2
\$\begingroup\$

I've recently learned about the existence of dot product and I was wanted to know how to determine the value range between two normalized vectors.

Improve this question
\$\endgroup\$
2
  • \$\begingroup\$ Not sure I understand the question. Is this what you're looking for: chortle.ccsu.edu/vectorlessons/vch09/vch09_6.html ? \$\endgroup\$
    – Sean Middleditch
    Jan 31, 2019 at 2:44
  • \$\begingroup\$ Normalized vectors are between -1:1, and the dot product is ax*bx+ay*by It's fairly straight forward. \$\endgroup\$
    – Sidar
    Jan 31, 2019 at 4:02

1 Answer 1

Reset to default
7
\$\begingroup\$

The dot product of two normalized (unit) vectors will be a scalar value between -1 and 1.

Common useful interpretations of this value are

  • when it is 0, the two vectors are perpendicular (that is, forming a 90 degree angle with each other)
  • when it is 1, the vectors are parallel ("facing the same direction") and
  • when it is -1, the vectors are anti-parallel ("facing opposite directions")

Because the dot product is, geometrically, the product of the magnitudes of the vectors and the cosine of the angle between them (\$\vert{a}\vert\vert{b}\vert \cos{\theta} \$) it also useful for finding \$\theta\$: it's just \$\arccos{(a \cdot b)}\$ since the magnitudes of the vectors are both one.

You may also be interested in interpretations of the dot product between non-normalized vectors, which is discussed in this related question.

Improve this answer
\$\endgroup\$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .