Calculate Displacement Vector from Camp to Summit - 2079m

[vihits13]

The summit of a mountain, 2079 m above base camp, is measured on a map to be 4577 m horizontally from the camp in a direction 32.4° west of north. Choose the x-axis east, y-axis north, and z axis up. What are the components of the displacement vector from camp to summit?

x= ? m
y= ? m
z= ? m

What is the magnitude?
? meters

Thanks if u r helping me out!

 

[berkeman]

from my replies to your first and second question threads:

Welcome to PF, vihits. As the rules say when you signed up just now, you must show your work up to now on these problems. We do not do your homework for you. We *do* help you when your are stuck on some concept, or making an error that we can spot.

So, can you show your work so far?

Like, what does your diagram look like? What did you choose for the direction of your x-axis?

 

[kyle8921]

I would approach this problem by first setting up a coordinate system with the x-axis pointing east, the y-axis pointing north, and the z-axis pointing up. Using this coordinate system, we can calculate the components of the displacement vector from camp to summit.

x = 4577 m * cos(32.4°) = 3818.98 m (east)
y = 4577 m * sin(32.4°) = 2407.76 m (north)
z = 2079 m (up)

To find the magnitude of the displacement vector, we can use the Pythagorean theorem:

magnitude = √(x^2 + y^2 + z^2) = √(3818.98^2 + 2407.76^2 + 2079^2) = 5561.69 m

Therefore, the components of the displacement vector from camp to summit are:
x = 3818.98 m (east)
y = 2407.76 m (north)
z = 2079 m (up)

And the magnitude of the displacement vector is 5561.69 m.