Further Problems 2

Exercise 1

Calculate the angle between $\vec{a}=\left(\begin{array}{r} 4 \\ -2\\ 1 \end{array}\right)$ and the three coordinate axes.
Find a value for $z$ such that the angle between $\vec{a}=\left(\begin{array}{r} 1 \\ 0\\ z \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} 0 \\ 1\\ z \end{array}\right)$ is $60^\circ$ .
Find possible values for $u$ such that $\vec{a}=\left(\begin{array}{r} 2 \\ -1\\ 7 \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} 5 \\ 3\\ u \end{array}\right)$ are orthogonal.
Consider the points $A(4\vert 2\vert 8)$ and $B(6\vert 5\vert -3)$ . Find a point $P$ on the y-axis such that the vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$ are orthogonal.
Consider the triangle $ABC$ where $A(2\vert -3 \vert 4)$ and $B(7\vert 9 \vert 6)$ . The vertex $C$ is on the straight line that passes through the points $P(-1\vert 1\vert 4)$ and $Q(-1\vert 1 \vert 5)$ . Calculate the coordinates of $C$ (and also the triangle area) such that
1. $ABC$ has a right angle at $C$ .
2. $ABC$ is an isosceles triangle.
Determine if the point $P(1 \vert -1 \vert3)$ is on the straight line $g$ , where
1. $g$ passes through the points $A(-1\vert 1\vert 2)$ and $B(5\vert -5\vert 5)$ .
2. $g$ passes through the point $A(2\vert -2\vert 2)$ and has direction vector $\vec{v}=\left(\begin{array}{r} 3 \\ -3\\ 3 \end{array}\right)$ .
3. $g$ passes through the point $A(1\vert 0 \vert 1)$ and is parallel to the straight line $h$ , where $h$ passes through the points $C(1\vert 1\vert 1)$ and $D(1\vert -1\vert 5)$ .
Determine the point of intersection and the angle of intersection between the straight lines $l_1$ and $l_2$ , where $l_1$ passes through the points $A_1(-3\vert 5\vert -2)$ and $B_1(0\vert 0\vert 7)$ , and $l_2$ passes through the points $A_2(-16\vert -6\vert 9)$ and $B_2(-8\vert -3\vert 8)$ .
Consider a straight line $g$ that passes through the points $A(5\vert 9\vert u)$ and $B(2\vert 6\vert -8)$ . Find a value for $u$ such that $g$ intersects the x-axis.
Mirror the point $P(5\vert 2\vert -1)$ about the straight line $g$ that passes through the points $A(2\vert -3\vert 6)$ and $B(6\vert 1\vert 8)$ .
Determine if the point $P(2\vert 2\vert 1)$ is on plane $E$ , where
1. $E$ contains the points $A(3\vert 1\vert 0)$ , $B(1\vert -4\vert -2)$ , and $C(10\vert 1\vert -3)$ .
2. $E$ contains point $A(0\vert 4\vert 3)$ and is orthogonal to the straight line $g$ , where $g$ passes through the points $U(-1\vert 6\vert -5)$ and $V(5\vert -2\vert 7)$ .
Determine the relative position between the straight line $g$ and plane $E$ , where $g$ passes through the points $A(2\vert 1\vert 0)$ and $B(-2\vert -2\vert -1)$ , and $E$ has the normal equation
$x-2y+2z=11$
The straight line $g$ passes through the points $A(0\vert 0\vert 5)$ and $B(2\vert 2\vert 1)$ . Plane $E$ contains the points $U(6\vert 0\vert 0), V(0\vert 2\vert 4)$ and $W(-1\vert -1\vert 3)$ . If it exists, determine the point of intersection between $g$ and $E$ , and also the angle of intersection (smallest angle between $g$ and $E$ ).
A plane $E$ contains the point $A(0\vert 3\vert 0)$ and has the normal vector $\vec{n}=\left(\begin{array}{r} 2 \\ 3\\ -8 \end{array}\right)$ . Find the smallest distance between the origin and $E$ .
The straight line $g$ has the following equation:
$\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 0 \\ 3\\ 0 \end{array}\right)+c\cdot \left(\begin{array}{ccc} 2 \\ 3\\ -8 \end{array}\right)$
Find the closest distance between point $P(1\vert -1\vert -1)$ and $g$ .
A plane $E$ passes through the point $A(0\vert 5\vert 0)$ and has the normal vector
$\vec{n}=\left(\begin{array}{r} 4 \\ -3\\ 0 \end{array}\right)$
Determine the smallest distance between point $P(4\vert 22\vert 3)$ and $E$ .
A plane $E$ passes through the points $A(2\vert 1\vert 1), B(-1\vert 3\vert 7)$ and $C(1\vert 2\vert 2)$ . Determine the trace points of $E$ , that is, the intersection points of the plane with the three coordinate axes. Use these points to indicate the position of the plane in a 3d-drawing.
Plane $E$ passes through the point $A(5\vert 0\vert 5)$ and has the normal vector
$\vec{n}=\left(\begin{array}{r} 8 \\ 4\\ 1 \end{array}\right)$
The point $P(-3\vert -3\vert 0)$ is reflected about $E$ (so that it is on the other side of the plane). Determine the coordinates of the reflected point $P^\prime$ .
An oblique (or a skew) pyramid has a triangular base $A(3\vert 0\vert 0)$ , $B(0\vert 6\vert 0)$ and $C(0\vert 0\vert 9)$ . The apex (or vertex, "Spitze") is at $P(13\vert 8\vert 7)$ . Determine the volume of the pyramid.
A right pyramid of height $h=9$ has a square base with corner points $A(3\vert 5\vert 5)$ , $B(1\vert 1\vert 1)$ , $C(5\vert 3\vert -3)$ and $D$ . Determine the coordinates of the apex $P$ .
An air traffic controller is sitting at the origin of a coordinate system. At $t=0\, min$ she observes an aeroplane $P_1$ at point $A(2\vert 1\vert 1)$ ( $km$ ). It is travelling in a straight line at a constant speed ( $km/min$ ) of
$\vec{v}=\left(\begin{array}{r} 2 \\ 2\\ 1 \end{array}\right)$
1. Determine the position of $P_1$ three minutes later.
2. At what time was $P_1$ closest to the origin?
3. The monitored space stretches from the origin over $50 km$ in all directions. When will the plane leave this region.
4. At time $t=0 min$ , when $P_1$ is at $A$ , a second plane $P_2$ is at $B(1\vert 2\vert 5)$ (in $km$ ). Three minutes later $P_2$ is found at $C(4\vert 5\vert 2)$ (in $km$ ). Is it possible that the two aeroplanes can crash into each other?
Show that the two straight lines $g$ and $h$ with the equations
$g:\quad\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 1 \\ -3\\ 5 \end{array}\right)+c\cdot \left(\begin{array}{ccc} 3 \\ -4\\ 2 \end{array}\right)$
and
$h:\quad\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 4 \\ 2\\ 9 \end{array}\right)+d\cdot \left(\begin{array}{ccc} -6 \\ 8\\ -4 \end{array}\right)$
are parallel. Then determine the shortest distance between them.
Show that the two planes $E$ and $F$ with the equations
$E:\quad 2x-4y+9z=12$
and
$F:\quad -3x+6y-13.5z=6$
are parallel. Then determine the shortest distance between them.

Solution

The $x$ -axis has direction vector $\vec{x}=\left(\begin{array}{ccc} 1 \\ 0\\ 0 \end{array}\right)$ thus the angle $\alpha$ between the $x$ -axis and the $\vec{a}$ is $\cos(\alpha)=\frac{\vec{a}\bullet\vec{x}}{\vec{a}\cdot \vec{x}}=\frac{4}{\sqrt{21}\sqrt{1}}=0.8728$ Thus, $\alpha=\arccos(0.8728)=29.2^\circ$ . A similar calculation shows that the angle for the $y$ -axis is $115.8^\circ$ , for the $z$ -axis it is $77.4^\circ$ .
The angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$ , thus we have $\underbrace{\cos(60^\circ)}_{=0.5}=\frac{\vec{a}\bullet\vec{b}}{\vec{a}\cdot \vec{b}}=\frac{z^2}{\sqrt{1+z^2}\sqrt{1+z^2}}=\frac{z^2}{1+z^2}$ It follows $z^2 = 0.5(1+z^2)=0.5+0.5z^2 \rightarrow z^2=1$ and therefore $z_1=1, z_2=-1$ .
Find $u$ such that $\vec{a}\bullet\vec{b}= 7+7u=0$ . It follows $u=-1$ .
$P$ on the $y$ -axis means, that $P(0|c|0)$ for some value $y$ . Of course, you can also use view the $y$ -axis as a straight line with the line equation $\vec{P}=\vec{O}+c\vec{v}$ where $O=(0|0|0)$ is a point on the line and $\vec{v}=\left(\begin{array}{ccc} 0 \\ 1\\ 0 \end{array}\right)$ is the direction vector of this line. So, find $c$ such that $\overrightarrow{PA}\bullet \overrightarrow{BP}=\left(\begin{array}{ccc} 4 \\ 2-c\\ 8 \end{array}\right)\bullet \left(\begin{array}{ccc} 6 \\ 5-c\\ -3 \end{array}\right) = (2-c)(5-c)=0$ It follows $c_1=2$ and $c_2=5$ , and thus $P_1(0\vert 2\vert 0), P_2(0\vert 5\vert 0)$
As $C(x|y|z)$ is on the line passing through $P$ and $Q$ , there is a scalar $c$ such that $\left(\begin{array}{ccc} x \\ x\\ z \end{array}\right) = \left(\begin{array}{ccc} -1 \\ 1\\ 4 \end{array}\right) + c\left(\begin{array}{ccc} 0 \\ 0\\ 1 \end{array}\right)$ and it follows $\begin{array}{ccc} x & = & -1\\ y & = & 1\\ z & = & c+4 \end{array}$ After calculating $C$ , the triangle area can then be calculated using the vector product, that is, $A=\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BC}|$
1. Right angle at $C$ means $\overrightarrow{AC}\bullet\overrightarrow{BC}=\left(\begin{array}{ccc} -3 \\ 4\\ c \end{array}\right)\bullet \left(\begin{array}{ccc} -8 \\ -8\\ c-2 \end{array}\right)=-8+c^2-2c=0$ Using the midnight formula, we get $c_1=-2$ and $c_2=4$ , thus $C_1(-1\vert 1\vert 8)$ (with area $36.78$ ) or $C_2(-1\vert 1 \vert 2)$ (with area $32.31$ ).
2. Find $C$ such that the sides $AC$ and $BC$ have the same length (as it is an isosceles triangle). So find $c$ with $|\overrightarrow{AC}|=|\overrightarrow{BC}|$ So $\sqrt{9+16+c^2}=\sqrt{64+64+(c-2)^2}$ If we square both sides to eliminate the root, we ultimately arrive at the linear equation $4c=107 \rightarrow c=26.75$ $C(-1\vert 1\vert 30.75)$ and the area is $173.6$ .
We have
1. Check if there is a value $c$ with $\overrightarrow{AP} = c\cdot\overrightarrow{AB}$ (collinear). There is, so $P \in g$ .
2. Check if there is a value $c$ with $\overrightarrow{AP}=c\cdot\vec{v}$ (collinear). There is not, so $P \not\in g$ .
3. Check if $\overrightarrow{AP}=c\cdot\overrightarrow{CD}$ (collinear). There is, so $P\in g$ .
Let $S(x|y|z)$ be the point of intersection between the lines. As $S$ is on both lines, we have for $l_1$ that $\overrightarrow{A_1S}=c\cdot \overrightarrow{A_1 B_1} \rightarrow \left(\begin{array}{ccc} x+3 \\ y-5 \\ z+2\end{array}\right) = c\left(\begin{array}{c} 3 \\ -5 \\ 9\end{array}\right) \rightarrow \begin{array}{ccr} x & = & 3c-3\\ y & =& -5c+5 \\ z & =& 9c-2\end{array}$ and for $l_2$ $\overrightarrow{A_2S}=d\cdot \overrightarrow{A_2 B_2} \rightarrow \left(\begin{array}{ccc} x+16 \\ y+6 \\ z-9\end{array}\right) = d\left(\begin{array}{c} 8 \\ 3 \\ -1\end{array}\right) \rightarrow \begin{array}{ccr} x & = & 8d-16\\ y & =& 3d-6 \\ z & =& -d+9\end{array}$ Thus, we have to solve the system of linear equations $\left|\begin{array}{rcl} 3c-3 &=& 8d-16\\ -5c+5 &=& 3d-6\\ 9c-2 &=& -d+9\end{array}\right|$ It follows $c=1$ and $d=2$ . We get $S_c(0\vert 0\vert 7)$ and $S_d(0\vert 0\vert 7)$ , that is, the two points are the same and the point of intersection exists. The angle of intersection can be calculated with the point of intersection, it is simply the angle between the two direction vectors of $l_1$ and $l_2$ . We have $\cos(\alpha)=\frac{\overrightarrow{A_1B_1}\bullet \overrightarrow{A_2B_2}}{|\overrightarrow{A_1B_1}|\cdot |\overrightarrow{A_2B_2}|}=0\rightarrow \alpha=90^\circ$
Let $S(x|y|z)$ be the intersection point between $g$ and the $x$ -axis. As $S$ is also on $g$ , there must be a value $c$ with $\overrightarrow{AS}=d\cdot \overrightarrow{AB} \rightarrow \left(\begin{array}{ccc} x-5 \\ y-9 \\ z-u\end{array}\right) = c\left(\begin{array}{c} -3 \\ -3 \\ -8-u\end{array}\right) \rightarrow \begin{array}{ccr} x & = & -3c+5\\ y & =& -3c+9 \\ z & =& (-8-u)c+u\end{array}$ Now, because $S$ is also on the $x$ -axis, we have $y=0$ and $z=0$ . So we see that from the equation in the middle of the system above, it follows $c=3$ and from the last equation we have $(-8-u)\cdot 3+u=0 \rightarrow u=-12$
We need to find the point $S(x|y|z)$ on $g$ such that $\overrightarrow{PS} \perp \overrightarrow{AB}$ . The reflected point is then $S^\prime=P+\overrightarrow{PS}$ . Let's start with $S$ being on $g$ . It follows that $\overrightarrow{AS}=c\cdot \overrightarrow{AB} \rightarrow \left(\begin{array}{ccc} x-2 \\ y+3 \\ z-6\end{array}\right) = c\cdot \left(\begin{array}{c} 4 \\ 4 \\ 2\end{array}\right) \rightarrow \begin{array}{ccr} x & = & 4c+2\\ y & =& 4c-3 \\ z & =& 2c+6\end{array}$ and because of $\overrightarrow{PS} \perp \overrightarrow{AB}$ if must be $\left(\begin{array}{ccc} x-5 \\ y-2 \\ z+1\end{array}\right)\bullet \left(\begin{array}{ccc} 4 \\ 4 \\ 2\end{array}\right) = 4(x-5)+4(y-2)+2(z+1)=0$ Inserting the values for $x, y$ and $z$ into this equation, we get the equation $4(4c-3)+4(4c-5)+2(2c+7)=0 \rightarrow c=0.5$ and thus $S(4|-1|7)$ . With $S^\prime=S+\overrightarrow{PS}$ it follows $S^\prime(3\vert -4\vert 15)$ .
To specify the plane we need a point in $E$ and a normal vector of $E$ .
1. $\vec{n}=\overrightarrow{AB} \perp \overrightarrow{AC}=\left(\begin{array}{ccc} 15 \\ -20 \\ 35\end{array}\right)$ . To check if $P$ is on $E$ , we can either write down the normal equation and see if $P$ fulfils this equation, or we can check if $\overrightarrow{AP}\perp \vec{n}$ . In both cases do we get that $P\in E$ .
2. As $g$ is orthogonal to $E$ , a normal vector $\vec{n}$ of $E$ is given by the direction vector of $g$ , that is, we have $\vec{n}=\overrightarrow{UV}$ . We then find that $P \not\in E$ .
The possible relative positions are that $g\parallel E$ ( $g$ not in the plane of $E$ ), that $g \in E$ (that is, $g$ is in $E$ , or in other words, that $g\parallel E$ and at least one point of $g$ is $E$ ), or that $g$ and $E$ intersect. Let's first check if $g$ and $E$ are parallel, which is the case if $\vec{v}\perp \vec{n}$ where $\vec{v}=\overrightarrow{AB}=\left(\begin{array}{ccc} -4 \\ -3 \\ -1\end{array}\right)$ is the direction vector of $g$ and $\vec{n}=\overrightarrow{AB}=\left(\begin{array}{ccc} 1 \\ -2 \\ 2\end{array}\right)$ is the normal vector of $E$ an can be obtained from the normal equation. As the scalar product $\vec{v}\bullet \vec{n}=0$ it follows that $g$ and $E$ are parallel. To check if $g$ is in $E$ , we check if $A$ in in $g$ by inserting the coordinates of $A$ into the normal equation: $2-2\cdot 1+2\cdot 0=0\neq 11$ and we see that $A$ and therefore $g$ is not in $E$ .
The line $g$ has the direction vector $\vec{v}=\overrightarrow{AB}=\left(\begin{array}{ccc} 2 \\ 2 \\ -4\end{array}\right)$ Let's take the direction vector $\vec{v}=\left(\begin{array}{ccc} 1 \\ 1 \\ -2\end{array}\right)$ Plane $E$ has the normal vector $\vec{n}=\overrightarrow{UV}\times \overrightarrow{UW}=\left(\begin{array}{ccc} 10 \\ -10 \\ 20\end{array}\right)$ Let's take the normal vector $\vec{n}=\left(\begin{array}{ccc} 1 \\ -1 \\ 2\end{array}\right)$ As $\vec{v}$ and $\vec{n}$ are not orthogonal (scalar product different from $0$ ), they are not parallel to each other and thus a point of intersection $S(x|y|z)$ will exist. As $S$ is on $g$ , it follows $\overrightarrow{AS}=c\cdot \overrightarrow{AB} \rightarrow \left(\begin{array}{ccc} x \\ y \\ z-5\end{array}\right) = c\cdot \left(\begin{array}{c} 1 \\ 1 \\ -2\end{array}\right) \rightarrow \begin{array}{ccl} x & = & c\\ y & =& c \\ z & =& -2c+5\end{array}$ (which we could have gotten directly form the line equation $\vec{S}=\vec{A}+c\vec{v}$ ). As $S$ is on $E$ we also have $\vec{n} \bullet \overrightarrow{US}=0 \rightarrow 1\cdot (x-6)+(-1)\cdot y+2\cdot z=0 \rightarrow x-y+2z=6$ (this is, by the way, the normal equation of the plane, which we could have gotten directly ...). Inserting the $x, y$ and $z$ from the line equation into the normal equation, we get $c-c+2(-2c+5)=6 \rightarrow c=1$ and thus $S(1\vert 1\vert 3)$ . To find the angle of intersection, we determine the angle between $\vec{v}$ and $\vec{n}$ , which we can derive using the formula $\cos(\alpha)=\frac{\vec{v}\bullet\vec{n}}{|\vec{v}|\cdot |\vec{n}|}=131.8^\circ$ We normally give the smaller angle, so this would be $180^\circ-\alpha=48.2^\circ$ . Note that this is the angle between $\vec{v}$ and $\vec{n}$ . To find the angle between the line and the plane, we have to subtract this angle form $90^\circ$ (make a figure to see this!). So the angle is $41.81^\circ$ .
We need to find the point of intersection $S$ between $E$ and the line $g$ which is orthogonal to $E$ and passes through the origin $O(0|0|0)$ . The smallest distance $d$ is then $d=|\overrightarrow{OS}|$ . As $g$ is orthogonal to $E$ , its direction vector is simply the normal vector of $E$ . Thus, as $S(x|y|z)$ is on $g$ , we have $\overrightarrow{OS}=c\cdot \vec{n} \rightarrow \left(\begin{array}{ccc} x \\ y \\ z\end{array}\right) = c\cdot \left(\begin{array}{c} 2 \\ 3 \\ -8\end{array}\right) \rightarrow \begin{array}{ccl} x & = & 2c\\ y & =& 3c \\ z & =& -8c\end{array}$ (line equation). And as $S$ is also in the plane, we have $\vec{n} \bullet \overrightarrow{AS}=0 \rightarrow 2x+3(y-3)-8z=0 \rightarrow 2x+3y-8z=9$ (normal equation of plane). Inserting the $x, y$ and $z$ from the line equation into the normal equation, we get $2(2c)+3(3c)-8(-8c)=9 \rightarrow c=\frac{9}{77}$ and thus $S(\frac{18}{77}\vert \frac{27}{77}\vert -\frac{72}{77})$ . Thus, $d=\frac{9}{\sqrt{77}}$ .
We need to find the point $S$ on $g$ with $\overrightarrow{PS}\perp \vec{v}$ , where $\left(\begin{array}{ccc} 2 \\ 3 \\ -8\end{array}\right)$ is the direction vector of $g$ . The smallest distance $d$ is then $d=|\overrightarrow{PS}|$ . As $S(x|y|z)$ is on $g$ , we have $\left(\begin{array}{ccc} x \\ y \\ z\end{array}\right) = \left(\begin{array}{ccc} 0 \\ 3 \\ 0\end{array}\right) + c\cdot \left(\begin{array}{c} 2 \\ 3 \\ -8\end{array}\right) \rightarrow \begin{array}{ccl} x & = & 2c\\ y & =& 3c+3 \\ z & =& -8c\end{array}$ And from $\overrightarrow{PS}\bullet \vec{v}=0$ follows $2(x-1)+3(y+1)-8(z+1)=0 \rightarrow 2x+3y-8z=7$ Thus, we get the equation $2(2c)+3(3c+3)-8(-8c)=7 \rightarrow c=-\frac{2}{77}$ Thus, $S(-\frac{4}{77}|\frac{225}{77}|\frac{16}{77})$ and therefore $d=|\overrightarrow{PS}|=4.23$
We need to find the point of intersection $S$ between $E$ and the line $g$ which is orthogonal to $E$ and passes through $P$ . The smallest distance $d$ is then $d=|\overrightarrow{PS}|$ . As $g$ is orthogonal to $E$ , its direction vector is simply the normal vector of $E$ . Thus, as $S(x|y|z)$ is on $g$ , we have $\overrightarrow{PS}=c\cdot \vec{n} \rightarrow \left(\begin{array}{ccc} x-4 \\ y-22 \\ z-3\end{array}\right) = c\cdot \left(\begin{array}{c} 4 \\ -3 \\ 0\end{array}\right) \rightarrow \begin{array}{ccl} x & = & 4c+4\\ y & =& -3c+22 \\ z & =& 3\end{array}$ (line equation). And as $S$ is also in the plane, we have $\vec{n} \bullet \overrightarrow{AS}=0 \rightarrow 4x-3(y-5)+0z=0 \rightarrow 4x-3y+0z=-15$ (normal equation of plane). Inserting the $x, y$ and $z$ from the line equation into the normal equation, we get $4(4c+4)-3(-3c+22)=-15 \rightarrow c=1.4$ and thus $S(9.6|17.8|3)$ . Thus, $d=|\overrightarrow{PS}|=7$ .
The normal vector of the plane is $\vec{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\left(\begin{array}{ccc} -4 \\ -3 \\ -1\end{array}\right)$ Let's take the normal vector $\vec{n}=\left(\begin{array}{ccc} 4 \\ 3 \\ 1\end{array}\right)$ Thus, the normal equation of the plane is therefore $4x+3y+z=d$ where $d$ is obtained by inserting a point on the plane, e.g. $A$ . Thus we get $d=4\cdot 2+3\cdot 1+1\cdot 1=12$ Thus the complete normal equation is $4x+3y+z=12$ The trace point on the $x$ -axis has the coordinates $S(x|0|0)$ for some $x$ . As this point is also in the plane, if must fulfil the normal equation: $4x+3\cdot 0+0=12 \rightarrow x=3$ Similar, the trace point on the $y$ -axis has the coordinates $S(0|y|0)$ from which follows $y=4$ , and for the trace point $S(0|0|z)$ on the $z$ -axis we get $z=12$ .
We need to find the point of intersection $S$ between $E$ and the line $g$ which is orthogonal to $E$ and passes through $P$ . The reflected point $P^\prime$ is then given by $P^\prime=S+\overrightarrow{PS}$ . As $g$ is orthogonal to $E$ , its direction vector is simply the normal vector of $E$ . Thus, as $S(x|y|z)$ is on $g$ , we have $\overrightarrow{PS}=c\cdot \vec{n} \rightarrow \left(\begin{array}{ccc} x+3 \\ y+3 \\ z\end{array}\right) = c\cdot \left(\begin{array}{c} 8 \\ 4 \\ 1\end{array}\right) \rightarrow \begin{array}{ccl} x & = & 8c-3\\ y & =& 4c-3 \\ z & =& c\end{array}$ (line equation). And as $S$ is also in the plane, we have $\vec{n} \bullet \overrightarrow{AS}=0 \rightarrow 8(x-5)+4y+1(z-5)=0 \rightarrow 8x+4y+z=45$ (normal equation of plane). Inserting the $x, y$ and $z$ from the line equation into the normal equation, we get $8(8c-3)+4(4c-3)+c=45 \rightarrow c=1$ and thus $S(5|1|1)$ . With $P^\prime=S+\overrightarrow{PS}$ we get $P^\prime(13\vert 5\vert 2)$ .
The volume is given by $V=\frac{hb}{3}$ , where $b$ is the area of the base and $h$ is the height of the pyramid. Let's start with the area of the base, which can be calculated as follows: $b=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\left|\left(\begin{array}{ccc} 54 \\ 27 \\ 18\end{array}\right)\right|=31.5$ Note that $\overrightarrow{AB}\times\overrightarrow{AC}$ is also a normal vector of the plane containing the triangle $ABC$ . Or, lets take the (collinear) normal vector $\vec{n}=\left(\begin{array}{ccc} 6 \\ 3 \\ 2\end{array}\right)$ We will need $\vec{n}$ in a moment. To find the height, we need to find the point of intersection $S$ between the plane $E$ (containing the base $B$ ) and the line $g$ which is orthogonal to $E$ and passes through $P$ . The height is then the smallest distance $h=|\overrightarrow{PS}|$ . As $g$ is orthogonal to $E$ , its direction vector is simply the normal vector of $E$ . Thus, as $S(x|y|z)$ is on $g$ , we have $\overrightarrow{PS}=c\cdot \vec{n} \rightarrow \left(\begin{array}{ccc} x-13 \\ y-8 \\ z-7\end{array}\right) = c\cdot \left(\begin{array}{c} 6 \\ 3 \\ 2\end{array}\right) \rightarrow \begin{array}{ccl} x & = & 6c+13\\ y & =& 3c+8 \\ z & =& 2c+7\end{array}$ (line equation). And as $S$ is also in the plane, we have $\vec{n} \bullet \overrightarrow{AS}=0 \rightarrow 6(x-3)+3y+2z=0 \rightarrow 6x+3y+2z=18$ (normal equation of plane). Inserting the $x, y$ and $z$ from the line equation into the normal equation, we get $6(6c+13)+3(3c+8)+2(2c+7)=18 \rightarrow c=-2$ and thus $S(1|2|3)$ . Thus, $h=|\overrightarrow{PS}|=14$ . Thus, the volume is $V=147$ .
As it is a right pyramid, the vertex or apex $P$ is "directly above" the middle point $M$ of the square. So let's find $M$ first. As $AC$ is a diagonal of the square, and $M$ is in the middle of this diagonal, we have $M=A+\frac{1}{2}\overrightarrow{AC} \rightarrow M(4|4|1)$ Thus, the vertex $P$ of the pyramid is at $M+c\vec{n}$ , where $\vec{n} = \overrightarrow{AB}\cdot\overrightarrow{AC} = \left(\begin{array}{c} 24 \\ -24 \\ 12\end{array}\right)$ is a vector that is orthogonal to the plane containing the square $ABCD$ , and the value $c$ has to be chosen such that the length of $c\cdot\vec{n}$ (the height) is $9$ . As $|\vec{n}=36|$ , it follows that $c=\frac{9}{36}=0.25$ . It follows from $P=M+c\vec{n}$ that $P_1(10\vert -2\vert 4)$ , or if we subtract, $P=M-c\vec{n}$ , we get $P_2(-2\vert 10\vert -2)$ .
We have
1. $U(8\vert 7\vert 4)$
2. $\Delta t=-\frac{7}{9} min$
3. $\Delta t=15.88 min$
4. the lines are skew, so it is not possible for the planes to collide!
The lines $g$ and $h$ are parallel because their direction vectors are collinear. To find the smallest distance between the two lines, choose a point on $g$ (e.g. $A(1|-3|5)$ ) and determine the smallest distance between this point and $h$ . So find the point $S(x|y|z)$ on $h$ such that $\vec{AS}\perp\vec{v}$ , where $B(4|2|9)$ is a point on $h$ and $v$ is a direction vector of $h$ . Now, because $S$ is on $h$ , there is a scaling factor $d$ with $h:\quad\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 4 \\ 2\\ 9 \end{array}\right)+d\left(\begin{array}{ccc} -6 \\ 8\\ -4 \end{array}\right)$ and it follows $\begin{array}{ccr} x & = & -6d+4\\ y & =& 8d+2 \\ z & =& -4d+9\end{array}$ And as $\vec{AS}\perp\vec{v}$ , it follows $-6(x-1)+8(y+3)-4(z-5) =0$ and we get the following equation for $d$ : $-6(-6d+4-1)+8(8d+2+3)-4(-4d+9-5) =0 \rightarrow d=-0.0517$ Thus, $S(4.31|1.59|9.21)$ and the smallest distance is $|\vec{AS}|=7.05$ .
The planes $E$ and $F$ are parallel because their normal vectors are collinear. To find the smallest distance between the two planes, pick a point on plane $E$ and find the smallest distance between this point and plane $F$ . Let's find a point $A(x|y|z)$ in $E$ : its coordinates must fulfil the normal equation $2x-4y+9z=12$ e.g., $x=6, y=0, z=0$ , thus it is $A(6|0|0)$ . To find the smallest distance between $A$ and $F$ , we intersect the straight line $g$ with $F$ , where $g$ passes through the point $A$ and has direction $\vec{n}=\left(\begin{array}{ccc} -3 \\ 6\\ -13.5 \end{array}\right)$ which is the normal vector of plane $F$ . Let's denote the intersection point with $S(x|y|z)$ . As $S$ is on $g$ , there must be a stretching factor $c$ with $\overrightarrow{AS}=c \vec{n} \rightarrow \left(\begin{array}{ccc} x-6 \\ y \\ z\end{array}\right) = c\left(\begin{array}{c} -3 \\ 6 \\ -13.5\end{array}\right) \rightarrow \begin{array}{ccr} x & = & -3c+6\\ y & =& 6c \\ z & =& -13.5c\end{array}$ As $S$ is on $F$ , $S$ must fulfil the normal equation of $F$ : $-3x+6y-13.5z=6 \rightarrow -3(-3c+6)+6(6c)-13.5(-13.5c)=6\rightarrow c=-0.105$ Thus we get $S(5.68|-0.63|1.43)$ . The smallest distance is $|\vec{AS}|=1.59$ .