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.
