## Solving Lesson 55 Page 100 Math 10 SBT – Kite>

Topic

For triangle ABC. Get the points EASY, E, USA, WOMEN satisfy $$\overrightarrow {AD} = \frac{1}{3}\overrightarrow {AB} ,\overrightarrow {AE} = \frac{2}{5}\overrightarrow {AC} ,\overrightarrow {BM} = \frac{1}{3}\overrightarrow {BC} ,\overrightarrow {AN} = k\overrightarrow {AM}$$

with k is real number. Express the vectors $$\overrightarrow {AN} ,\overrightarrow {DE} ,\overrightarrow {EN}$$ in terms of the vectors $$\overrightarrow a = \overrightarrow {AB} ,\overrightarrow b = \overrightarrow {AC} \ ) And find k to three points EASY, E, WOMEN straight. Solution method – See details Step 1: Locate the points EASY, E, USA, WOMEN on the edge AB, AC, BC, AM Step 2: Use the rules to represent vectors in terms of \(\overrightarrow {AB}$$ and $$\overrightarrow {AC}$$

Step 3: Use the condition $$\overrightarrow {EN} = t\overrightarrow {DE}$$ to prove EASY, E, WOMEN straight.

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Detailed explanation

According to the assumption EASY, E, USA, WOMEN lies between the two endpoints of the corresponding edges AB, AC, BC, AM

a) We have: $$\overrightarrow {AD} = \frac{1}{3}\overrightarrow {AB} = \frac{1}{3}\overrightarrow a$$; $$\overrightarrow {AE} = \frac{2}{5}\overrightarrow {AC} = \frac{2}{5}\overrightarrow b$$;

$$\overrightarrow {BM} = \frac{1}{3}\overrightarrow {BC} \Leftrightarrow \overrightarrow {AM} – \overrightarrow {AB} = \frac{1}{3}\left( {\overrightarrow { AC} – \overrightarrow {AB} } \right) \Leftrightarrow \overrightarrow {AM} = \overrightarrow {AB} + \frac{1}{3}\left( {\overrightarrow {AC} – \overrightarrow {AB} } \right) \Leftrightarrow \overrightarrow {AM} = \frac{2}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b$$

+ $$\overrightarrow {AN} = k\overrightarrow {AM} = k\left( {\frac{2}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b } \right) = \ frac{{2k}}{3}\overrightarrow a + \frac{k}{3}\overrightarrow b$$

+ $$\overrightarrow {DE} = \overrightarrow {AE} – \overrightarrow {AD} = – \frac{1}{3}\overrightarrow a + \frac{2}{5}\overrightarrow b$$

+ $$\overrightarrow {EN} = \overrightarrow {AN} – \overrightarrow {AE} = k\left( {\frac{2}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b } \right) – \frac{2}{5}\overrightarrow b = \frac{{2k}}{3}\overrightarrow a + \frac{{5k – 6}}{{15}}\overrightarrow b$$

b) EASY, E, WOMEN aligns if and only if $$\overrightarrow {EN} = t\overrightarrow {DE}$$ $$\Leftrightarrow \frac{{2k}}{3}\overrightarrow a + \frac{{5k – 6}}{ {15}}\overrightarrow b = t\left( { – \frac{1}{3}\overrightarrow a + \frac{2}{5}\overrightarrow b } \right)$$

$$\Leftrightarrow \left\{ \begin{array}{l}\frac{{2k}}{3} = – \frac{t}{3}\\\frac{{5k – 6}}{{15) }} = \frac{{2t}}{5}\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}\frac{2}{3}k + \frac{1} {3}t = 0\\\frac{1}{3}k – \frac{2}{5}t = \frac{2}{5}\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}k = \frac{6}{{17}}\\t = – \frac{{12}}{{17}}\end{array} \right.$$

So for $$k = \frac{6}{{17}}$$ then EASY, E, WOMEN straight.