ihatealgebrasomuch 2 years ago i really need help with vectors! can someone please explain to me how to go about solving this problem? The pilot of an airplane flies at a constant speed of 200 km/h at a bearing of N 25degrees S. There is a 40 km/h crosswind blowing southeast (S 45degrees E). What are the plane's actual speed and direction? ***will draw picture***

1. ryan123345

i love algebra <3

2. ihatealgebrasomuch

|dw:1360291818800:dw|

3. stamp

Your picture is a bit all over the place. Let us try again, using vector notation.$V_{plane}=200\frac{km}{h}<cos25°,sin25°>$Can you find the other vector?

4. ihatealgebrasomuch

would it be 40km/hr <cos45,sin45> ?? i dont understand how you got that equation

5. stamp

The velocity doesn't have direction, it is a scalar that varies. The direction is determined by the angle it forms with the axis. Observe:|dw:1360292754973:dw|

6. stamp

Your vector is ok except southeast means that your y vector is negative$V_{wind}=40\frac{km}{h}<cos45°, -sin45°>|dw:1360292917561:dw|$

7. stamp

Please note that my vector drawing should have arrows on the tails, and my axis should be labeled x and y appropriately

8. stamp

$V_{wind}=40\frac{km}{h}<cos45°, -sin45°>$

9. stamp

The planes actual speed and direction will be$V_{actual}=V_{plane}+V_{wind}$

10. ihatealgebrasomuch

so i would do 200km/h <cos25, sin25> + 40km/h <cos45,-sin45> in order to get the magnitude of the vector i am looking for?

11. stamp

Not to get the magnitude of the vector you are looking for, but to get the resultant vector. The resultant vector is how fast the plane is actually going, because even though the airplane travels at 200km/h in a direction, the wind offsets it by blowing against. The planes true course is determined by the added vectors. To find the magnitude of the vector, you would use pythagoream theorem.