# Vector and scalars

## Homework Statement

A plane ﬂies from base camp to Lake A, 280 km away inthe direction 20.0°north of east. After dropping off sup-plies, it ﬂies to Lake B, which is 190 km at 30.0°west of north from Lake A. Graphically determine the distance and direction from Lake B to the base camp.

## The Attempt at a Solution

I've drawn a line from base camp to lake A with an angle of 20°, another line from lake A to lake with an angle of 30°.
I have calculated the components of vector A, I already know the $$|A|=280km, |B|=190km$$, so:
$$A_{x}=Acos(\frac{pi}{9})=263km$$,
$$A_{y}=Asin(\frac{pi}{9})=95.8km$$,
$$B_{x}=Bsin(\frac{-pi}{6})=-95km$$,
$$B_{y}=Bcos(\frac{-pi}{6})=165km$$;
$$R^→=A^→+B^→$$;
$$|R|=sqrt((168^2)+(261^2))=310km$$,
$$cosσ=\frac{168}{310}→σ=57.2°$$,
$$sinσ=\frac{261}{310}→σ=57.4°$$;
I get the exact result because:
$$B_{x}=Bsin(\frac{-pi}{6})=-95km$$,
$$B_{y}=Bcos(\frac{-pi}{6})=165km$$;
i don't know why,
In which quadrant are the vector B and R?

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HallsofIvy
Homework Helper
"30 degrees West of North" is in the second quadrant (the x-component is negative ("West") and the y component is positive ("North")). Surely that is easy to see.

Which quadrant R, the return route, is in is a little harder. Assuming your calculations are correct, since the x and y coordinate are both positive, it is in the first quadrant. However, it is hard to tell whether your R is correct because you don't say where you got the "168" and "261" that you use in your calculations. I suspect they are A+ B but, if so, that is incorrect. The location after the second leg is A+ B. The path necessary to fly back to the base camp is -(A+ B). That would be in the third quadrant.

yes I've got them from A+B=R,
so R=-(A+B)->|R|=sqrt((-168^2)+(-261^2))=310km is the distance from lake B to base camp;
and if i calculate σ = arctan Ry/Rx->σ=57.2° south of west.
OK 30° west of north are 30° beetween y axis and vector B^->, 30°+90°= 120°=2pi/3 in radiant.

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