Dot Product and the Normal Vector?

Question

Just a quick question about the dot product and normal vector....

Say we have a plane with 3 points
Q(a, b, c)
R(d, e, f)
S(x, y, z)

Say I take the cross product of vectors QR and QS to find the normal vector to the plane.

Now, what I'm wondering is why the dot product of n (the normal vector to the plane) and the vector RS can yield a number OTHER than 0. Isn't the normal vector of QR and QS also the normal vector of RS (since they're all in the same plane), and if so, why isn't the dot product always 0?

I hope that made sense, haha. Thanks in advance!

Answer 1

I disagree.



Let u = QR X QS



then u • QR = u • QS = 0 as you say.



But since RS = QS - QR, then



u • RS = u • (QS - QR) = u • QS - u • QR = 0

Answer 2

i think of the others have been assisting yet didnt answer your one question what's theta once you dont have an attitude given think of of the vector as triangles yet for this one its even much less puzzling you notice that the two vectors are multiples of one yet another meaning one vector may be larger by making use of a variety to offer the different this implies the two are an identical attitude and for this reason the attitude theta = 0 yet while we had different angles unwell %. some (a million,2,3) and (a million,2,a million) the attitude would be formed between the two vectors yet once you think of this additionally makes 2 factors of a triangle the triangle it makes the two factors of is merely no longer ninety deg triangle so we would desire to apply the COSINE regulation c^2 = a^2 + b^2 -2abcos theta if we resolve for 2abcos theta 2abcos theta = a^2 + b^2 - c^2 now you have 3 factors to be conscious of remember in case you have 2 vectors you could subtract one from the different to advantageous the vector pointing from one end to the different vector a = (a million,2,3) vector b = (a million,2,a million) vector c will then equivalent b - a = (a million,2,a million) - (a million,2,3) = (0,0,-2) or may be a - b = (0,0,2) it doesnt count one will element one way the different any incorrect way now we choose the distances of all of them we use distances interior the cosine regulation length vector a = (a million,2,3) making it uncomplicated we want a^2 it quite is distance squared a^2 = (a million^2 + 2^2 + 3^2) a^2 = 14 vector b = (a million,2,a million) b^2 = a million^2 + 2^2 + a million^2 b^2 = 6 vector c will then equivalent b - a = (a million,2,a million) - (a million,2,3) = (0,0,-2) c^2 = 0^2 + 0^2 + 2^2 c^2 = 4 so we've 14 6 and four we now resolve the equation cos theta = (a^2 + b^2 - c^2)/ 2ab all of us be conscious of a = sqrt(a^2) = sqrt(14) b = sqrt ( 6) something ar e above cos theta = (14 + 6 - 4)/ (2*sqrt(14) * sqrt(6)) theta = inverse cosine (sixteen/ (2*sqrt(14) * sqrt (6))) theta = inverse cosine (8/ (sqrt(14) *sqrt(6)) it quite is the way you come across theta the attitude between the two angles once you in basic terms be conscious of two vectors ... and in the previous you think of it quite is annoying it took me over 30 minutes to keep in mind what that became into and that i've got been engaged on math for 2 many years you will get used to remembering yet alot of situations you be conscious of toooo plenty and each so often it is going to take a couple of minutes to keep in mind the trick to unravel some thing..

Answer 3

The dot product of two orthogonal vectors is always 0. So, if you are doing your math correcting, the dot product of any vector on the plane and any vector orthogonal to that plane should yield a result of 0. Always.



I'd say check the math and try again