# Math Help - Vectors!

1. ## Vectors!

Hi! I'm new to the forums but I'm in IB Calc and we're reviewing for the upcoming SL exams. I practically have no vectors experience, though, so I'm looking for some help on one of the practice problems I found. Here it goes...

There's a diagram showing the positions of towns O, A, B, and X.

Town A is 240 km E and 70 km N of O.
Town B is 480 km E and 250 km N of O.
Town X is 339 km E and 238 km N of O.

An airplane flies at a constant speed of 300 km/h from O towards A.

ai) Show that a unit vector in the direction of OA is <0.96, 0.28>. I know how to answer this part.
aii) Write down the velocity vector for the airplane in the form <v1, v2>. Wha? Say again?
aiii) How long does it take for the airplane to reach A? Got this one, too.

At A the airplane changes direction so it now flies towards B. The angle between the original direction and the new direction is theta as shown in a second diagram. This diagram also shows the point Y, between A and B, where the airplane comes closest to X.

b) Use the scalar product of two vectors to find the value of theta in degrees. I'm guessing that I should use the vectors OA and AB but I don't know how I'd actually carry that out.

ci) Write down the vector AX. ???
cii) Show that the vector n = <-3, 4> is perpendicular to AB. I think I know how to do this one.
cii) Calculate the distance XY. ???

d) How far is the airplane from A when it reaches Y? I guess I need to know some of the previous answers to get this one!

Thank you to anyone who can help ahead of time!

2. Originally Posted by Macy
...
There's a diagram showing the positions of towns O, A, B, and X.

Town A is 240 km E and 70 km N of O.
Town B is 480 km E and 250 km N of O.
Town X is 339 km E and 238 km N of O.

An airplane flies at a constant speed of 300 km/h from O towards A.

ai) Show that a unit vector in the direction of OA is <0.96, 0.28>. I know how to answer this part.
aii) Write down the velocity vector for the airplane in the form <v1, v2>. Wha? Say again?
The velocity vector has the direction (0.96, 0.28) and the magnitude (or length) of 300 km. Therefore the velocity vector is:

$\vec v = 300 \cdot (0.96, 0.28) = (288, 84)$

aiii) How long does it take for the airplane to reach A? Got this one, too.

At A the airplane changes direction so it now flies towards B. The angle between the original direction and the new direction is theta as shown in a second diagram. This diagram also shows the point Y, between A and B, where the airplane comes closest to X.

b) Use the scalar product of two vectors to find the value of theta in degrees. I'm guessing that I should use the vectors OA and AB but I don't know how I'd actually carry that out.

ci) Write down the vector AX. ???
cii) Show that the vector n = <-3, 4> is perpendicular to AB. I think I know how to do this one.
cii) Calculate the distance XY. ???

d) How far is the airplane from A when it reaches Y? I guess I need to know some of the previous answers to get this one!
to b) You have to calculate the angle $\angle{(OAB)}$:

1. Vector $\overrightarrow{AO}=(-240, -70)$

2. Vector $\overrightarrow{AB}=(480, 250) - (240, 70) = (240, 180)$

3. Use the formula $\cos(\theta) = \frac{\overrightarrow{AO} \cdot \overrightarrow{AB}}{|\overrightarrow{AO} | \cdot |\overrightarrow{AB} |}$ .... That means: $\cos(\theta) = \frac{(-240, -70) \cdot (240, 180)}{250 \cdot 300} = \frac{-70200}{75000}$ $= -0.936~\implies~\theta \approx 159.39^\circ$

to c)

1. Vector $\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = (339, 238) - (240, 70) = (99, 168)$

2. $\overrightarrow{AB} = (240, 180) = 60 \cdot (4, 3)$

Since $(4, 3) \cdot (-3, 4) = 0$ the vectors $\vec n$ and $\overrightarrow{AB}$ are perpendicular.

3. The points A, X and Y are vertices of a right triangle with the right angle at Y. The distance $\overline{AX}$ is the hypotenuse of this right triangle.
The scalar product $\overrightarrow{AB} \cdot \overrightarrow{AX}$ yields the projection of $\overrightarrow{AX}$ perpendicular to $\overrightarrow{AB}$ multiplied by the length of $\overrightarrow{AB}$. That means you could calculate the length of $\overline{AY}$ if you divide the scalar product by the length of $\overrightarrow{AB}$.
Now you can calculate the length of $\overrightarrow{XY}$:

$|\overrightarrow{AX}| = \sqrt{38025}$

$(240, 180) \cdot (99, 168) = 54000$ and therefore $|\overrightarrow{AY}| = \frac{54000}{300} = 180$ (Btw that's the answer to d))

Now calculate $\overline{XY} = \sqrt{38025 - 180^2}=\sqrt{5625} = 75$

3. Wow, thank you so much! Hopefully I won't have any more trouble with vectors...