Question

Walk fifty meters at 30o north or east from the old oak tree. (2) Turn 45o to your left (you should now be facing 75o north of east) and walk another fifty meters. You will find the treasure buried under a rock. What straight-line path would take you directly from the old oak tree to the rock with the treasure buried under it

Answer 1

Answer:

A straight line of approximately 75 meters, 1.4º north

Step-by-step explanation:

Hi, let's make it step by step to make it clearer

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in $R^{2}$. To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

3) Finally, the last vector leads to the treasure from the Old Oak Tree, i.e. the resultant.

So, let's calculate the norm which is the length of the each vector.

1) Graphing them we can find the points, then the components and then calculate the norm, the length of each vector.  

Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

$\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle$

Now, let's calculate each vector length by calculating the norm.

$\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86$

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.