1^° Find all scalar products of unit coordinate vectors…

$1^\circ$ Find all scalar products of unit coordinate vectors (orths) in space. $2^\circ$ Find $(\vec{i}-\vec{j}+\vec{k})\cdot(2\vec{i}-3\vec{j}-\vec{k})$.

Explanation of Method

Scalar products of the unit coordinate vectors and an application.

Step 1: Part $1^\circ$

$|\vec i|=|\vec j|=|\vec k|=1$ and the three are pairwise perpendicular. Hence

\[ \vec{i}\cdot\vec{i} = \vec{j}\cdot\vec{j} = \vec{k}\cdot\vec{k} = 1\cdot1\cdot\cos0^\circ = 1, \qquad \vec{i}\cdot\vec{j} = \vec{j}\cdot\vec{k} = \vec{k}\cdot\vec{i} = 1\cdot1\cdot\cos90^\circ = 0, \] and by commutativity $\vec j\cdot\vec i=\vec k\cdot\vec j=\vec i\cdot\vec k=0$.

$\cdot$	$\vec{i}$	$\vec{j}$	$\vec{k}$
$\vec{i}$	$\mathbf 1$	$0$	$0$
$\vec{j}$	$0$	$\mathbf 1$	$0$
$\vec{k}$	$0$	$0$	$\mathbf 1$

\[\vec e_i\cdot\vec e_j = \delta_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases} \qquad (\vec e_1=\vec i,\ \vec e_2=\vec j,\ \vec e_3=\vec k)\]

$O$ $\vec i$ $\vec j$ $\vec k$

The orths are unit vectors, pairwise perpendicular (right-angle marks).

Step 2: Part $2^\circ$

Expand with the distributive law; only the products $\vec i\cdot\vec i$, $\vec j\cdot\vec j$, $\vec k\cdot\vec k$ survive:

	$2\vec{i}$	$-3\vec{j}$	$-\vec{k}$	Row sum
$+\vec{i}$	$2$	$0$	$0$	$2$
$-\vec{j}$	$0$	$3$	$0$	$3$
$+\vec{k}$	$0$	$0$	$-1$	$-1$

\[(\vec{i}-\vec{j}+\vec{k})\cdot(2\vec{i}-3\vec{j}-\vec{k}) = 2+3-1 = 4\]

Coordinate check: $(1)(2)+(-1)(-3)+(1)(-1)=4$ $\checkmark$

Angle between the two vectors

\[ |\vec a|=\sqrt3, \qquad |\vec b|=\sqrt{14}, \qquad \cos\theta = \frac{4}{\sqrt{42}} \approx 0.617, \qquad \theta \approx 51.9^\circ \]

$\theta$ $O$ $\vec a$ $\vec b$

$\vec a=\vec i-\vec j+\vec k$
$\vec b=2\vec i-3\vec j-\vec k$

$\vec a\cdot\vec b=4$, $\theta=\arccos\tfrac{4}{\sqrt{42}}\approx51.9^\circ$. Dotted lines drop the tips to the $xy$-plane.

Result

\[\boxed{\,\vec e_i\cdot\vec e_j = \delta_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases} \qquad (\vec e_1=\vec i,\ \vec e_2=\vec j,\ \vec e_3=\vec k)\,}\]

\[\boxed{\,(\vec{i}-\vec{j}+\vec{k})\cdot(2\vec{i}-3\vec{j}-\vec{k}) = 2+3-1 = 4\,}\]