Vectors - Operations and relations

Q1

Consider the vectors

\vec{a} = \left(\begin{array}{ccc} 0 \\ -3 \\ 1\end{array}\right) \quad \text{and}\quad \vec{b}=\left(\begin{array}{ccc} 0 \\ 4 \\ 2\end{array}\right)

and the point $A(1|2|3)$ .

Indicate point $A$ in a 3d-coordinate system.
Draw the position vector of $A$ and determine its components.
Draw vectors $\vec{a}$ and $\vec{b}$ .
Draw the following vectors by construction: $1.5 \vec{a}$ , $-2 \vec{a}$ , $\vec{a}+2\vec{b}$ , $2\vec{a}-\vec{b}$ .
Draw a vector which is:
- equal to $\vec{b}$
- collinear to $\vec{b}$
- orthogonal to $\vec{b}$
Determine the magnitude of $\vec{a}$ and $\vec{b}$ .
Determine the scalar product of $\vec{a}$ and $\vec{b}$ .
Determine the angle between $\vec{a}$ and $\vec{b}$ .
Determine the vector product (or cross product) of $\vec{a}$ and $\vec{b}$ . Show that this vector is orthogonal to $\vec{a}$ und $\vec{b}$ .

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A1

Q2

Consider the vectors

\vec{a}=\left(\begin{array}{ccc} 1 \\ 2 \\ -1\end{array}\right) \quad \text{and}\quad \vec{b}=\left(\begin{array}{ccc} 2 \\ -3 \\ 3\end{array}\right)

and the points $A(1|0|3)$ and $B(0|1|-2)$ .

Determine the components of the following vectors: $1.5 \vec{a}$ , $-2 \vec{a}$ , $\vec{a}+2\vec{b}$ , $2\vec{a}-\vec{b}$ .
Determine the magnitude of the vectors $3 \vec{a}$ , and $\vec{a}+\vec{b}$ .
Point $P(1|2|-3)$ is moved along the vector $\vec{a}$ . What are the coordinates of the moved point?
Determine the angle between $\vec{a}$ and $\vec{b}$ .
Are the vectors $\vec{a}$ and $\vec{b}$ equal, collinear, or orthogonal?
Determine a vector that is collinear to $\vec{a}$ .
Determine a vector that is orthogonal to $\vec{a}$ and $\vec{b}$ .
Is $\vec{a}$ a unit vector? If not, find a unit vector of $\vec{a}$ .
Find all vectors that are parallel to $\vec{b}$ and have length $10$ .
Determine the vector from $A$ to $B$ .
Determine the shortest distance from $A$ to $B$ .

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A2

Q3

Consider the points $A(1|3|2)$ , $B(0,-1,4)$ , and $C(1|1|2)$ , and the vectors

\vec{u}=\left(\begin{array}{ccc} 1 \\ 0 \\ 1\end{array}\right) \quad \text{and}\quad \vec{v}=\left(\begin{array}{ccc} -1 \\ 1 \\ 2\end{array}\right)

Find the direction vector of the straight line passing through the points $A$ and $C$ .
Find another point on the straight line passing through $A$ and has direction $\vec{u}$ .
Find the normal vector of the plane containing the points $A$ , $B$ , and $C$ .
Find two other points on the plane containing the point $A$ and has normal vector $\vec{u}$ .
Is the point $P(0|2|2)$ on the straight line given by $A$ and $B$ ?
Is the point $P(-1|-5|6)$ on the straight line given by $A$ and $B$ ?
Is the point $P\left(\frac{2}{3}\middle|1\middle|\frac{8}{3}\right)$ on the plane given by the points $A$ , $B$ , and $C$ ?
Is the point $P\left(\frac{1}{3}\middle|1\middle|\frac{4}{3}\right)$ on the plane given by the points $A$ , $B$ , and $C$ ?
Find the point $P$ which is in the middle of the segment from $B$ to $C$ .
SKIP Determine the centre of gravity of the triangle $ABC$ .
Determine the angle at $A$ of the triangle $ABC$ .
Determine the circumference and area of the triangle $ABC$ .

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A3

Q4

Consider the three points $A(0|1|0), B(4|-2|1)$ and $C(-2|3|2)$ .

The straight line $g$ passes through $A$ and $B$ . Determine the equation of the straight line.
Use the equation to show that $C$ is not on $g$ .
Find the normal equation of the plane $E$ which contains the points $A$ , $B$ and $C$ .
Use the normal equation to show that point $P(1|1|1)$ is not in the plane.

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A4

$g$ passes through $A(0|1|0)$ and has a direction vector $\vec v = \overrightarrow{AB}=\left(\begin{array}{ccc} 4 \\ -3 \\ 1\end{array}\right)$ Thus, the equation of the straight line is $\underline{\left(\begin{array}{ccc} x \\ y \\ z\end{array}\right)=\left(\begin{array}{ccc} 0 \\ 1 \\ 0\end{array}\right)+c\cdot \left(\begin{array}{ccc} 4 \\ -3 \\ 1\end{array}\right)}$ That is, any point $P(x|y|z)$ is on $g$ if there is a $c$ such that the equation above is satisfied.
Is there a $c$ with $\left(\begin{array}{ccc} -2 \\ 3 \\ 2\end{array}\right)=\left(\begin{array}{ccc} 0 \\ 1 \\ 0\end{array}\right)+c\cdot \left(\begin{array}{ccc} 4 \\ -3 \\ 1\end{array}\right) ?$ That is, is there a $c$ with $\begin{array}{rll} -2 &=&0+4c\\ 3 &=&1-3c\\ 2 &=&0+c\\ \end{array}$ Clearly not, so $\underline{C\not\in g}$ .
A normal vector of $E$ is $\vec n = \overrightarrow{AB} \times \overrightarrow{AC}= \left(\begin{array}{ccc} -8 \\ -10 \\ 2\end{array}\right)$ Thus, the normal equation is $-8x-10y+2z=d$ To find $d$ , insert any point into the equation which is in $E$ , e.g. $A(0|1|0)$ . We get $-8\cdot 0-10\cdot 1+2\cdot 0 = d \rightarrow d=-10$ Or what about inserting $B(4|-2|1)$ : $-8\cdot 4-10\cdot (-2)+2\cdot 1 = d \rightarrow d=-10$ Or $D(-2|3|2)$ $-8\cdot (-2)-10\cdot 3+2\cdot 2 = d \rightarrow d=-10$ In fact, any point $(x|y|z)$ in $E$ satisfies the normal equation $\underline{-8x-10y+2z=-10}$
Use the normal equation to show that point $P(1|1|1)$ is not in the plane, $\underline{P\not\in E}$ , because if we insert the coordinates into the normal equation we get not $6$ : $-8\cdot 1-10\cdot 1+2\cdot 1=-16 \neq -10$