My Math Forum Plane Coordinates

 Linear Algebra Linear Algebra Math Forum

December 13th, 2016, 06:59 PM   #1
Newbie

Joined: Dec 2016
From: Natal - Brazil

Posts: 10
Thanks: 0

Plane Coordinates

The attached figure illustrates a rectangular coordinate system xy generated by the unit vectors of the canonical base $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$ and a system of coordinates non-rectangular $\displaystyle {x}'{y}'$ generated by the unit vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ ($\displaystyle \left \| \vec{u} \right \| = \left \| \vec{v} \right \| = 1$), where the axis $\displaystyle {y}'$ coincides with the y-axis, while $\displaystyle {x}'$ is obtained from its anti-clockwise rotation by an angle $\displaystyle \theta = \frac{\pi}{6}$. Find the $\displaystyle {x}'{y}'$ coordinates of the points whose coordinates $\displaystyle xy$ are $\displaystyle (10,-3)$. Use $\displaystyle cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\displaystyle sin(\frac{\pi}{6}) = \frac{1}{2}$.( Remember that $\displaystyle (x,y) = x\vec{i}+y\vec{j} = {x}'\vec{u}+{y}'\vec{v}$)
Attached Images
 Capturarex.PNG (11.5 KB, 1 views)

December 13th, 2016, 08:15 PM   #2
Math Team

Joined: May 2013
From: The Astral plane

Posts: 1,797
Thanks: 715

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by mosvas The attached figure illustrates a rectangular coordinate system xy generated by the unit vectors of the canonical base $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$ and a system of coordinates non-rectangular $\displaystyle {x}'{y}'$ generated by the unit vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ ($\displaystyle \left \| \vec{u} \right \| = \left \| \vec{v} \right \| = 1$), where the axis $\displaystyle {y}'$ coincides with the y-axis, while $\displaystyle {x}'$ is obtained from its anti-clockwise rotation by an angle $\displaystyle \theta = \frac{\pi}{6}$. Find the $\displaystyle {x}'{y}'$ coordinates of the points whose coordinates $\displaystyle xy$ are $\displaystyle (10,-3)$. Use $\displaystyle cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\displaystyle sin(\frac{\pi}{6}) = \frac{1}{2}$.( Remember that $\displaystyle (x,y) = x\vec{i}+y\vec{j} = {x}'\vec{u}+{y}'\vec{v}$)
What have you been able to do so far?

-Dan

December 14th, 2016, 02:01 AM   #3
Newbie

Joined: Dec 2016
From: Natal - Brazil

Posts: 10
Thanks: 0

Quote:
 Originally Posted by topsquark What have you been able to do so far? -Dan
I have not done anything yet, I do not know where to start.

 December 14th, 2016, 04:57 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,192 Thanks: 871 Seeing that this involves angles and you get coordinates of a points in a given coordinate system by drawing perpendiculars from the point to the coordinate axes, I would start by drawing those lines and using trigonometry. (Do you suppose that's why they gave you the sine and cosine of $\displaystyle \pi/6$?) Thanks from topsquark

 Tags coordinates, plane

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Addez123 Calculus 4 November 12th, 2016 06:51 AM extreme112 Linear Algebra 2 October 13th, 2015 07:34 AM Eppo Algebra 5 January 9th, 2008 01:35 PM dex2l Algebra 4 November 2nd, 2007 04:30 AM blazingclaymore Computer Science 4 July 18th, 2007 05:59 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top