Video Transcript
Given that 𝐚 is equal to the
vector one, five and 𝐛 is equal to the vector negative seven, three, what are the
components of two 𝐚 plus two 𝐛 and two lots of 𝐚 plus 𝐛?
The two vectors in the question are
𝐚, which is equal to one, five, and 𝐛, which is equal to negative seven,
three. We have been asked to calculate two
things: two 𝐚 plus two 𝐛 and two lots of 𝐚 plus 𝐛. We will start by finding the first
of these.
In order to carry out this sum, we
need to find two 𝐚 and two 𝐛. This will require scalar
multiplication of vectors, so let’s recall how we do this. If we multiply the vector 𝑐, 𝑑 by
the scalar 𝑘, then we’ll get the vector 𝑘𝑐, 𝑘𝑑. Essentially, we multiply each of
the components of the vector by the scalar. We have that the vector 𝐚 is equal
to one, five, and we need to find two 𝐚. So we simply multiply each of the
components by two. We get that two 𝐚 is equal to two,
10. Similarly, since 𝐛 is equal to
negative seven, three, we get that two 𝐛 is equal to negative 14, six.
Now we have found two 𝐚 and two
𝐛, we simply need to add them together. Let’s quickly remind ourselves how
we add two vectors together. If we add the vectors 𝑐, 𝑑 and
𝑚, 𝑛, then we will obtain the vector 𝑐 plus 𝑚, 𝑑 plus 𝑛. We simply add the corresponding
components of the two vectors. Now we can apply this to the
vectors two 𝐚 and two 𝐛. We have that two 𝐚 plus two 𝐛 is
equal to two, 10 plus negative 14, six. Now, we add the corresponding
components of these vectors. So that’s two plus negative 14, 10
plus six. Simplifying this, we obtain
negative 12, 16. Hence, the solution to the first
part of this question is that two 𝐚 plus two 𝐛 is equal to the vector negative 12,
16.
Next, we can move on to find two
lots of 𝐚 plus 𝐛. There are two ways we can find
this. In the first method, we will start
by finding 𝐚 plus 𝐛 using vector addition. We have that 𝐚 plus 𝐛 is equal to
one, five plus negative seven, three. We add the corresponding components
of the two vectors and simplify. Hence, we have that 𝐚 plus 𝐛 is
equal to negative six, eight. In order to find two 𝐚 plus 𝐛, we
need to multiply this vector by the scalar two. Using the scalar multiplication of
a vector, we multiply each of the components by the constant. In doing this, we reach the
solution to the second part of the question, which is that two lots of 𝐚 plus 𝐛 is
equal to negative 12, 16.
There is, however, another method
we could’ve used to reach this solution. This method involves the
distributive property of scalar multiplication over vector addition. This property tells us that the
scalar multiple of a sum of vectors is equal to the sum of the scalar multiple
multiplied by each of the vectors. In other words, 𝑘 multiplied by 𝐚
plus 𝐛 is equal to 𝑘 times 𝐚 plus 𝑘 times 𝐛. We can apply this to two lots of 𝐚
plus 𝐛. This gives us two 𝐚 plus two 𝐛,
which we can see is what we found in the first part of the question. Hence, we reach the same solution
of negative 12, 16.
