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### Vector Analysis

```
Date: 7/6/96 at 23:22:39
From: Anonymous
Subject: Vector Analysis

Hi,

1) Given the three points A(2,-1,2), B(-1,1,4), and C(4,3,-1), find
(a) the angle between R(AB) and R(AC); (b) the (scalar) area of
triangle ABC; (c) a unit vector perpendicular to ABC.

2) Given two points, M(2,5,-3) and N(-3,1,4); (a) find their
separation; (b) find the distance from the origin to the midpoint of
the line MN; (c)find the unit vector in the direction of R(MN);
(d) find the point of intersection of the line MN and the x=0 plane.
```

```
Date: 7/23/96 at 15:33:29
From: Doctor Jim
Subject: Re: Vector Analysis

Hi Ivan,

Question 1:

For part (a) remember that the cosine of the angle between the two
vectors is given by dividing the dot product of the vectors by the
product of their magnitudes and that R(AB)=B-A and R(AC)=C-A.

I think the easiest way to do part (b) is Hero's Formula: given a
triangle with sides of lengths a, b, and c, the area is given by the
square root of s(s-a)(s-b)(s-c) where s is half of the perimeter.
Remember that the distance between two points is the square root of
(x1-x2)^2+(y1-y2)^2+z1-z2)^2.

In part (c), to find a unit vector perpendicular to ABC, you just need
to find a vector perpendicular to all three of the vectors in the
triangle, at the same time.  That is, find a vector so that its dot
product with R(AB) and R(AC) and R(BC) is 0.  Once you know that
vector, divide by its magnitude to make it a unit vector.

Question 2:

Part (a): the distance between two points is the square root of
(x1-x2)^2+(y1-y2)^2+z1-z2)^2.

Part (b): the midpoint of a segment with endpoints (x1,y1.z1) and
(x2,y2,z2) is ((x1-x2)/2, (y1-y2)/2,(z1-z2)/2), then use the distance
formula again.

Part (c): divide the vector R(MN) by its magnitude.

-Doctor Jim,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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