𝑂𝐴𝐵 is a triangle. 𝑀 is the point on 𝐴𝐵 such that
𝐵𝑀 to 𝑀𝐴 is equal to two to three. 𝑂𝐴 is equal to three 𝑢. 𝑂𝐵 is equal to 𝑣. 𝑂𝑀 is equal to 𝑘 multiplied by
two 𝑢 plus 𝑣, where 𝑘 is a scalar quantity. Find the value of 𝑘.
I like to think of vectors as being
a bit like the tube system in London. Sometimes, I want to take a tube
from one place to another. But there won’t be a tube that does
that exact journey. Instead, I’ll have to take two
separate tubes. I’ll still end up getting to my
destination, just not in a single straight line.
In the case of this question, the
journey I want to take is from 𝑂 to 𝑀. The journey I’ll need to actually
take is to go from 𝑂 to 𝐵 and then 𝐵 to 𝑀 which is part of the journey from 𝐵
to 𝐴. We’ll start then by considering the
journey from 𝐵 to 𝐴.
There are no vectors to describe
that journey. So we’ll need to travel on
alternative way. We’ll follow the lines which are
described with vectors. The first line we’ll follow takes
us from 𝐵 to 𝑂. Then, the second line takes us from
𝑂 to 𝐴. That gives us the entire journey
from 𝐵 to 𝐴.
Let’s begin with the journey from
𝐵 to 𝑂. We can see that the arrow going
from 𝑂 to 𝐵 is marked with the vector 𝑣. If we want to travel from 𝐵 to 𝑂,
this is the opposite direction. So we change the sign of the
vector. The journey from 𝐵 to 𝑂 is,
therefore, negative 𝑣.
Next, we want to go from 𝑂 to
𝐴. That’s given as three 𝑢. Our total journey then is given by
𝐵 to 𝑂 and then 𝑂 to 𝐴. It’s negative 𝑣 plus three 𝑢,
which could be written as three 𝑢 minus 𝑣. Remember though we didn’t want to
go all the way from 𝐵 to 𝐴. We actually only wanted to travel
part of this journey to 𝑀. We’re given that 𝐵𝑀 to 𝑀𝐴 is
two to three.
That means the journey of 𝐵 to 𝐴
can be split into five total parts. 𝐵𝑀 is two of these five
parts. It’s two-fifths of the journey from
𝐵 to 𝐴. Since 𝐵 to 𝐴 is given by three 𝑢
minus 𝑣, 𝐵𝑀 must be given then by two-fifths of three 𝑢 minus 𝑣. Now that we have the journey from
𝐵 to 𝑀, we can write down the entire journey from 𝑂 to 𝑀.
To get from 𝑂 to 𝑀, we’ll travel
from 𝑂 to 𝐵 and then 𝐵 to 𝑀. The vector 𝑂𝐵 is given as 𝑣. And we calculated the vector 𝐵𝑀
to be two-fifths of three 𝑢 minus 𝑣. So the journey 𝑂𝑀 is given by 𝑣
plus two-fifths of three 𝑢 minus 𝑣.
Next, we’ll expand this
bracket. Two-fifths multiplied by three is
the same as two-fifths multiplied by three over one. Two times three is six and five
times one is five. So two-fifths multiplied by three
is six-fifths. That means then that two-fifths
multiplied by three 𝑢 is six-fifths 𝑢. Two-fifths multiplied by negative
𝑣 is negative two-fifths 𝑣.
Finally, we want to simplify this
expression by collecting like terms. To do this, we need to make the
denominators for this vector 𝑣 the same. 𝑣 is the same as five-fifths
𝑣. Five-fifths 𝑣 minus two-fifth 𝑣
is three-fifths 𝑣. Our expression for the vector 𝑂𝑀
simplifies then to three-fifths 𝑣 plus six-fifths 𝑢.
The problem is this doesn’t look
much like the expression given in the question. To make the vector three-fifths 𝑣
plus six-fifth 𝑢 look like what we’ve got in the question, first, we’ll change the
order. Three-fifths 𝑣 plus six-fifths 𝑢
is exactly the same as six-fifths 𝑢 plus three-fifths 𝑣.
Next, to find the value of 𝑘, we
need to factorize this expression. The highest common factor of
six-fifths and three-fifths is three-fifths. So three-fifths goes on the outside
of our bracket. Three-fifths multiplied by two 𝑢
is six-fifths 𝑢 and three-fifths multiplied by 𝑣 is three-fifths 𝑣. So the inside of our brackets
becomes two 𝑢 plus 𝑣.
Notice now that our expression is
the same form as the one given to us in the question. Since 𝑘 is the number on the
outside of the bracket, it’s the number multiplying two 𝑢 plus 𝑣, 𝑘 must be equal