# how to find a normal vector?

10 ビュー (過去 30 日間)
Sierra 2022 年 6 月 25 日
コメント済み: Sierra 2022 年 6 月 25 日
for example, there are 2 points. P0(4,3,2) ,P1(8,5,4) and the vector ->P0P1
I know that In three dimension, there are infinite number of vectors perpendicular to a given vector.
but i know the point which is on the plane.
To use this function, I need to find a normal vector of the plane.
In my case, P1 point wil be the V0 and P1 for this function.
[I,check]=plane_line_intersect(n,V0,P0,P1)
% n: normal vector of the Plane
% V0: any point that belong s to the Plane
% P0: end point 1 of the segment P0P1
% P1: end point 2 of the segment P0P1 サインインしてコメントする。

### 採用された回答

Matt J 2022 年 6 月 25 日

In my case, P1 point wil be the V0 and P1 for this function.
You need 3 distinct, non-colinear points in a the plane to calculate its normal. If V0,P0,V1 are such points, then you would do,
normal=cross(P1-P0,V-P0)
##### 10 件のコメント表示非表示 9 件の古いコメント
Sierra 2022 年 6 月 25 日
To Matt J
I thought P1 is V0, P1 in this function.

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