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BS in relation to BF and CS in relation to CE

mathmari

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Apr 14, 2013
4,440
Hey!! :giggle:

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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
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
Apr 14, 2013
4,440
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
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 $AB$? (Worried)

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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,440

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,440
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
$$\vec{CS}=y\cdot \vec{CE}=y\cdot (\vec{AE}-\vec{AE})=y\cdot b-y\cdot (\vec{c}-\vec{EB})=y\cdot \vec{b}-y\cdot \vec{c}+y\cdot \vec{EB}$$
So we have to write somehow $ \vec{EB}$.

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

mathmari

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MHB Site Helper
Apr 14, 2013
4,440
Isn't $\vec{AE}-\vec{AE}=0$? (Worried)
Perhaps you meant to refer to $\vec{AB}$. If so, isn't that one equal to $\vec c$ instead of $\vec b$? (Sweating)
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
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 $\vec{CE} = \vec{AE}-\vec{AC}$. (Sweating)

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
Apr 14, 2013
4,440
A vector is equal to a vector to its end point minus a vector to its starting point.
So we have $\vec{CE} = \vec{AE}-\vec{AC}$. (Sweating)

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?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
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 $\vec{AE}=\frac 15 \vec{AB}=\frac 15\vec c$? :unsure:
 

mathmari

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Apr 14, 2013
4,440

Klaas van Aarsen

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Mar 5, 2012
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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
Apr 14, 2013
4,440
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,162
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

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MHB Site Helper
Apr 14, 2013
4,440
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? :unsure:

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

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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

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MHB Site Helper
Apr 14, 2013
4,440
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? :unsure:

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? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
9,162
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. (Sun)

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
Apr 14, 2013
4,440
Nice!
That works as well. (Sun)

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!! (Sun)
 

mathmari

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Apr 14, 2013
4,440
Do you maybe have/know where I can find some similar exercices like above with vectors? :unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,162
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. 🤔