Vector problem

**s3a** · April 23rd 2009, 07:47 PM

The only part I get is the 2km [NE]. When it says to go 120° towards the right for the 1.5 km vector, I don't know how to manage that exactly or how to connect it to the 2km vector.

Any help would be GREATLY appreciated!
Thanks in advance!

**earboth** · April 23rd 2009, 10:29 PM

Originally Posted by s3a

The only part I get is the 2km [NE]. When it says to go 120° towards the right for the 1.5 km vector, I don't know how to manage that exactly or how to connect it to the 2km vector.

Any help would be GREATLY appreciated!
Thanks in advance!

Use a coordinate system with the x-axis pointing East and the y-axis pointing North. Then the direction NE is described by the vector $\vec v = (1,1)$ .

Make a sketch.

You are dealing with a triangle where you know the length of two adjacent sides and the included angle.

**earboth** · April 24th 2009, 11:00 AM

Originally Posted by s3a

The only part I get is the 2km [NE]. When it says to go 120° towards the right for the 1.5 km vector, I don't know how to manage that exactly or how to connect it to the 2km vector.

Any help would be GREATLY appreciated!
Thanks in advance!

2nd part.

I assume that you are asked to use vectors. According to my previous post the first part of the trip is described by a vector which has the direction (1, 1). This vector should have the length of 2. Therefore the first part is represented by the vector:

$\vec v = \left(\sqrt{2}\ ,\sqrt{2}\right)$

The second part of the trip is described by a vector which points at an angle of -75° to the East direction (=positive x-axis). Therefore the second vector is

$\vec u = 1.5\cdot (\cos(-75^\circ)\ ,\ \sin(-75^\circ)) = (0.388229\ ,\ -1.44889)$

The endpoint of the trip is detremined by the vector

$\vec e = (\sqrt{2}+0.38829\ ,\ \sqrt{2}+(-1.44889)) = (1.80244\ ,\ -0.03468)$

The distance between start point and endpoint is nothing but the length of $\vec e$

$|\vec e|\approx 1.80277$

Use the coordinates of $\vec e$ to calculate the direction.

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