# Calc iii question/analytic geometry

• jaejoon89
In summary, the problem is to prove that vector OA + OB + OC is perpendicular to plane ABC for any given vertices of a regular tetrahedron. The conversation discusses the difficulty in finding a regular tetrahedron that satisfies this condition and suggests using a variable length for the sides in order to generalize the proof for all regular tetrahedrons.
jaejoon89
Given vertices of regular tetrahedron OABC, prove that vector OA + OB + OC is perpendicular to plane ABC...

I've been racking my brain on this one, can't figure it out... would appreciate some help

I thought I'd center it on the origin:
(0, 0, 0)
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)

A + B + C = (1, 0, 0) + (0, 1, 0) +(0, 0, 1) = (1, 1, 1)
(not sure if that's right - seems like it would be outside the tetrahedron)

No idea where to go from there.

The problem here is that your tetrahedron isn't regular. Consider triangle OAB. Two of the sides have length 1, but the hypoenuse has length $\sqrt{2}$. You'll have to choose different vectors.

Is there an easier way to do it? Showing that this is true for one regular tetrahedron doesn't necessarily/formally mean it holds for all of them, does it?

Also, even when I switch coordinates, it doesn't work. For example, a regular tetrahedron (... at least wikipedia says this one is regular)

O (1, 1, 1)
A (-1, -1, 1)
B (-1, 1, -1)
C (1, -1, -1)

vector OA = A - O = (-2, -2, 0)
Similarly, vector OB = (-2, 0, -2)
OC = (0, -2, -2)

AB = (0, 2, -2)
AC = (2, 0, -2)

AB x AC = (-4, -4, -4)
OA + OB + OC = (-4, -4, -4)

And when you take the dot product of those it should be 0 (if they are perpendicular, as they are supposed to be)... but it isn't 0

What's wrong? Also, again is showing it holds for one regular tetrahedron sufficient?

Last edited:
jaejoon89 said:
Is there an easier way to do it?

If there is, I don't see it.

Showing that this is true for one regular tetrahedron doesn't necessarily/formally mean it holds for all of them, does it?

No, but you're trying to do hear is easily generalized. Instead of trying to come up with a tetrahedron whose sides are of unit length just let the length be some variable, say $a$. Then *presto* you're working with any possible regular tetrahedron.

## 1. What is Calculus III?

Calculus III is a branch of mathematics that focuses on multivariable calculus, including topics such as vectors, partial derivatives, multiple integrals, and vector calculus. It is typically taken after Calculus I and II and is considered a more advanced level of calculus.

## 2. What topics are covered in Calculus III?

Topics covered in Calculus III include vectors, vector-valued functions, partial derivatives, multiple integrals, line integrals, surface integrals, and vector calculus. These topics build upon the concepts learned in Calculus I and II, but with a focus on functions of multiple variables.

## 3. How is Calculus III different from Calculus I and II?

Calculus III is different from Calculus I and II in that it deals with functions of multiple variables, rather than just one variable. This adds another level of complexity to the problems and requires a deeper understanding of concepts such as vectors and partial derivatives.

## 4. How does analytic geometry relate to Calculus III?

Analytic geometry is a branch of mathematics that deals with the study of geometric shapes using algebraic equations. It is an important tool in Calculus III as it helps visualize and solve problems involving vectors and curves in three-dimensional space.

## 5. What are some real-life applications of Calculus III?

Calculus III has many real-life applications, including in physics, engineering, computer graphics, economics, and statistics. It is used to model and analyze systems with multiple variables, such as fluid flow, electric fields, and optimization problems.

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