Analytical Geometry I

Geometry Level 2

Let $\vec{(u+v)}$ and $\vec{(2u-v)}$ be Euclidean vectors, and $\theta$ the angle between them. If :

$|\vec{u}|^2=4$
$|\vec{v}|^2=4$
$\vec{u} \cdot \vec{v}=-1$

π/2 2π/3 π π/6

1 solution

Dimitris Kouroupos
Sep 14, 2017

- $\vec{(u+v)} \cdot \vec{(2u-v)}=$ $< \vec{u}+\vec{v},\vec{2u}-\vec{v}$ > = < $\vec{u},\vec{2u}-\vec{v}$ > + < $\vec{v},\vec{2u}-\vec{v}$ > =

2 < $\vec{u},\vec{u}$ > - < $\vec{u},\vec{v}$ > + 2 < $\vec{v},\vec{u}$ > - < $\vec{v},\vec{v}$ >= $-3$

- $|\vec{u}+\vec{v}|= \sqrt{<\vec{u}+\vec{v},\vec{u}+\vec{v}>} =\sqrt{<\vec{u},\vec{u}+\vec{v}>+<\vec{v},\vec{u}+\vec{v}>}=\sqrt{<\vec{u},\vec{u}>+<\vec{u},\vec{v}>+<\vec{v},\vec{u}>+<\vec{v},\vec{v}>}=\sqrt{<\vec{u},\vec{u}>+2<\vec{u},\vec{v}>+<\vec{v},\vec{v}>}=\sqrt{1-2+4}=\sqrt{3}$

- $|\vec{2u}-\vec{v}|= \sqrt{<\vec{2u}-\vec{v},\vec{2u}-\vec{v}>}=\sqrt{<\vec{2u},\vec{2u}-\vec{v}>-<\vec{v},\vec{2u}-\vec{v}>}=\sqrt{4<\vec{u},\vec{u}>-2<\vec{u},\vec{v}>-2<\vec{v},\vec{u}>+<\vec{v},\vec{v}>}=\sqrt{12}=2\sqrt{3}$

$\cos \theta = \frac{\overrightarrow{u+v} \cdot \overrightarrow{2u-v}}{\left | \overrightarrow{u+v} \right | \left | \overrightarrow{2u-v} \right | }. =$ $\frac{-1}{2}$

that means $\theta$ $=2π/3$

This is wrong because |2u-v|=root(24) (not 12) and the scalar product (u+v)(2u-v) is +3, not -3. cos = 1/4.

lovro cupic - 1 month, 2 weeks ago