# Taking the Dot product of two normalized vectors (unit vectors) vs the dot product of two non-normalized vectors

I know that taking the dot product of two vectors, $a$ and $b$ represents the projection of $a$ to $b$, and the resulting scalar is the length of that projection. But I am confused on the concept of normalizing vectors and the dot product.

If we take the dot product of two normalized vectors, $a$ and $b$, does the resulting scalar still represent the projection of $a$ to $b$?

What about taking the dot product of two NON-normalized vectors? What does this represent?

• taking the dot product of two vectors, a and b represents the projection of a to b, and the resulting scalar is the length of that projection That is only true iff $\,b\,$ is a unit vector. – dxiv Dec 26 '17 at 19:50
• what if b isn't a unit vector? then what does the dot product represent? – Soon_to_be_code_master Dec 26 '17 at 19:53
• $a \cdot b = |a| \,|b|\,\cos \theta$ where $|a| \cos \theta$ is the length of the projection of $a$ onto $b$ and $|b|$ is the length of $b$. – dxiv Dec 26 '17 at 19:54
• Would the same logic apply with $|b|cos \theta$? is $|b|cos \theta$ the projection of $b$ onto $a$? – Soon_to_be_code_master Dec 26 '17 at 19:56
• Indeed it does. – dxiv Dec 26 '17 at 19:57

Since the fundamental identity of the dot product is involving a product of norms of the vectors:

$u \cdot v = \|u\| \|v\| \cos \theta$

then YES, the normalization plays a role for this particular interpretation of the dot product as projection length.

Yes of course, if both vectors are normalized the dot product is still representing a projection, it just does not matter which vector projects on which, since the result is the same.

If one of the vectors, let say $v$, is normalized, the picture to imagine in this case is a circle of radius $u$ with a right triangle inscribed on it with $u$ as hypotenuse and the unit vector $v$ is representing an axis of projection, in which the adjacent cathetus of the triangle lies. The projection is $u \cdot v = \|u\| \cos \theta$. Since the cosine is the ratio of adjacent cathetus over hypotenuse $\cos \theta = a / \|u\|$, the projection reduces to $u \cdot v = \frac{\|u\| a}{\|u\|} = a$, where $a$ is the length of the adjacent cathetus i.e., the projection length.

When $u$ and $v$ are both non-normalized the dot product is still encoding the projection, well actually it is encoding two projections at the same time, so it cannot tell the exact length without a division by either $\|u\|$ or $\|v\|$, but it still can tell something abut the projection. The dot product will be zero if vectors are orthogonal (no projection possible) and will be exactly $\pm \|u\| \|v\|$ when vectors lie on parallel axis. The sign will be positive if their angle is less than 180° or negative if it is more than 180°.

• The dot product of normalized vectors when calculated numerically is a stable way of detecting how equal two vectors are in relation to direction and sense. This is very useful for computer graphics for example. – QBziZ Jan 13 at 9:38