Vectors

[EmberP]

I have no problem working with vectors, but I'm having trouble converting this word problem into equations or a picture. I don't see what it's asking:

"A mountaineer uses a global positioning system receiver to measure his displacement from base camp to the top of Mount McKinley. In terms of horizontal coordinates x and y, the base camp is at x = 0, y = 0, and the base of Mount McKinley is at x = 1600 m, y = 4200 m. If the altitude of Mount McKinley is 6200 m and the base camp is at 4024 m, what is the displacement in going from base camp to the top?"

Because of the program my homework is on, I know that if you substitute the altitude of the base camp for 4131 m, the answer would be 4950 m. I just have no idea how to get there.

Thanks in advance for any help!

 

[tkhunny]

Traingulation, My Friend.

Draw a right triangle with the top of Mt. McKinley at one acute vertex (T), base camp at the other actute vertex (B), and the right angle beneath the summit, inside the mountain, at the same elevation as the Base (R).

Calculate segment BR with the Pythagorean Theorem. \(\displaystyle BR = \sqrt{1600^{2}+4200^{2}}\)

Calculate TR by subtraction. \(\displaystyle TR = 6200 - 4024\)

Now what?

 

[EmberP]

Oh, so then you just use Pythagorean Theorem to get the remaining side. It's all on one triangle - I see now. Here I am with this insane picture involving two triangles and a rectangle. Thanks so much!

 

[soroban]

Hello, EmberP!

A mountaineer uses a global positioning system receiver to measure his displacement from base camp to the top of Mount McKinley.
In terms of horizontal coordinates x and y, the base camp is at x = 0, y = 0, and the base of Mount McKinley is at x = 1600 m, y = 4200 m.
If the altitude of Mount McKinley is 6200 m and the base camp is at 4024 m, what is the displacement in going from base camp to the top?"

Because of the program my homework is on, I know that if you substitute the altitude of the base camp for 4131 m, the answer would be 4950 m.
I have no idea what this means . . .

Think of it this way . . .

We have a rectangular box: \(\displaystyle 1600 \times 4200 \times 2176\)

         *---------o T
        /:        /|
       / :       / |2176
      *---------*  |
      |  * - - -|- * R
      | .       | /
      |.        |/ 4200
    B o---------*
         1600

And we want the length of the diagonal \(\displaystyle BT.\)


There is a formula for a "space diagonal".

. . \(\displaystyle BT \;=\;\sqrt{1600^2 + 4200^2 + 2176^2} \;=\;\sqrt{24,\!934,\!976} \;=\;4993.493366\text{ m.}\)

 

[shawnarae]

Hello, I need help with vector analysis. Having a test next week and I missed lecture, here is a problem I cant seem to get going on ...Please help
A plane traveling southwest has a vector heading of 180mi/hr at 240 deg.
A northwest head wind has a vector heading of 80 mi/hr and 150 deg.
What is the planes final vector heading and what is the theta and new displacement angle......... :roll: :?

 

[Deleted member 4993]

Hello, I need help with vector analysis. Having a test next week and I missed lecture, here is a problem I cant seem to get going on ...Please help
A plane traveling southwest has a vector heading of 180mi/hr at 240 deg.
A northwest head wind has a vector heading of 80 mi/hr and 150 deg.
What is the planes final vector heading and what is the theta and new displacement angle......... :roll: :?

Duplicate Post:

viewtopic.php?f=10&t=44218