Notice the the result of the dot product is not another vector – it's just another number or a scalar. That's why we often call the dot product the scalar product. We can easily extend the definition of the dot product to vectors of any dimension, as long as the two have the same dimension.

Finally, the dot product can be thought of as a row vector (or a 1 × n matrix) multiplied by a column vector (an n × 1 matrix). Here's a picture of that for 2-D vectors:

$$ \begin{align} v_1 &\cdot v_2 = (x_1, y_1)\left( \begin{matrix} x_2 \\ y_2 \end{matrix} \right) = x_1 x_2 + y_1 y_2 \\[5pt] &\phantom{0000000} \color{#E90F89}{(1 \times 2)(2 \times 1) = 1 \times 1 \; \text{(scalar)}} \end{align}$$

$$|\vec v_1 - \vec v_2|^2 = \left(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\right)^2$$

Here we're just using the distance formula (which is really just the Pythagorean theorem) with those coordinates. Now we have the square of a square root, so that right side is just

$$= (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Now for the right side of our LOC equation. If we remember that the square of the length of a vector is

$$|v_1|^2 = \left(\sqrt{x_1^2 + y_1^2} \right)^2$$

we get

$$ \begin{align} |v_1|^2 = x_1^2 + y_1^2 \\[5pt] |v_2|^2 = x_2^2 + y_2^2 \end{align}$$

Now we can rewrite both sides, and expand the binomials on the left to get:

$$ \begin{align} (x_2 &- x_1)^2 + (y_2 - y_1)^2 \\[5pt] &= x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2|\vec v_1||\vec v_2| cos(\theta) \end{align}$$

If we expand the squared binomials on the left side, we notice that the squared terms containing purely $x$'s or $y$'s are represented on both sides of the equation, so we can eliminate them:

$$ \require{cancel} \begin{align} \cancel{x_2^2} &- 2x_1x_2 + \cancel{x_1^2} + \cancel{y_2^2} - 2y_1y_2 + \cancel{y_1^2} \\ &= \cancel{x_1^2} + \cancel{y_1^2} + \cancel{x_2^2} + \cancel{y_2^2} - 2|\vec v_1||\vec v_2| cos(\theta) \end{align}$$

Finally, we can divide all of those $-2$'s away:

$$\cancel{-2}x_1x_2 + \cancel{-2}y_1y_2 = \cancel{-2}|\vec v_1||\vec v_2| cos(\theta)$$

to give

$$x_1x_2 + y_1y_2 = |\vec v_1||\vec v_2| cos(\theta)$$

Now all that remains is to recognize that $x_1 x_2 + y_1 y_2$ is just the definition of the dot product, so we've arrived at the relationship between the dot product of two vectors and the angle between them:

$$\vec v_1 \cdot \vec v_2 = |\vec v_1||\vec v_2| cos(\theta)$$

$$\vec v_1 \cdot \vec v_2 = (-1)(2) + 2(5) = 8$$

The lengths of the vectors are found using the distance formula:

$$ \begin{align} |\vec v_1| &= ((-1)^2 + 2^2)^{1/2} = \sqrt{5} \\[5pt] |\vec v_2| &= (2^2 + 5^2)^{1/2} = \sqrt{29} \end{align}$$

And finally the angle:

$$\theta = cos^{-1}\left( \frac{8}{\sqrt{5\cdot 29}} \right) = 48.4˚$$

All of our dot products are zero, and the angle with $cos(\theta) = 0$ is $\theta = 90^{\circ}$.

So the angles between these three unit vectors are all 90˚, just what we'd expect for the alignment of the axes of a rectilinear coordinate system.

3. $(1, 2, -1), \; (2, 2, -3)$

   Solution

   Let $\vec a = (1, 2, -1)$ and $\vec b = (2, 2, -3).$ Then

   $$\theta = cos^{-1}\left( \frac{a\cdot \vec b}{|\vec a||\vec b|} \right)$$

   $$ \begin{align} |\vec a| &= \sqrt{1 + 4 + 1} = \sqrt{6} \\[5pt] |\vec b| &= \sqrt{4 + 4 + 9} = \sqrt{17} \\[5pt] \vec a \cdot \vec b &= 1(2) + 2(2) + (-1)(-3) = 9 \\[5pt] \theta &= cos^{-1}\left( \frac{9}{\sqrt{6}\sqrt{17}} \right) \\[5pt] &= cos^{-1} \left( \frac{9}{\sqrt{102}} \right) \approx 27˚ \end{align}$$




4. $(1, 1, 2, 1), \; (-2, 1, -1, 1)$

   Solution

   Let $\vec a = (1, 1, 2, 1)$ and $\vec b = (-2, 1, -1, 1).$ Then

   $$\theta = cos^{-1}\left( \frac{a\cdot \vec b}{|\vec a||\vec b|} \right)$$

   $$ \begin{align} |\vec a| &= \sqrt{1 + 1 + 4 + 1} = \sqrt{7} \\[5pt] |\vec b| &= \sqrt{4 + 1 + 1 + 1} = \sqrt{7} \\[5pt] \vec a \cdot \vec b &= 1(-2) + 1(1) + 2(-1) + 1(1)= -2 \\[5pt] \theta &= cos^{-1}\left( \frac{-2}{\sqrt{7}\sqrt{7}} \right) \\[5pt] &= cos^{-1} \left( \frac{-2}{7} \right) \approx 107˚ \end{align}$$

   It's difficult to visualize these vectors in a 4-dimensional space, but the angle between them is 107˚.

So all that is required of this vector is that $y = 0$ and $x = -z$.

$$\vec v_{\perp} = (1, \, 0, \, -1)$$

The vectors $(3, 0, -3)$ and $(4.5, 0, 4.5)$ would work just as well. They differ only in length, but they're perpendicular to $(1, 0, 1$). What we need to do now is to adjust the length of $(1, 0, 1)$ to $1$. First we need to find its current length:

$$|(1, 0, -1)| = (1^2 + 0^2 + (-1)^2)^{1/2} = \sqrt{2}$$

Now to adjust the length we just divide each component of our vector by that length. You can check for yourself that the length of this new vector is $1$.

$$\vec v_{\perp} = \left( \frac{1}{\sqrt{2}}, \, 0, \, \frac{-1}{\sqrt{2}} \right)$$

The process of adjusting the length of a vector to 1, or making it a unit vector, is called normalization.

Finally, if $(a, 0, -a)$ is perpendicular to $(1, 0, 1)$, then so is $(-a, 0, a)$. It also satisfies the requirement that $x = -z$, it just points in the opposite direction.

The diagram above shows our two vectors. They lie in the x-z plane. Notice that there are many other vectors perpendicular to $(1, 0, 1)$ outside of that plane.

4. $(2, -2, 1)$ and $(1, 0, 0)$

   Solution

   In this case we'll need to set up and solve two equations for our three unknowns. They are

   $$ \begin{align} (2, -2, 1)\cdot (x, y, z) &= 0 \rightarrow 2x - 2y + z = 0 \\[5pt] (1, 0, 0) \cdot (x, y, z) &= 0 \rightarrow x = 0 \end{align}$$

   Given that $x = 0,$ we have that $2y = z,$ or $y = \frac{1}{2}z,$ so if we let $z = 1,$ then our orthogonal vector is

   $$v = (0, \frac{1}{2}, 1)$$

   Now the length of this vector is $\sqrt{5}/2,$ so to normalize it, we divide all vector components by that length to get:

   $$v = \left( 0, \frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5} \right)$$

   You can confirm that the dot products of this vector with the original two are zero, and that its length is 1.




5. $(1, -1, -2, 1), \; (-2, -1, 1, 2), \; (2, 3, 4, 1)$

   Solution

   In this case we'll need to set up and solve three equations for our four unknowns. They are

   $$ \begin{align} (1, -1, 2, 1)\cdot (w, x, y, z) &= 0 \rightarrow w - x + 2y + z = 0 \\[5pt] (2, 3, 4, 1)\cdot (w, x, y, z) &= 0 \rightarrow 2w + 3x + + 4y + z = 0 \\[5pt] (2, -1, 1, 2) \cdot (w, x, y, z) &= 0 \rightarrow 2w - x + y + 2z = 0 \end{align}$$

   If we multiply the top equation by -2, we can set the three equations up to eliminate w:

   $$ \begin{align} -2w + 2x - 4y - 2z &= 0 \\[4pt] 2w + 3x + 4y + z &= 0 \\[4pt] 2w - x + y + 2z &= 0 \end{align}$$

   Adding the first equation to the second and to the third gives two new equations, reduced a bit:

   $$ \begin{align} 5x - z = 0 \\[4pt] x - 3y = 0 \end{align}$$

   Now we can find x, y and z in terms of one of those. It works out nicely if we let $y = 1,$ then $x = 3$ and $z = 5x = 15.$

   Then w can be found from the first of the original equations:

   $$ \begin{align} w &= x - 2y - z = 0 \\[4pt] &= 3 - 2 - 15 = 15 \end{align}$$

   So one version of our orthogonal vector (the differ only in length) is

   $$v = (w, x, y, z) = (-14, 3, 1, 15)$$

   In order to normalize this vector, we need its length:

   $$|v| = \sqrt{14^2 + 3^2 + 1 + 15^2} = \sqrt{431}$$

   Our normalized orghogonal vector is

   $$v_n = \left( \frac{-14}{20.76}, \frac{3}{20.76}, \frac{1}{20.76}, \frac{15}{20.76},\right),$$

   where $20.76 \approx \sqrt{431}.$




6. $(2, 1, -2, -1), \; (1, 1, -1, 2), \; (0, 1, 1, 1)$

   Solution

   In this case we'll need to set up and solve three equations for our four unknowns. They are

   $$ \begin{align} (2, 1, -2, -1)\cdot (w, x, y, z) &= 0 \rightarrow 2w + x - 2y - z = 0 \\[5pt] (1, 1, -1, 2)\cdot (w, x, y, z) &= 0 \rightarrow w + x - y + 2z = 0 \\[5pt] (0, 1, 1, 1) \cdot (w, x, y, z) &= 0 \rightarrow x + y + z = 0 \end{align}$$

   Now let's use the last equation to eliminate z in the first two, adding it to the first, subtracting 2 × it from the second to get:

   $$ \begin{align} 2w + 2x - y &= 0 \\[5pt] -2w + 2x + 6y &= 0 \end{align}$$

   Now add these to eliminate w and find a direct relationship between x and y:

   $$4x + 5y = 0 \rightarrow x = \frac{-5}{4}$$

   If we let $x = 1,$ then $y = \frac{-5}{4}.$

   Now we can use one of the equations above to find w:

   $$w = x + 3y = -\frac{5}{4} + 3 = \frac{7}{4}$$

   Finally, we can use the last of our original equations to find z:

   $$z = -x - y = \frac{5}{4} - 1 = \frac{1}{4},$$

   So our vector is

   $$v = \left( \frac{7}{4}, \frac{-5}{4}, 1, \frac{1}{4} \right)$$

   Its length is $\sqrt{91}/4,$ so we obtain the normalized (unit length) vector by dividing each component of our orthogonal vector by this length.

   $$ \begin{align} v_n &= \left( \frac{7}{4}\frac{4}{\sqrt{91}}, \frac{-5}{4}\frac{4}{\sqrt{91}}, 1\frac{4}{\sqrt{91}}, \frac{1}{4}\frac{4}{\sqrt{91}} \right) \\[5pt] &= \left( \frac{7}{\sqrt{91}}, \frac{-5}{\sqrt{91}}, \frac{4}{\sqrt{91}}, \frac{1}{\sqrt{91}} \right) \end{align}$$

Now if we multiply the two matrices, we're just taking a series of dot products like this:

$$ = \begin{bmatrix} a \cdot x & a \cdot y & a \cdot z \\ b \cdot x & b \cdot y & b \cdot z \\ c \cdot x & c \cdot y & c \cdot z \end{bmatrix}$$

Now take a look at that matrix. The element in the second row, second column is just the dot product of the second row vector, b, and the second column vector, y. It's the same for each element in the result matrix. Here is the expanded result:

$$ \begin{bmatrix} a_1x_1 + a_2x_2 + a_3x_3 & a_1y_1 + a_2y_2 + a_3y_3 & a_1z_1 + a_2z_2 + a_3z_3 \\ b_1x_1 + b_2x_2 + b_3x_3 & b_1y_1 + b_2y_2 + b_3y_3 & b_1z_1 + b_2z_2 + b_3z_3 \\ c_1x_1 + c_2x_2 + c_3x_3 & c_1y_1 + c_2y_2 + c_3y_3 & c_1z_1 + c_2z_2 + c_3z_3 \end{bmatrix}$$

Just as an aside, we can write the transpose of a matrix, too. Just consider a matrix as a row of column vectors or a column of row vectors. Here's a 2×2 example:

$$A = \left( \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right) \phantom{000} A^T = \left( \begin{matrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{matrix} \right)$$

Now we make two slight changes to our thinking about vectors and the dot product.

1. We write all vectors, by default, as column vectors. Let's let the column vector be our natural way to write any vector.
2. The dot product of vectors $\vec a$ and $\vec b$ is then $\vec a^T \vec b$.

That is, if $\vec a = \left( \begin{matrix} a_1 \\ a_2 \end{matrix} \right)$ and $\vec b = \left( \begin{matrix} b_1 \\ b_2 \end{matrix} \right)$, then the dot product $\vec a \cdot \vec b$ is

$$\vec a \cdot \vec b = \vec a^T \vec b = (a_1, a_2) \left( \begin{matrix} b_1 \\ b_2 \end{matrix} \right) = a_1 b_1 + a+2 b_2$$.

This form of the dot product works for any two vectors of any dimension as long as they're the same dimension.

Matrix multiplication works in the same way: we simply multiply each row vector in the left-side matrics by each column vector in the right-side matrix.