$1 \times 4 + 2 \times 5 + 3 \times 6 = 32$

$(1,2,3) \cdot {(4,5,6) \over |(4,5,6)|} = {32 \over \sqrt{77}}$

$(1, 0, 2) \cdot (4, 10, -2) = 0$. This means that the vectors are perpendicular.

$(1,2,2) \cdot (1,2,2) = 9$.

$(0,1,2) \times (3,4,5) = \left[ \begin{array}{c} 1 {\small \times} 5 - 2 {\small \times} 4 \\ 2 {\small \times} 3 - 0 {\small \times} 5 \\ 0 {\small \times} 4 - 1 {\small \times} 3 \\ \end{array} \right] = \begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}$

When $v$ is not unit length.

When the projection of $u$ onto the line of $v$ is in the opposite direction to $v$. Equivalently: When the angle between $u$ and $v$ is in $[{\pi \over 2},{3 \pi \over 2}]$.

$w$ is unit length ($|w| = |u| \; |v| \; \sin {\pi \over 2}$) and is perpendicular to both $u$ and $v$. So $\langle u, v, w \rangle$ forms an orthogonal basis.

Find a vector such that its dot product with $(1,2,3,4,5)$ is zero. This can be done by swapping two of the coordinates, negating one of those, and setting the rest to zero. For example: $(2,-1,0,0,0)$.

$(1,2,3)$

Project onto each axis: $({-2 \over \sqrt{2}}, \; 2, \; {4 \over \sqrt{2}} ) = (-\sqrt{2}, \; 2, \; 2 \sqrt{2} )$.

Rotates the vector by $-45^o$ about the $(0,1,0)$ axis.

In 2D, the rotation by $45^o$ is done with the matrix $$\displaystyle \left[\begin{array}{cc} \cos 45 & -\sin 45 \\ \sin 45 & \cos 45 \end{array}\right] = \left[\begin{array}{cc} {1 \over \sqrt{2}} & -{1 \over \sqrt{2}} \\ {1 \over \sqrt{2}} & {1 \over \sqrt{2}} \end{array}\right]$$

Implicitly, this rotation is being done in a 2D $x y$ plane that sits in a 3D space, with the $z$ axis pointing perpendicularly out of the $x y$ plane.

So this is a rotation by $45^o$ around the $z$ axis.  If we were using 3D coordinates, it would be written as $$\left[\begin{array}{ccc} {1 \over \sqrt{2}} & -{1 \over \sqrt{2}} & 0\\ {1 \over \sqrt{2}} & {1 \over \sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right]$$

since that matrix does not change the $z$ coordinate.

To do the same rotation in the $x z$ plane, instead, we need to rotate around the $y$ axis. Swapping the roles of $y$ and $z$ in the matrix above yields $$\left[\begin{array}{ccc} {1 \over \sqrt{2}} & 0 & -{1 \over \sqrt{2}} \\ 0 & 1 & 0 \\ {1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}}  \end{array}\right]$$

All vectors are mutually perpendicular:

$(0,1,0) \cdot ({1 \over \sqrt{2}},0,{-1 \over \sqrt{2}}) = 0$

$(0,1,0) \cdot ({1 \over \sqrt{2}},0,{1 \over \sqrt{2}}) = 0$

$({1 \over \sqrt{2}},0,{-1 \over \sqrt{2}}) \cdot ({1 \over \sqrt{2}},0,{1 \over \sqrt{2}}) = 0$

All vectors are unit length:

$(0,1,0) \cdot (0,1,0) = 1$

$({1 \over \sqrt{2}},0,{1 \over \sqrt{2}}) \cdot ({1 \over \sqrt{2}},0,{1 \over \sqrt{2}}) = 1$

$({1 \over \sqrt{2}},0,{-1 \over \sqrt{2}}) \cdot ({1 \over \sqrt{2}},0,{-1 \over \sqrt{2}}) = 1$

First, find the parallel vector that is the projection of $(1,0,2)$ to $(0,2,1)$:

$u = \left( { (1,0,2) \; \cdot \; (0,2,1) \over (0,2,1) \; \cdot \; (0,2,1) } \right) \; (0,2,1) = ( 0, { 4 \over 5}, {2 \over 5} )$.

Second, get the perpendicular vector by subtracting that parallel vector from $(1,0,2)$:

$v = (1,0,2) - ( 0, { 4 \over 5}, {2 \over 5} ) = (1, { -4 \over 5}, {8 \over 5})$.

Then $(1,0,2) = u + v = ( 0, { 4 \over 5}, {2 \over 5} ) + (1, { -4 \over 5}, {8 \over 5})$.

Note that u and v are perpendicular: $u \cdot v = ( 0, { 4 \over 5}, {2 \over 5} ) \; \cdot \; (1, { -4 \over 5}, {8 \over 5}) = 0$.

First, find vectors $u$ and $v$ along the triangle edges:

$u = (1,1,1) - (0,1,1) = (1,0,0)$

$v = (-2,2,-1) - (0,1,1) = (-2,1,-2)$

Second, take their cross product:

$u \times v = (1,0,0) \times (-2,1,-2) = \left[ \begin{array}{c} (0 {\small \times} -2) - (0 {\small \times} 1) \\ (0 {\small \times} -2) - (1 {\small \times} -2) \\ (1 {\small \times} 1) - (0 {\small \times} -2) \\ \end{array} \right] = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$

Third, normalize the cross product:

$n = {u \times v \over |u \times v|} = { (0,2,1) \over |(0,2,1)| } = (0,{2 \over \sqrt{5}}, {1 \over \sqrt{5}})$

There are two such normal vectors: $n$ and $-n$.