need help - 3d vectors

hello guys :slight_smile:

i believe there are a lot of mathematical souls among blenderheads, and i need help.

actually, i have a problem with the ‘algorithm’ of solving a problem like this:

for example - let’s have a vector v = 5i + 8j -4k . find a point R, which lies on the vector r and the vector r is perpendicular to the v and passes through origin O(0,0,0). for more advanced people, the r passes through the point P(2,3,-6)

thank you all who tried to help me :slight_smile:

PS: if you did not understand what i need help with, please tell me :slight_smile: thx

i remember my teacher at school teaching this. i didn’t nothing though.

Well, if v = f(i,j,k) where i,j,k each are a distance and offset, then recall the slope of line m for a perpendicular is -m, and a vector is just a line with a length. That means that the -m defines a 3d plane, and the vector r must lie on that plane. So, all you need to do is define the line that lies on M and passes through point P. Recall the slope-intercept form where you know a point and the plane.What you haven’t said is how large vector r is - just that it starts from origin (so all the offsets are 0) and passes thru a point - what is the magnitiude?

For a vector r to be perpendicular to v, the inner product (or dot product) of v and r must equal zero.From looking at the equation, you can see that

r = -4i + 5j + 5k

is a solution. I’m not sure what you mean by passing through the origin. Technically that vector doesn’t pass through the origin, we only visualize it as doing so, its really just a point in R3. If you assume its a line (or ray if t >= 0) you parameterize it so it passes through the origin as:

r(t) = tr

where r was previously given.

In a more general case, if you project any point P onto your original vector v, you can then take that point and the original point P and form a vector that will pass through the point and be perpendicular to v, by vector subtraction of Pproj - Poriginal (I think I have that subtraction right, but it could be the other way around).

whoa, thx for answers.

well, i’ve mistaken: the vector v is a line, and passes through (for example) A(1,2,3)

CyborgDragon: is the slope intercept form usable in 3D? because i’ve seen it in 2D working, but i[m not sure about the 3d.

forTe: “by looking at the equation we can see that…” how did you get that euqtion? :smiley: probably that’s the thing i don’t understand quite well. i know that the dot product must be zero for the lines to be perpendicular.

First off, a Vector is not a “line”. You can imagine it as a line, but it actually is an element of a Vectorspace and an object on its own, more like a point than a line. You’re just thinking of the geometrical interpretation of a vector :wink:

However, to answer the question: the (usual) inner product in R³ is defined as (write x = (x1,x2,x3), y = (y1,y2,y3) )
<x,y> := x1y1 + x2y2 + x3*y3

Hence, in this case we have the equation:
x15 + x28 + x3*-4 = 0
and need a solution for x = (x1,x2,x3). The one given by forte is one and therefore the resulting vector is perpendicular to the given vector v.

:slight_smile: thank you :slight_smile:

Glad that you have been helped!~