐁 will be negative 15, negative 21. Once again, there is no value of 𝑘
for which vector 𝐀 is equal to 𝑘 times vector 𝐁. So, we can eliminate option (C)
because when ℎ is equal to negative five, these two vectors are not parallel. Finally then, in option (D), we’re
checking the value of ℎ is equal to negative seven. When we plug this value into the
vectors 𝐀 and 𝐁, once again, there is no value of 𝑘 which makes negative seven,
negative five and negative 21, negative 29 parallel. We have therefore eliminated
options (B), (C), and (D), leaving us with the value of seven.

We’ll now look at one final
question.

Fill in the blank: vectors 𝐀
equals one, two and vector 𝐁 equals negative two, one are what.

The first thing we might consider
checking is if vectors 𝐀 and 𝐁 are parallel. We can recall that any two vectors
𝐀 and 𝐁 are parallel if we can write that vector 𝐀 is equal to 𝑘 times vector 𝐁
for any scalar 𝑘, where 𝑘 is not equal to zero. So, let’s check if this true for
the given vectors. Is there a value 𝑘 for which the
vector one, two is equal to 𝑘 times negative two, one? We can simplify the vector on the
right-hand side as negative two 𝑘, 𝑘. We can then evaluate the 𝑥- and
𝑦-components separately, so the 𝑥-components would give us one is equal to
negative two 𝑘. Dividing this equation by negative
two on both sides would give us that negative one-half is equal to 𝑘. So, 𝑘 is equal to negative
one-half.

Let’s have a look at the
𝑦-components. This gives us the equation two is
equal to 𝑘 or, alternatively, 𝑘 is equal to two. However, we have two different
values of 𝑘, which means that there is no value of 𝑘 for which vector 𝐀 is equal
to 𝑘 times vector 𝐁. This means that these two vectors
are not parallel, so let’s see what else they might be. Let’s check if they’re
perpendicular. We recall that if the dot product
of two vectors is equal to zero, then those vectors are perpendicular. For any two vectors 𝐮 is equal to
𝑥 one, 𝑦 one and 𝐯 is equal to 𝑥 two, 𝑦 two, then the dot product 𝐮𝐯 is equal
to the product 𝑥 one 𝑥 two plus the product 𝑦 one 𝑦 two.

To calculate the dot product of our
two vectors 𝐀 and 𝐁, then we work out one times negative two plus two times
one. And we can simplify this to
negative two plus two which is equal to zero. For any two vectors 𝐮 and 𝐯, they
are perpendicular if the dot product 𝐮 dot 𝐯 is equal to zero. We have established that the dot
product of vectors 𝐀 and 𝐁 here is equal to zero. This means that we can fill in the
blank. Vectors 𝐀 is equal to one, two and
𝐁 is equal to negative two, one are perpendicular.

We can now summarize the key points
of this video. We saw that two vectors 𝐮 and 𝐯
are parallel if vector 𝐮 is equal to 𝑘 times vector 𝐯 for some scalar quantity
𝑘, where 𝑘 is not equal to zero. Finally, we saw that vectors 𝐮 and
𝐯 are perpendicular if the dot product of vectors 𝐮 and 𝐯 is equal to zero.