C2
(a) Find the distance from the point $(1,0)$ to the line connecting the origin and the point $(0,1)$.
(b) Find the distance from the point $(1,0)$ to the line connecting the origin and the point $(1,1)$.
(c) Find the distance from the point $(1,0,0)$ to the line connecting the origin and the point $(1,1,1)$.
(a) Answer: 1
(a) Solution: The line connecting the origin and the point $(0,1)$ is the $y$-axis. We can see that the closest point to $(1,0)$ on the $y$-axis is the origin, so the distance from $(1,0)$ to the $y$-axis is 1 .
(b) Answer: $\frac{\sqrt{2}}{2}$
(b) Solution: Let $A=(1,0), B=(1,1)$, and $O$ be the origin. Let $C$ be the closest point on line $B O$ to $A$. We know that $A C$ and $B O$ are perpendicular.

Since $B O$ has slope $1, A C$ must have slope -1 . Thus, the line equation for $A C$ is $x+y=1$, and this intersects line $B O$, given by the equation $y=x$, at the point $\left(\frac{1}{2}, \frac{1}{2}\right)$. The distance $A C$ is thus
$$
\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}-1\right)^{2}}=\frac{\sqrt{2}}{2} \text {. }
$$

Alternatively, note that the three given points are vertices of a unit square. The required distance is the distance between a vertex of a unit square and its centre (the point of intersection of its diagonals), that is a half of a diagonal's length, which is $\frac{\sqrt{2}}{2}$.

One may also notice that the line $O B$ is the bisect of the first quadrant and that we have to find the leg of the right triangle with equal legs, the other angles of $45^{\circ}$ with the hypotenuse equal to one. Then either the equation $2 x^{2}=1$ or $x=\cos 45^{\circ}=\sin 45^{\circ}$ leading immediately Note that triangle $A B O$ is a right triangle with $A B=\sqrt{2}, B O=\sqrt{3}$, and $A O=1$. Thus, we can write the area of triangle $A B O$ in two different ways:
$$
\frac{\sqrt{2}}{2}=\frac{A O \cdot A B}{2}=\frac{B O \cdot A C}{2}=\frac{\sqrt{3}}{2} A C,
$$
and so we get
$$
A C=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3} .
$$
to the correct answer.
(c) Answer: $\frac{\sqrt{6}}{3}$
(c) Solutions: Let $A=(1,0,0), B=(1,1,1)$, and $O$ be the origin. Let $C$ be the point on line $B O$ closest to $A$.
Solution 1.1: We can parameterize the point $C$ with the coordinates $(t, t, t)$; then, lines $A C$ and $B O$ are perpendicular; thus, the dot product
$$
(1,1,1) \cdot(t-1, t, t)=0 .
$$

Solving for $t$, we get that $t=\frac{1}{3}$, and hence $C=\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ and the distance AC is
$$
\sqrt{\left(\frac{1}{3}-1\right)^{2}+\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}}=\frac{\sqrt{6}}{3} \text {. }
$$

Solution 1.2: First, find the projection of the vector OA on the vector OB which is
$$
\mathrm{OC}=\frac{\mathrm{OA} \cdot \mathrm{OB}}{\|\mathrm{OB}\|^{2}} \mathrm{OB}=\frac{1 \cdot 1+0 \cdot 1+0 \cdot 1}{1^{2}+1^{2}+1^{2}}(1,1,1)=\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) .
$$

The end of the projection vector $C(1 / 3,1 / 3,1 / 3)$ is the closest point on the line to $A$, and the distance AC is found as in Solution 1.1.

Solution 2: We require the altitude from point $A$ to line $B O$, hitting $B O$ at $C$.