Points and Vectors

1003040201

Level: 
C
We are given the vectors $\vec{a}=(-1; 2;3)$, $\vec{b}=(3; 1; -2)$ and $\vec{c}=(1; 2;-1)$. Find the coordinates of a vector $\vec{v}$, such that $\vec{v}$ is perpendicular to both vectors $\vec{a}$ and $\vec{b}$, while $\vec{v}\cdot\vec{c}=12$ holds.
$\vec{v}=(-6;6;-6)$
$\vec{v}=(6;-6;6)$
$\vec{v}=(-7;7;-7)$
$\vec{v}=(7;-7;7)$

1103030505

Level: 
B
The vectors \( \vec{u} \) and \( \vec{v} \) are given by the figure. Find cosine of the angle \( \varphi \) between \( \vec{u} \) and \( \vec{v} \). Help: Use the dot product of the given vectors.
\( \cos\varphi=-\frac9{17} \)
\( \cos\varphi=\frac9{17} \)
\( \cos\varphi=\frac{\sqrt{17}}{2\sqrt{13}} \)
\( \cos\varphi=-\frac{\sqrt{17}}{2\sqrt{13}} \)

1103030504

Level: 
B
The vectors \( \vec{u} \) and \( \vec{v} \) are given by the figure. Find cosine of the angle \(\varphi \) between \( \vec{u} \) and \( \vec{v} \). Help: Use the dot product of the given vectors.
\( \cos\varphi=\frac{13\sqrt{10}}{50} \)
\( \cos\varphi=\frac{970}{50} \)
\( \cos\varphi=\frac{3\sqrt{10}}{10} \)
\( \cos\varphi=\frac{\sqrt{10}}{5} \)

1103030503

Level: 
B
Find the coordinates of the vectors \( \vec{u} \) and \( \vec{v} \) given by the picture and evaluate their dot product.
\( \vec{u}=(-8;-7;9);\ \ \vec{v} =(8;7;9);\ \ \vec{u}\cdot\vec{v} = -32 \)
\( \vec{u}=(-8;-7;9);\ \ \vec{v} =(8;7;9);\ \ \vec{u}\cdot\vec{v} = 0 \)
\( \vec{u}=(-8;-7;9);\ \ \vec{v} =(8;7;9);\ \ \vec{u}\cdot\vec{v} = (-64;-49;81) \)
\( \vec{u}=(8;7;-9);\ \ \vec{v} =(-8;-7;-9);\ \ \vec{u}\cdot\vec{v} = (-64;-49;81) \)

1103030502

Level: 
B
Find the coordinates of the vectors \( \vec{u} \) and \( \vec{v} \) given by the picture and evaluate their dot product.
\( \vec{u}=(-3;6);\ \ \vec{v} =(-9;-6);\ \ \vec{u}\cdot\vec{v} = -9 \)
\( \vec{u}=(3;-6);\ \ \vec{v} =(9;6);\ \ \vec{u}\cdot\vec{v} = -9 \)
\( \vec{u}=(-3;6);\ \ \vec{v} =(-9;-6);\ \ \vec{u}\cdot\vec{v} = 9 \)
\( \vec{u}=(3;-6);\ \ \vec{v} =(9;6);\ \ \vec{u}\cdot\vec{v} = 0 \)

1103030501

Level: 
B
The vectors \( \vec{u} \), \( \vec{v}\), \( \vec{w} \), \( \vec{z} \) are indicated in a cube shown in the figure. The cube edge length is \( 1 \). Find the dot products of: \[ \vec{v}\cdot\vec{z}\text{ ,}\ \ \vec{u}\cdot\vec{v} \text{ ,}\ \ \vec{w}\cdot\vec{u}\]
\( \vec{v}\cdot\vec{z}=1 \), \( \vec{u}\cdot\vec{v}=0 \), \( \vec{w}\cdot\vec{u}=1 \)
\( \vec{v}\cdot\vec{z}=\frac{\sqrt2}2 \), \( \vec{u}\cdot\vec{v}=1 \), \( \vec{w}\cdot\vec{u}=\sqrt3 \)
\( \vec{v}\cdot\vec{z}=\sqrt2 \), \( \vec{u}\cdot\vec{v}=0 \), \( \vec{w}\cdot\vec{u}=1 \)
\( \vec{v}\cdot\vec{z}=1 \), \( \vec{u}\cdot\vec{v}=1 \), \( \vec{w}\cdot\vec{u}=\sqrt3 \)

1103020804

Level: 
A
In the parallelogram \( ABCD \) shown in the picture, \( G \) is the midpoint of \( CD \), \( F \) is the midpoint of \( BC \) and \( \vec{u}=\overrightarrow{CG} \), \( \vec{v}=\overrightarrow{CF} \), \( \vec{a}=\overrightarrow{AD} \) and \( \vec{b}=\overrightarrow{AC} \). Express vectors \( \vec{a} \) and \( \vec{b} \) as a linear combination of vectors \( \vec{u} \) and \( \vec{v} \).
\( \vec{a}=-2\vec{v};\ \vec{b}=-2\vec{u}-2\vec{v} \)
\( \vec{a}=\vec{b}+2\vec{u};\ \vec{b}=-2\vec{u}+2\vec{v} \)
\( \vec{a}=\vec{b}-2\vec{u};\ \vec{b}=-\sqrt2\vec{u}-\sqrt2\vec{v} \)
\( \vec{a}=-2\vec{v};\ \vec{b}=2\vec{u}+2\vec{v} \)