My Math Forum Find a vector?

 Calculus Calculus Math Forum

 October 11th, 2018, 10:15 PM #1 Member   Joined: Oct 2012 Posts: 65 Thanks: 0 Find a vector? Let u , v a vectors of 2-dim such that u=<4,-2> ,|v|=10 the angle between u and v is 45 degree. Find a coordinate for v satisfies these conditions.
October 12th, 2018, 08:58 AM   #2
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,770
Thanks: 1424

$\cos(45^\circ) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}|\cdot|\vec{v}|}$

$\vec{u} = <4,-2>$, let $\vec{v} = <a,b>$

$\dfrac{1}{\sqrt{2}} = \dfrac{4a + (- 2)b}{\sqrt{20} \cdot 10} \implies 4a-2b = 10\sqrt{10}$

also, $a^2+b^2=100$

Quote:
 Find a coordinate for v satisfies these conditions.
can solve the system of equations to finish?

 October 13th, 2018, 05:35 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,700 Thanks: 1804 The points (0, 0), (4, -2), (6, 2), and (2, 4) are the vertices of a square whose diagonals have length 2√10. Hence the desired vector is (1/2)<6, 2> or (1/2)<2, -6>, i.e. <3, 1>, or <1, -3>.
October 13th, 2018, 06:50 AM   #4
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,770
Thanks: 1424

Quote:
 Originally Posted by skipjack The points (0, 0), (4, -2), (6, 2), and (2, 4) are the vertices of a square whose diagonals have length 2√10. Hence the desired vector is (1/2)<6, 2> or (1/2)<2, -6>, i.e. <3, 1>, or <1, -3>.
original post said $|v| = 10$, not $\sqrt{10}$ ...

 October 13th, 2018, 08:54 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,700 Thanks: 1804 Thanks. The coordinates I gave should therefore be multiplied by √10.

 Tags find, vector

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post blacklisted Calculus 1 October 22nd, 2014 03:34 PM OriaG Calculus 1 May 22nd, 2013 09:14 AM miran97 Algebra 4 March 11th, 2013 12:14 PM rnck Advanced Statistics 1 August 21st, 2011 08:12 PM al1850 Calculus 4 May 3rd, 2008 04:04 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top