Points and vectors

9000108808

Level: 
B
Find the angle between the altitude \(v_{c}\) and side \(b\) in the triangle \(ABC\) for \(A = [1;2]\), \(B = [7;-2]\) and \(C = [6;1]\). Round to the nearest degree. Hint: In geometry, the altitude \(v_{c}\) of the triangle \(ABC\) is the line segment through the vertex \(C\) and perpendicular to the line containing the opposite side of the triangle.
\(68^{\circ }\)
\(75^{\circ }\)
\(44^{\circ }\)
\(61^{\circ }\)

1003020901

Level: 
C
Let there be vectors: \(\overrightarrow{a}=(1;3;-1)\), \(\overrightarrow{b}=(0;3;1)\), \(\overrightarrow{c}=(-1;2;2)\). Find \(\overrightarrow{a}\times\overrightarrow{b}\) and \(\left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{c}\).
\(\overrightarrow{a}\times\overrightarrow{b}=(6;-1;3); \left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{c}=-2\)
\(\overrightarrow{a}\times\overrightarrow{b}=8; \left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{c}=(-8,16,16)\)
\(\overrightarrow{a}\times\overrightarrow{b}=(-6;1;-3); \left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{c}=2\)
\(\overrightarrow{a}\times\overrightarrow{b}=\sqrt{46}; \left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{c}=2\)

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.
$\vec{v}=(-6;6;-6)$
$\vec{v}=(6;-6;6)$
$\vec{v}=(-7;7;-7)$
$\vec{v}=(7;-7;7)$

1003040207

Level: 
C
Given the points $A = [2;0;3]$ and $B = [-1;2;0]$, specify all the points $C$ lying on the $z$-axis, such that the area of the triangle $ABC$ is $2\sqrt2$. Hint: Use a cross product of vectors.
$C_1=[0;0;1];\ C_2=\left[0;0;\frac{29}{13}\right]$
$C_1=[0;0;1];\ C_2=\left[0;0;-1\right]$
$C_1=[0;0;-1];\ C_2=\left[0;0;\frac{13}{29}\right]$
$C_1=[0;0;-1];\ C_2=\left[0;0;\frac{29}{13}\right]$

1103040206

Level: 
C
Given the points $A = [1;5]$ and $B = [-4;2]$, specify all the points $C$ lying on the $x$-axis, such that the area of the triangle $ABC$ is $14$. Hint: Use a cross product of vectors.
$C_1=[2;0];\ C_2=\left[-\frac{50}3;0\right]$
$C_1=[1;0];\ C_2=\left[-\frac{47}3;0\right]$
$C_1=[2;0];\ C_2=\left[-\frac{47}3;0\right]$
$C_1=[1;0];\ C_2=\left[-\frac{50}3;0\right]$

1103040208

Level: 
C
We are given the points $A = [4;5;-1]$, $B = [-2;-1;2]$, $C = [-1;-3;0]$ and $D = [0;m;2]$. Find the missing coordinate of the point $D$ such that the point $D$ lies in the plane determined by the points $A$, $B$ and $C$. Hint: Use a linear combination of vectors shown in the picture or use their mixed product.
$m=3$
$m=-3$
$m=1$
$m$ does not exist