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 October 11th, 2018, 10:15 PM #1 Member   Joined: Oct 2012 Posts: 78 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
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$\cos(45^\circ) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}|\cdot|\vec{v}|}$

$\vec{u} = <4,-2>$, 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: 20,927 Thanks: 2205 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
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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: 20,927 Thanks: 2205 Thanks. The coordinates I gave should therefore be multiplied by √10.

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