Video Transcript
In this video, we will learn how to
add and subtract vectors in two dimensions. We will begin by recalling what we
mean by a vector and showing how we can draw them on the coordinate plane.
A vector is a quantity that has
both a magnitude and direction. A two-dimensional vector can be
written in the form 𝑥𝐢 plus 𝑦𝐣; using triangular brackets, 𝑥, 𝑦; or as a
column vector, 𝑥 over 𝑦. They can be drawn on the coordinate
plane.
Let’s consider the vector four,
three or four 𝐢 plus three 𝐣. This vector has an 𝐢-component of
four, so it moves four units in the 𝑥-direction. It has a 𝐣-component of three, so
it moves three units in the 𝑦-direction. We can then draw the vector on the
coordinate plane as shown. Its direction will be going away
from the origin, and its magnitude will be the length. Whilst it will not be required in
this video, we could calculate the magnitude of the vector using the Pythagorean
theorem. Four squared plus three squared is
equal to five squared. This means that the magnitude of
our vector is five.
We will now consider what happens
when we need to add and subtract vectors. Let’s imagine we want to add two
vectors 𝐚 and 𝐛. We can show this visually by first
drawing vector 𝐚. We can then draw vector 𝐛 from the
end of this vector. The vector 𝐚 plus 𝐛 will go from
the start of vector 𝐚 to the end of vector 𝐛. If vector 𝐚 was equal to five, two
and vector 𝐛 was equal to three, negative three, we could find vector 𝐚 plus 𝐛 by
adding the individual components separately. We add five and three and then
separately two and negative three. This means that 𝐚 plus 𝐛 would be
equal to eight, negative one.
We can follow a similar process
when subtracting vectors. To work out vector 𝐚 minus 𝐛, we
would subtract three from five and negative three from two. This would give us the vector two,
five.
Let’s consider what this would look
like in a sketch. Once again, we can begin by drawing
vector 𝐚. We know from our top diagram that
vector 𝐛 looks as shown. The vector negative 𝐛 would have
the same magnitude but go in the opposite direction. This gives us the vector 𝐚 plus
negative 𝐛, which is the same as 𝐚 minus 𝐛. We will now look at some specific
examples where we need to add and subtract vectors.
Given that vector 𝐀 is equal to
negative five 𝐢 plus 10𝐣 and vector 𝐁 is equal to negative four 𝐢 minus five 𝐣,
where 𝐢 and 𝐣 are two perpendicular unit vectors, find vector 𝐀 minus vector
𝐁.
We recall that in order to add or
subtract two or more vectors, we simply add or subtract the corresponding
components. In this question, we need to
subtract negative four 𝐢 minus five 𝐣 from negative five 𝐢 plus 10𝐣. We can then subtract the 𝐢- and
𝐣-components separately. Negative five minus negative four
is equal to negative one. This is because this is the same as
negative five plus four.
Subtracting negative five from 10
gives us 15, as once again this is the same as adding five to 10. Vector 𝐀 minus vector 𝐁 is
therefore equal to negative 𝐢 plus 15𝐣.
In our next question, we all need
to add and subtract three vectors.
Given that vector 𝐀 is equal to
nine, five; vector 𝐁 is equal to negative 10, three; and vector 𝐂 is equal to
negative three, six, find vector 𝐀 plus vector 𝐁 minus vector 𝐂.
In order to add or subtract two or
more vectors, we need to add or subtract the corresponding components. In this question, we have nine,
five plus negative 10, three minus negative three, six. If we consider the 𝑥-components,
we have nine plus negative 10 minus negative three, Adding negative 10 to nine is
the same as subtracting 10 from nine. We then need to add three to this,
as subtracting negative three is the same as adding three. Nine minus 10 equals negative
one. And adding three gives us two. The 𝑥-component of our answer is
therefore two.
We can then repeat this for the
𝑦-components. We have five plus three minus
six. This is also equal to two.
The vector 𝐀 plus 𝐁 minus 𝐂 is
equal to two, two.
In our next question, we will also
need to multiply vectors by scalars.
Given that vector 𝐀 is equal to
negative one, two and vector 𝐁 is equal to three, negative six, find six 𝐀 plus
two 𝐁.
In this question, we’re multiplying
vectors 𝐀 and 𝐁 by a scalar. We’re multiplying vector 𝐀 by six,
and we’re multiplying vector 𝐁 by two. To multiply any vector by a scalar,
we simply multiply each of the components by that scalar. Six multiplied by negative one is
negative six, and six multiplied by two is equal to 12. Therefore, six 𝐀 is equal to
negative six, 12.
We need to multiply vector 𝐁 by
two. Two multiplied by three is six, and
two multiplied by negative six is 12. Therefore, two 𝐁 is equal to six
negative 12.
We need to add these two
vectors. When adding or subtracting two
vectors, we simply add or subtract the corresponding components. In this case, we need to add
negative six and six and then also 12 and negative 12. Negative six plus six is equal to
zero, and 12 plus negative 12 is also equal to zero. This means that six 𝐀 plus two 𝐁
is equal to zero, zero. The sum of six multiplied by vector
𝐀 and two multiplied by vector 𝐁 is zero, zero, the origin.
In our next question, we need to
determine a vector that satisfies an equation.
Given that vector 𝐁 equals
negative nine, negative three; vector 𝐂 equals negative four, negative two; and
vector 𝐃 equals negative two, nine, determine the vector 𝐀 that satisfies the
equation 𝐀 is equal to negative four 𝐁 plus two 𝐂 minus six 𝐃.
In the equation given, we’ve
multiplied vectors 𝐁, 𝐂, and 𝐃 by a scalar. Firstly, vector 𝐁 has been
multiplied by negative four. To calculate this, we multiply each
of the components by negative four. Multiplying two negative numbers
gives a positive answer. Therefore, negative four multiplied
by negative nine is 36. And negative four multiplied by
negative three is 12.
We can repeat this process by
multiplying vector 𝐂 by two. This gives us negative eight,
negative four. Finally, we need to multiply vector
𝐃 by six. Six multiplied by negative two is
negative 12, and six multiplied by nine is 54. We need to add two 𝐂 to negative
four 𝐁 and then subtract six 𝐃 from this. Vector 𝐀 is therefore equal to 36,
12 plus negative eight, negative four minus negative 12, 54.
When adding or subtracting vectors,
we simply add or subtract the individual components. For the 𝑥-components in this case,
we have 36 plus negative eight minus negative 12. This is the same as 36 minus eight
plus 12, which gives us an answer of 40. The 𝑥-component of vector 𝐀 is
40.
We can repeat this for the
𝑦-components. We need to add negative four to 12
and then subtract 54. This is equal to negative 46.
The vector 𝐀 that satisfies the
equation negative four 𝐁 plus two 𝐂 minus six 𝐃 is 40, negative 46.
Our next question will briefly deal
with what happens when our vectors are in three dimensions.
If vector 𝐀 is equal to six,
negative four, seven and vector 𝐁 is equal to five, six, four, determine vector 𝐀
plus vector 𝐁.
We recall that when adding or
subtracting vectors in two dimensions, we simply add or subtract the corresponding
components. The same is true, as in this case
when we have vectors in three dimensions, we simply add the three different
components.
Firstly, we add the 𝑥- or
𝐢-component. We need to add six and five. This is equal to 11. Next, we need to add negative four
and six. This is equal to two. Finally, we need to add seven and
four. This is also equal to 11.
If vector 𝐀 is equal to six,
negative four, seven and vector 𝐁 is equal to five, six, four, then vector 𝐀 plus
vector 𝐁 is equal to 11, two, 11.
This means that when adding,
subtracting, or multiplying vectors in three dimensions by a scalar, we follow the
same process as with two dimensions.
In our final question, we will look
at a more practical problem involving geometry.
𝐴𝐵𝐶𝐷𝐸𝐹 is a regular
hexagon. 𝐴𝐹 is parallel to 𝐺𝐵, and 𝐸𝐹
is parallel to 𝐺𝐷. Express vector 𝐀𝐄 in terms of 𝐮
and 𝐯.
There are some lines missing on our
hexagon from 𝐴 to 𝐵, 𝐴 to 𝐺, and 𝐸 to 𝐺. As the hexagon is regular, we have
lots of parallel and equal-length lines. The line segments on our diagram
from 𝐷 to 𝐶, 𝐺 to 𝐵, and 𝐹 to 𝐴 will all be equal to vector 𝐮. This is because they all have the
same direction and magnitude. The same would be true of 𝐸𝐺. 𝐺𝐶, 𝐹𝐺, and 𝐸𝐷 will all be
equal to vector 𝐯. Once again, this would also be true
for 𝐴𝐵.
We also know that 𝐹𝐸, 𝐺𝐷, and
𝐵𝐶 are all of the same magnitude and direction. Let’s consider what these would be
equal to in terms of 𝐮 and 𝐯. We can get from the center of the
hexagon 𝐺 to the vertex 𝐷 via vertex 𝐶. This means that 𝐺𝐷 is equal to
𝐺𝐶 plus 𝐶𝐷. We know that 𝐺𝐶 is equal to
vector 𝐯. 𝐷𝐶 is equal to vector 𝐮. This means that 𝐶𝐷 is equal to
negative 𝐮. It has the same magnitude but is in
the opposite direction. We can therefore say that 𝐺𝐷 is
equal to 𝐯 minus 𝐮. The same is true for 𝐹𝐸 and
𝐵𝐶.
We are asked to express 𝐴𝐸 in
terms of 𝐮 and 𝐯. The quickest way to get from vertex
𝐴 to vertex 𝐸 is via vertex 𝐹. This means that 𝐴𝐸 is equal to
𝐴𝐹 plus 𝐹𝐸. As 𝐹𝐴 is equal to vertex 𝐮, 𝐴𝐹
will be equal to negative 𝐮. We know that 𝐹𝐸 is equal to 𝐯
minus 𝐮. 𝐴𝐸 is therefore equal to negative
𝐮 plus 𝐯 minus 𝐮.
This can be simplified to negative
two 𝐮 plus 𝐯. Alternatively, we could write it as
𝐯 minus two 𝐮. We can use this method to work out
the vector from any vertex to another vertex on the hexagon.
We will now summarize the key
points from this video.
We found out in this video that to
add or subtract two or more vectors, we simply add or subtract the corresponding
components. To multiply a vector by a scalar,
we multiply each component individually. In our final question, we saw that
we can use these skills to answer problems involving geometry. We also saw that the above rules
apply to vectors in both two and three dimensions.
