## Solving Lesson 46 Page 83 SBT Math 10 – Kite>

Topic

There are two subways A and REMOVE Running in the inner city also starts from two stations, moving in a straight line. On the radar screen of the control station (referred to as the coordinate plane Oxygen with units on axes in kilometers), after starting t (hour) (t 0), position of the ship A whose coordinates are determined by the formula: $$\left\{ \begin{array}{l}x = 7 + 36t\\y = – 8 + 8t\end{array} \right.$$ , position of ships REMOVE whose coordinates are (9 + 8t ; 5 – 36t).

a) Calculate the cosine of the angle between the paths of the two trains A and REMOVE

b) How long after departure are the two closest trains?

Solution method – See details

Step 1: Find the direction vectors of the two lines that are the paths of the two trains A and REMOVE based on the train’s path PT A and the ship’s coordinates REMOVE

Step 2: Calculate the cosine between the two direction vectors found in step 1 and take the positive value to calculate the cosine of the angle between the two paths of the two ships A and REMOVE

Step 3: Find the coordinates of 2 points USA, WOMEN (parameterize the coordinates of 2 points USA, WOMEN) in 2 positions where the ship A and ship REMOVE arrive after departure t hour. Find t to MN achieve GTN

Detailed explanation

a) Train A vector motion $$\overrightarrow {{u_1}} = (36;8)$$; ship REMOVE vector motion $$\overrightarrow {{u_2}} = (8; – 36)$$

We have: $$\overrightarrow {{u_1}} .\overrightarrow {{u_2}} = 36.8 + 8.( – 36) = 0$$$$\Rightarrow \overrightarrow {{u_1}} \bot \overrightarrow {{) u_2}} \Rightarrow \cos \left( {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right) = 0$$

Let $$\alpha$$ be the angle between the paths of the two ships. Then $$\cos \alpha = \left| {\cos \left( {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right)} \right| = 0$$

b) After t time: train A at position $$M(7 + 36t; – 8 + 8t)$$; ship REMOVE at the point position $$N(9 + 8t;5 – 36t)$$

We have: $$\overrightarrow {MN} = ( – 28t + 2; – 44t + 13) \Rightarrow MN = \sqrt {{{( – 28t + 2)}^2} + {{( – 44t + 13) }^2}}$$

$$\Leftrightarrow M{N^2} = {( – 28t + 2)^2} + {( – 44t + 13)^2} = 2720{t^2} – 1256t + 173$$

According to the assumption, MN achieve GTNN $$\Leftrightarrow M{N^2}$$ gain GTNN

Consider $$M{N^2} = 2720{t^2} – 1256t + 173 = 2720{\left( {t – \frac{{157}}{{680}}} \right)^2} + \ frac{{4761}}{{170}}$$$$\ge \frac{{4761}}{{170}}$$ $$\Rightarrow MN \ge \sqrt {\frac{{4761}}{{ 170}}}$$

The sign “=” occurs if and only if t = $$\frac{{157}}{{680}}$$

So after $$\frac{{157}}{{680}}$$ hours, the two trains are closest to each other and are separated by a distance of 5.29 km