Did I just find an error in one of the textbook problems? chapter 1 problem 20 of MATLAB An Introduction with Applications, 6th edition.pdf

They used the formula to find the area of a parallelogram instead of a triangle.

S=sqrt(s1^2+s2^2+s3^2) finds the area of a parallelogram while S=sqrt(s1^2+s2^2+s3^2)/2 is for a triangle. further more I graphed the 3 points out on a 3d grapher and measured the shortest distance from point p to line ab and it is exactly half the answer you would get if you used the book's formula.

my code:

%% 20)

%define variables

xp=2

yp=6

zp=-1

xa=-2

ya=-1.5

za=-3

xb=-2.5

yb=6

zb=4

r=sqrt((xb-xa)^2+(yb-ya)^2+(zb-za)^2) %solve for r

s1=xp*ya+xa*yb+xb*yp-(yp*xa+ya*xb+yb*xp) % solve for s1,s2,s3; this was fun to type

s2=yp*za+ya*zb+yb*zp-(zp*ya+za*yb+zb*yp)

s3=xp*za+xa*zb+xb*zp-(zp*xa+za*xb+zb*xp)

s=sqrt(s1^2+s2^2+s3^2) %solve for s

d=(2*s)/r %distance can now be calculated

%voila

%d=11.3310 (distance)

%s=58.1920 (triangle area)

graph with distance measured:

11.331/2=5.6655

Answer 1

[Priysha LNU]

It seems like this is not an official MathWorks text book. You may contact the author of book

MATLAB An Introduction with Applications, 6th edition.pdf

for further clarifications.

DISCLAIMER: These are my own views and in no way depict those of MathWorks.