{"id":373,"date":"2011-10-05T19:09:39","date_gmt":"2011-10-05T18:09:39","guid":{"rendered":"http:\/\/devnot.wordpress.com\/?p=373"},"modified":"2022-03-24T19:04:11","modified_gmt":"2022-03-24T18:04:11","slug":"calculate-a-point-at-a-distance-perpendicular-to-a-vector-offset-2d","status":"publish","type":"post","link":"https:\/\/thomas-jansen.eu\/?p=373","title":{"rendered":"Calculate a point at a distance perpendicular to a vector offset (2D)"},"content":{"rendered":"\n\n\n

To calculate a point at a distance perpendicular to a vector offset<\/p>\n\n\n\n

\"\"
Point perpendicular to offset
<\/figcaption><\/figure>\n\n\n\n

First the point on the vector \"P'\" at the required offset needs to be calculated. This easy to do:<\/p>\n\n\n\n

<\/p>\n\n\n\n

\"P' = A + (B-A) * offset\"<\/p>\n\n\n\n

\"P'_x = A_x + (B_x - A_x) * offset\"
\"P'_y = A_y + (B_y - A_y) * offset\" <\/p>\n\n\n\n

The vector to move the point P’ on is perpendicular to \"\overline{AB}\". This vector can be calculated using a 90 degree rotation matrix:<\/p>\n\n\n\n

\"\begin{bmatrix}<\/pre>\n\n\n\n
The matrix shows in the the first row \"x'= -y\" and the second row shows \"y'= x\"<\/p>\n\n\n\n
You can safely ignore the stuff above if you are only interested in getting the job done. The final formula is:<\/p>\n\n\n\n
\"\begin{pmatrix}x'\\y' \end{pmatrix}= \begin{pmatrix}-y\\x \end{pmatrix}\"
 <\/p>\n\n\n\n
<\/p>\n\n\n\n
We will call the vector (\"\overline{P'P}\")  vector \"C\".  <\/p>\n\n\n\n
\"C =\begin{pmatrix}- (B_y - A_y)\\   B_x - A_x \end{pmatrix}\"
<\/p>\n\n\n\n
 Vector \"C\" needs to be normalized (divided by it’s length \/ magnitude). <\/p>\n\n\n\n
 \"\hat{C} =\dfrac{C}{\left \|C \right \|}\" <\/p>\n\n\n\n
The resulting formula becomes:<\/p>\n\n\n\n
 \"P = (A + (B-A) * offset) + \hat {C} * distance\"<\/p>\n\n\n\n
A positive distance will get the point above the vector, a negative below the vector.<\/p>\n","protected":false},"excerpt":{"rendered":"
To calculate a point at a distance perpendicular to a vector offset First the point on the vector $P’$ at the required offset needs to be calculated. This easy to do: $P’ = A + (B-A) * offset$ $P’_x = A_x + (B_x – A_x) * offset$ $P’_y = A_y + (B_y – A_y) * […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[55,56,82],"class_list":["post-373","post","type-post","status-publish","format-standard","hentry","category-geometry","tag-perpendicular","tag-point","tag-vector"],"_links":{"self":[{"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/posts\/373","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=373"}],"version-history":[{"count":5,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/posts\/373\/revisions"}],"predecessor-version":[{"id":681,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=\/wp\/v2\/posts\/373\/revisions\/681"}],"wp:attachment":[{"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=373"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=373"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/thomas-jansen.eu\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=373"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}