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In triangle ABC, S is a point on side BC such that BS:SC = 1:2, and T is a point on side AC such that AT:TC = 4:3. Let U be the intersection of AS and BT.

We can write \(\overrightarrow{T} = w \overrightarrow{A} + x \overrightarrow{C}, \overrightarrow{S} = y \overrightarrow{B} + z \overrightarrow{C}\)for some real values of w, x, y, and z. Find w, x, y, and z.

 

We can also write \(\overrightarrow{U} = x \overrightarrow{A} + y \overrightarrow{B} + z \overrightarrow{C}\) for some real values of x, y, z. Find x, y, z.

 

Find\(\frac{AU}{US}\)

 Aug 4, 2019
 #1
avatar+23788 
+2

 

1.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

We can write \( \vec{T} = w \vec{A} + x \vec{C}\), \(\vec{S} = y \vec{B} + z \vec{C}\) for some real values of w, x, y, and z.
Find w, x, y, and z.

 

\(\begin{array}{|rcll|} \hline \vec{T} &=& \vec{A}+ \dfrac{4}{7}\left(\vec{C}-\vec{A} \right) \\ \vec{T} &=& \vec{A}+ \dfrac{4}{7}\vec{C} -\dfrac{4}{7}\vec{A} \\ \vec{T} &=& \dfrac{3}{7}\vec{A}+ \dfrac{4}{7}\vec{C} \quad &|\quad \vec{T} = w \vec{A} + x \vec{C}\\ \hline \mathbf{w}&=& \mathbf{\dfrac{3}{7}} \\ \mathbf{x}&=& \mathbf{\dfrac{4}{7}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \vec{S} &=& \vec{C}+ \dfrac{2}{3}\left(\vec{B}-\vec{C} \right) \\ \vec{S} &=& \vec{C}+ \dfrac{2}{3}\vec{B} -\dfrac{2}{3}\vec{C} \\ \vec{S} &=& \dfrac{2}{3}\vec{B}+ \dfrac{1}{3}\vec{C} \quad &|\quad \vec{S} = y \vec{B} + z \vec{C}\\ \hline \mathbf{y}&=& \mathbf{\dfrac{2}{3}} \\ \mathbf{z}&=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

 

laugh

 Aug 4, 2019
 #2
avatar+23788 
+2

2.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

 

Find \(\dfrac{AU}{US}\)

 

\(\begin{array}{|rcll|} \hline \vec{u} &=& \vec{A}-\vec{C} \\ \vec{v} &=& \vec{B}-\vec{C} \\ \hline \vec{U}-\vec{C} &=& \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) \quad | \quad \lambda \text{ is a real value} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \dfrac{US}{AS} \text{ is a real value} \\ \hline \vec{U}-\vec{C} = \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda \vec{v}-\dfrac{3}{7}\lambda\vec{u} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\vec{u}-\dfrac{2}{3}\dfrac{US}{AS}\vec{v} \\ \vec{u}\left(\underbrace{\dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} }_{=0}\right) &=& \vec{v}\left(\underbrace{\dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda }_{=0} \right) \quad | \quad \vec{u} \text{ and } \vec{v}\text{ are independent} \\ \hline \dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} &=& 0 \\ \quad |\quad \cdot \dfrac{7}{3} \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} &=& 0 \qquad (1) \\\\ \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda &=& 0 \qquad (2) \\ \hline (1)-(2): \qquad 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \left( \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda \right) &=& 0 \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \dfrac{2}{3}+\dfrac{2}{3}\dfrac{US}{AS}+\lambda &=& 0 \\ \dfrac{1}{3} -\dfrac{5}{3}\dfrac{US}{AS} &=& 0 \\ \dfrac{5}{3}\dfrac{US}{AS} &=& \dfrac{1}{3} \quad | \quad \cdot \dfrac{3}{5} \\ \mathbf{ \dfrac{US}{AS} } &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{AU}{AS} &=& \left( 1-\dfrac{US}{AS} \right) \\ \dfrac{AU}{AS} &=& \left( 1-\dfrac{1}{5} \right) \\ \mathbf{ \dfrac{AU}{AS} } &=& \mathbf{\dfrac{4}{5}} \\ \hline \dfrac{AU}{US} &=& \dfrac{ \dfrac{AU}{AS} }{ \dfrac{US}{AS} } \\ \dfrac{AU}{US} &=& \dfrac{ \dfrac{4}{5} }{ \dfrac{1}{5} } \\ \mathbf{ \dfrac{AU}{US} } &=& \mathbf{4} \\ \hline \end{array}\)

 

laugh

 Aug 4, 2019
edited by heureka  Aug 4, 2019
 #3
avatar+23788 
+2

3.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,
and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

 

We can also write \(\vec{U} = x \vec{A} + y \vec{B} + z \vec{C}\)
for some real values of x, y, z. Find x, y, z.

 

\(\begin{array}{|rcll|} \hline \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \vec{u} = \vec{A}-\vec{C},\ \vec{v} = \vec{B}-\vec{C} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\left(\vec{B}-\vec{C}\right)+ \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\left(\vec{B}-\vec{C}\right)\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\vec{B}+\dfrac{2}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \quad | \quad \dfrac{US}{AS}=\dfrac{1}{5} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \vec{C}+\dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \dfrac{1}{5}\vec{A} +\dfrac{8}{15}\vec{B}+\dfrac{4}{15}\vec{C} \quad |\quad \vec{U} = x \vec{A} + y \vec{B} + z \vec{C} \\ \hline \mathbf{x}&=& \mathbf{\dfrac{1}{5}} \\ \mathbf{y}&=& \mathbf{\dfrac{8}{15}} \\ \mathbf{z}&=& \mathbf{\dfrac{4}{15}} \\ \hline \end{array}\)

 

laugh

 Aug 4, 2019

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