# BS in relation to BF and CS in relation to CE

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

We have a triangle $ABC$. A point $E$ is on side $c$ such that $\overline{AE}=\frac{1}{5}\cdot \overline{AB}$, a point $F$ is on side $b$ such that $\overline{FC}=\frac{1}{4}\cdot \overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$.

Can we solve that using vectors?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We have a triangle $ABC$. A point $E$ is on side $c$ such that $\overline{AE}=\frac{1}{5}\cdot \overline{AB}$, a point $F$ is on side $b$ such that $\overline{FC}=\frac{1}{4}\cdot \overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$.

Can we solve that using vectors?
Hey mathmari !!

Let's first make a drawing:
\begin{tikzpicture}[scale=2]
%preamble \usetikzlibrary {calc}

\coordinate[label=left:A] (A) at (0,0);
\coordinate[label=right:B] (B) at (0:5);
\coordinate[label=C] (C) at (70:4);
\coordinate[label=below:E] (E) at (0:1);
\coordinate[label=left:F] (F) at (70:3);
\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)});
\draw (A) -- (B) -- (C) -- cycle;
\draw (C) -- (E);
\draw (B) -- (F);
\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\mathbf{\vec c}$} (B);
\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\mathbf{\vec b}$} (C);

\path (A) -- (B) -- node[ above right ] {a} (C) -- (A);
\path (A) -- node[ below ] {$\frac 15 c$} (E) -- node[ below ] {$\frac 45 c$} (B);
\path (A) -- node[ left] {$\frac 34 b$} (F) -- node[ left ] {$\frac 14 b$} (C);
\end{tikzpicture}

Let's define $x=\frac{BS}{BF}$ and $y=\frac{CS}{CE}$. So we want to find $x$ and $y$.
If we want to solve it with vectors, I think we need to express the relationships with $\mathbf{\vec b}$, $\mathbf{\vec c}$, $x$, and $y$.

#### mathmari

##### Well-known member
MHB Site Helper
Let's define $x=\frac{BS}{BF}$ and $y=\frac{CS}{CE}$. So we want to find $x$ and $y$.
If we want to solve it with vectors, I think we need to express the relationships with $\mathbf{\vec b}$, $\mathbf{\vec c}$, $x$, and $y$.

We have that $\vec{AF}-\vec{AB}=\vec{BF}$ and $\vec{AE}-\vec{AC}=\vec{CE}$, right?

Do these relations help us?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We have that $\vec{AF}-\vec{AB}=\vec{BF}$ and $\vec{AE}-\vec{AC}=\vec{CE}$, right?

Do these relations help us?
Isn't $F$ on side $AC$ instead of side $AB$? And $E$ on side $AB$?

The proper relations should be a step to find the relevant vectors and relationships yes.

#### mathmari

##### Well-known member
MHB Site Helper
Isn't $F$ on side $AC$ instead of side $AB$? And $E$ on side $AB$?
Do you mean the relation $\vec{AF}+\vec{FC}=\vec{AC}$ ?

Staff member

#### mathmari

##### Well-known member
MHB Site Helper
We have the following relations :
$$\vec{AF}+\vec{FC}=\vec{AC}=\vec{b} \\ \vec{AE}+\vec{EB}=\vec{AB}=\vec{c} \\ \vec{AC}+\vec{CE}=\vec{AE}\Rightarrow \vec{CE}=\vec{AC}-\vec{AE} \\ \vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AE})=y\cdot b-y\cdot (\vec{c}-\vec{EB})=y\cdot \vec{b}-y\cdot \vec{c}+y\cdot \vec{EB}$$
So we have to write somehow $\vec{EB}$.

Is everything correct so far?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AE})=y\cdot b-y\cdot (\vec{c}-\vec{EB})=y\cdot \vec{b}-y\cdot \vec{c}+y\cdot \vec{EB}$$
So we have to write somehow $\vec{EB}$.

Is everything correct so far?
Isn't $\vec{AE}-\vec{AE}=0$?
Perhaps you meant to refer to $\vec{AB}$. If so, isn't that one equal to $\vec c$ instead of $\vec b$?

#### mathmari

##### Well-known member
MHB Site Helper
Isn't $\vec{AE}-\vec{AE}=0$?
Perhaps you meant to refer to $\vec{AB}$. If so, isn't that one equal to $\vec c$ instead of $\vec b$?
Ahh I meant $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AC}-\vec{AE})=y\cdot b-y\cdot (\vec{c}-\vec{EB})=y\cdot \vec{b}-y\cdot \vec{c}+y\cdot \vec{EB}$$ Is that wrong? Does this help us?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ahh I meant $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AC}-\vec{AE})=y\cdot b-y\cdot (\vec{c}-\vec{EB})=y\cdot \vec{b}-y\cdot \vec{c}+y\cdot \vec{EB}$$ Is that wrong? Does this help us?
A vector is equal to a vector to its end point minus a vector to its starting point.
So we have $\vec{CE} = \vec{AE}-\vec{AC}$.

I believe we need to get a set of equations that refers only to $x, y, \vec b,$ and $\vec c$.
That is, $\vec{EB}$ should be eliminated if we have it.

#### mathmari

##### Well-known member
MHB Site Helper
A vector is equal to a vector to its end point minus a vector to its starting point.
So we have $\vec{CE} = \vec{AE}-\vec{AC}$.

I believe we need to get a set of equations that refers only to $x, y, \vec b,$ and $\vec c$.
That is, $\vec{EB}$ should be eliminated if we have it.
So we have $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AC})=y\cdot (\vec{c}-\vec{EB})-y\cdot b=-y\cdot \vec{c}+y\cdot \vec{EB}-y\cdot \vec{b}$$ right? I don't see how we can write $\vec{EB}$ as a function of $x,y,\vec{b}, \vec{c}$. Could you give me a hint?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we have $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AC})=y\cdot (\vec{c}-\vec{EB})-y\cdot b=-y\cdot \vec{c}+y\cdot \vec{EB}-y\cdot \vec{b}$$ right? I don't see how we can write $\vec{EB}$ as a function of $x,y,\vec{b}, \vec{c}$. Could you give me a hint?
Don't we have $\vec{AE}=\frac 15 \vec{AB}=\frac 15\vec c$?

#### mathmari

##### Well-known member
MHB Site Helper
Don't we have $\vec{AE}=\frac 15 \vec{AB}=\frac 15\vec c$?
Ahh so we have $\vec{CS} =y\frac{1}{5}\vec{c}-y\vec{b}$, right?

Do we have to solve for y?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ahh so we have $\vec{CS} =y\frac{1}{5}\vec{c}-y\vec{b}$, right?

Do we have to solve for y?
The vectors $\vec b$ and $\vec c$ form a basis of the plane.
If we express every vector as a linear combination of these 2, we can get the simplest system possible.

We have the unknowns $x$ and $y$.
So we're looking for a system of equations that contains only $x, y, \vec b$, and $\vec c$, which we want to solve for $x$ and $y$.
As it is you still have $\vec CS$ meaning we haven't reduced the system to those yet.
And we are also missing another equation with respect to $x$.

#### mathmari

##### Well-known member
MHB Site Helper
The vectors $\vec b$ and $\vec c$ form a basis of the plane.
If we express every vector as a linear combination of these 2, we can get the simplest system possible.

We have the unknowns $x$ and $y$.
So we're looking for a system of equations that contains only $x, y, \vec b$, and $\vec c$, which we want to solve for $x$ and $y$.
As it is you still have $\vec CS$ meaning we haven't reduced the system to those yet.
And we are also missing another equation with respect to $x$.
So far we have $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AC})=y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right ) \\ \vec{BS}=x\cdot \vec{BF}=x\cdot (\vec{AF}-\vec{AB})=x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )$$
So we have to write $\vec{CS}$ and $\vec{BS}$ in terms of $x, y, \vec b$, and $\vec c$, right? But how?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So far we have $$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AC})=y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right ) \\ \vec{BS}=x\cdot \vec{BF}=x\cdot (\vec{AF}-\vec{AB})=x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )$$
So we have to write $\vec{CS}$ and $\vec{BS}$ in terms of $x, y, \vec b$, and $\vec c$, right? But how?
We have $\vec{CS}=\vec{AS}-\vec{AC}=\vec{AS}-\vec b$ and $\vec{BS}=\vec{AS}-\vec{AB}=\vec{AS}-\vec c$.

#### mathmari

##### Well-known member
MHB Site Helper
We have $\vec{CS}=\vec{AS}-\vec{AC}=\vec{AS}-\vec b$ and $\vec{BS}=\vec{AS}-\vec{AB}=\vec{AS}-\vec c$.
So we get $$\vec{CS}=y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right ) \Rightarrow \vec{AS}-\vec{b}=y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right ) \Rightarrow \vec{AS}=y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} \\ \vec{BS}=x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right ) \Rightarrow \vec{AS}-\vec{c}=x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right ) \Rightarrow \vec{AS}=x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )+\vec{c}$$ So we get $$y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} =x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )+\vec{c}$$ We need a second equation, right?

I don't see which second equation we should take...

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we get $$y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} =x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )+\vec{c}$$ We need a second equation, right?

I don't see which second equation we should take...
These are already 2 equations since $\vec b$ and $\vec c$ each have 2 dimensions.
We can solve it for $x$ and $y$ with e.g. Cramer's rule.

#### mathmari

##### Well-known member
MHB Site Helper
These are already 2 equations since $\vec b$ and $\vec c$ each have 2 dimensions.
We can solve it for $x$ and $y$ with e.g. Cramer's rule.
How do we use Cramer's rule?

Do we maybe do the following?
\begin{align*}y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} =x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )+\vec{c} &\Rightarrow y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} -x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )-\vec{c}=\vec{0} \\ & \Rightarrow \frac{y}{5}\vec{c}-y\vec{b}+\vec{b} -\frac{3x}{4}\vec{b}+x\vec{c}-\vec{c}=\vec{0} \\ & \Rightarrow \left (-y+1-\frac{3x}{4}\right )\vec{b}+\left (\frac{y}{5}+x -1\right )\vec{c}=\vec{0}\end{align*} Since $\vec{b}$ and $\vec{c}$ are linearly independent the coefficients must be zero, so we get the system \begin{align*}-y+1-\frac{3x}{4}= & 0 \\ \frac{y}{5}+x -1= & 0\end{align*} The solution is $x=\frac{16}{17}$ and $y=\frac{5}{17}$.

Is that correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How do we use Cramer's rule?

Do we maybe do the following?
\begin{align*}y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} =x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )+\vec{c} &\Rightarrow y\cdot \left (\frac{1}{5}\vec{c}-\vec{b}\right )+\vec{b} -x\cdot \left (\frac{3}{4}\vec{b}-\vec{c}\right )-\vec{c}=\vec{0} \\ & \Rightarrow \frac{y}{5}\vec{c}-y\vec{b}+\vec{b} -\frac{3x}{4}\vec{b}+x\vec{c}-\vec{c}=\vec{0} \\ & \Rightarrow \left (-y+1-\frac{3x}{4}\right )\vec{b}+\left (\frac{y}{5}+x -1\right )\vec{c}=\vec{0}\end{align*} Since $\vec{b}$ and $\vec{c}$ are linearly independent the coefficients must be zero, so we get the system \begin{align*}-y+1-\frac{3x}{4}= & 0 \\ \frac{y}{5}+x -1= & 0\end{align*} The solution is $x=\frac{16}{17}$ and $y=\frac{5}{17}$.

Is that correct?
Nice!
That works as well.

According to Cramer's rule we have:
$y\left(\frac 15\vec c-\vec b\right)+\vec b=x\left(\frac 34 \vec b - \vec c\right) + \vec c \implies \\ x\left(\frac 34 \vec b - \vec c\right) + y\left(\vec b-\frac 15\vec c\right)=\vec b - \vec c \implies \\ \begin{cases}x=\dfrac{\det\left(\vec b-\vec c, \vec b-\frac 15\vec c\right)}{\det\left(\frac 34\vec b-\vec c, \vec b-\frac 15\vec c\right)} \\ y=\dfrac{\det\left(\frac 34\vec b-\vec c,\vec b-\vec c\right)}{\det\left(\frac 34\vec b-\vec c, \vec b-\frac 15\vec c\right)}\end{cases}$

We can evaluate the determinants with its calculation rules.
For instance:
\begin{aligned}\det\left(\vec b-\vec c, \vec b-\frac 15\vec c\right)
&=\det\left(\vec b-\vec c, \vec b-\frac 15\vec c-(\vec b-\vec c)\right) \\
&=\det\left(\vec b-\vec c, \frac 45\vec c\right) \\
&=\frac 45\det\left(\vec b-\vec c, \vec c\right) \\
&=\frac 45\det\left(\vec b-\vec c+\vec c, \vec c\right) \\
&=\frac 45\det(\vec b, \vec c)\end{aligned}

#### mathmari

##### Well-known member
MHB Site Helper
Nice!
That works as well.

According to Cramer's rule we have:
$y\left(\frac 15\vec c-\vec b\right)+\vec b=x\left(\frac 34 \vec b - \vec c\right) + \vec c \implies \\ x\left(\frac 34 \vec b - \vec c\right) + y\left(\vec b-\frac 15\vec c\right)=\vec b - \vec c \implies \\ \begin{cases}x=\dfrac{\det\left(\vec b-\vec c, \vec b-\frac 15\vec c\right)}{\det\left(\frac 34\vec b-\vec c, \vec b-\frac 15\vec c\right)} \\ y=\dfrac{\det\left(\frac 34\vec b-\vec c,\vec b-\vec c\right)}{\det\left(\frac 34\vec b-\vec c, \vec b-\frac 15\vec c\right)}\end{cases}$

We can evaluate the determinants with its calculation rules.
For instance:
\begin{aligned}\det\left(\vec b-\vec c, \vec b-\frac 15\vec c\right)
&=\det\left(\vec b-\vec c, \vec b-\frac 15\vec c-(\vec b-\vec c)\right) \\
&=\det\left(\vec b-\vec c, \frac 45\vec c\right) \\
&=\frac 45\det\left(\vec b-\vec c, \vec c\right) \\
&=\frac 45\det\left(\vec b-\vec c+\vec c, \vec c\right) \\
&=\frac 45\det(\vec b, \vec c)\end{aligned}
Ahh I see!! Thank you very much!!

#### mathmari

##### Well-known member
MHB Site Helper
Do you maybe have/know where I can find some similar exercices like above with vectors?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Do you maybe have/know where I can find some similar exercices like above with vectors?
Not really... but most geometric problems can be solved in multiple ways...
Typically we can choose to use vectors, use trigonometry, or use geometric propositions.
If we want to, we can also try to solve the same exercise multiple times with different approaches.