Vectors - Points of intersection and shortest distance

A1

1. Let's first find the normal equation of the $xy$-plane. A possible normal vector of the $xy$-plane is $$\vec{n} =\left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right)$$ and the point $O(0|0|0)$ is in this plane. Thus, the normal equation is $$0x+0y+1z=d$$ and to find $d$ we insert the point $O$, resulting in $d=0$. Thus, the normal equation is $$0x+0y+1z=0$$ (or in short, z=0, which makes sense as every point in the $xy$-plane must have a zero $z$-coordinate). Let $S(x|y|z)$ be the point of intersection. Because $S\in g$ follows there is a constant $c$ such that $$\vec{S}=\vec{A}+c\vec{n}$$ that is $$\left(\begin{array}{c} x \\ y \\ z \end{array}\right) = \left(\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right) +c\cdot \left(\begin{array}{c} 4 \\ 3 \\ 1 \end{array}\right) \rightarrow \begin{array}{rr} x & = & 2+4c \\ y & = & 1+3c \\ z & = & 1 + c \end{array}$$ As $S$ is also in the plane, these $x$, $y$, and $z$ must fulfil the normal equation: $$0(2+4c)+0(1+3c)+1(1+c)=0$$ from which follows $c=-1$, and therefore $\underline{S(-2|-2|0)}$.
2. Let's first find the normal equation of $E$: $$1x-2y+2z=d$$ and inserting $B(11|0|0)$ into the equation we get $d=11$. Thus, the normal equation is $$1x-2y+2z=11$$ Let's now find the intersection of this plane with the $x$-axis, which is a straight line containing the point $O(0|0|0)$ and has direction vector $$\vec{v} =\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right)$$ Let $S(x|y|z)$ be the point of intersection. As $S$ is on the $x$-axis, there must be a $c$ with $$\vec{S}=\vec{O}+c\vec{v} \rightarrow \begin{array}{rr} x & = & c \\ y & = & 0 \\ z & = & 0 \end{array}$$ (which makes sense, as every point on the $x$-axis must have zero $y$- and $z$-coordinates). As $S$ is also in $E$, it must fulfil the normal equation from above: $$1c-2\cdot 0 + 2\cdot 0=11$$ and it follows $c=11$. It follows $\underline{S(11|0|0)}$. Observe that this is just point $B$, which we could have seen already at the beginning: $B$ is on the $x$-axis and also in plane $E$, so it must be the intersection point. Calculating the intersection point with the $y$-axis in a similar way we get $\underline{S(0|-5.5|0)}$, with the $z$-axis we get $\underline{S(0|0|5.5)}$.
3. $g$ is parallel to $E$ if $\vec v \perp \vec u$, which is indeed the case (because of $\vec v \bullet \vec u=0$). To show that $g$ is (or is not) in $E$, take a point on $g$ and show that this point is (or is not) in $E$. For example, take the point $A$, which is on $g$. As $\overrightarrow{BA} \not\perp \vec u$, $A$ is not in $E$, and thus $g$ is not in $E$.

A2

Try to find the point of intersection $S(x|y|z)$ between the two lines. From $S\in g$ it follows that there must be a constant $s$ with $\vec{S}=\vec{A}+s\overrightarrow{AB}$:

and because $S$ is also in $h$, there must be some constant $t$ with $\vec{S}=\vec{C}+t\overrightarrow{CD}$

Thus we have to solve the system of equations

To solve this equation, take the first one and solve for $s$, that is, $s=\frac{7t+1}{10}$. In the second (or third) equation replace the $s$ with the expression in $t$ above. You get an equation in $t$, and you can find a value for $t$. And with this value you can now calculate the value for $s$. We get $t=-0.419$ and $s=-0.193$.

Now we can calculate the coordinates of $S$ using $t$ or $s$. With $s=-0.193$, we get $S_s(-0.93|2.77|0.158)$, and with $t=-0.419$ we get $S_t(-0.93| 2.77|-5.514)$. Note that the two points are not the same. This indicates that the lines do not intersect. And because the lines are also not parallel (there is no $c$ with $\overrightarrow{AB}=c\cdot \overrightarrow{CD}$), they have to be skew.

If $S_t=S_s$, we would conclude that there is an intersection point. This will be the case for A3.

A3

Try to find the point of intersection $S(x|y|z)$ between the two lines. From $S\in g$ it follows that there must be a constant $s$ with $\vec{S}=\vec{A}+s\overrightarrow{AB}$, and from $S\in h$ follows that there must be a constant $t$ with $\vec{S}=\vec{C}+t\overrightarrow{CD}$. We arrive at the system of equations

To solve this, take for example the first equation, which we solve for $t$:

If we replace in the second equation the $t$ with the expression above, we get an equation in $s$ only, and we can solve for $s$. We get $s=-2$. Now we can calculate $t$, which is $t=3$.

We can now calculate $S$ using $t=3$, and we get $S_t(8|-2|3)$. We can also calculate $S$ using $s=-2$, and this time (compared to A2) we also get $S_s(8|-2|3)$. Because we get the same intersection points for $s$ and $g$, we conclude that the lines intersect and the intersection point is $\underline{S(8|-2|3)}$.

For the angle of intersection $\alpha$ we have

and thus $\alpha=\underline{155.16^\circ}$.

Note that we do not need the intersection point $S$ to calculate the angle.

A4

A direction vector of $g$ is

A normal vector of $E$ is

For any point $P \in E$ it is

Alternatively, $P$ has to satisfy the normal equation

We find $d$ by inserting into the equation any known point which is in $E$, e.g. let's insert $U(0|0|-8)$:

Let $P(x|y|z)$ be the intersection point. Because of $P\in g$ it follows

And because of $P\in E$ it follows that

(note that the last equation is simply the normal equation from above).

Inserting $x=-6s+4, y=-6s-2, z=-s+1$ into the equation above, we get $s=-\frac{1}{183}$. Thus, the point of intersection is $\underline{P(4.03 |-1.97 | 1.01)}$.

The intersection angle is the angle between the normal vector of the plane, and the direction vector of the straight line, i.e.

It follows $\alpha=\cos^{-1}(0.798)=\underline{142.9^\circ}$ (or $37.1^\circ$).

A5

1. Denote by $S(x|y|z)$ the point on $g$ which is closest to $P$. This is the case if $\overrightarrow{PS}\perp \overrightarrow{AB}$, that is, if $$-(x-2)-(y+2)+(z-2)=0$$ From $S \in g$ follows $$\begin{array}{lll} x&=&-t+1\\ y&=&-t+1\\ z&=&t\\ \end{array}$$ Inserting $x, y$ and $z$ into the first equation, we get $t=\frac{4}{3}$ and therefore $S(-\frac{1}{3}|-\frac{1}{3}| \frac{4}{3})$. The shortest distance is $$d(P,g)=|\overrightarrow{PS}|=\underline{\sqrt{\frac{26}{3}}}$$
2. A normal vector of $E$ is $$\vec{n}=\left(\begin{array}{ccc} 1 \\ -1 \\ 0\end{array}\right)$$ Let's find the point $S(x|y|z)$ on $E$ which is closest to $P$. We can find $S$ be intersecting $E$ with the straight line $g$ which is orthogonal to $E$ and passes through $P$. Thus, a direction vector of $g$ is $\vec n$ ($\vec v=\vec n$). From $S\in g$ follows $$\begin{array}{lll} x&=&t+2\\ y&=&-t-2\\ z&=&2\\ \end{array}$$ From $S \in E$ follows (because the origin $0$ is in $E$) that $\overrightarrow{0S} \perp \vec n$, that is $$x-y+0z=0$$ (the same equation we get by using the normal equation). Inserting $x=t+2, y=-t-2, z=2$ into the equation above, we get $t=-2$ and therefore $S(0|0|2)$. The shortest distance is $$d(P,E)=|\overrightarrow{PS}|=\underline{\sqrt{8}}$$