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Answers without the blur. Sign up and see all textbooks for free! Q. 39

Expert-verified Found in: Page 824 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises 36–41 use the given sets of points to find:(a) A nonzero vector N perpendicular to the plane determined by the points.(b) Two unit vectors perpendicular to the plane determined by the points.(c) The area of the triangle determined by the points.$P\left(2,-5,1\right),Q\left(-4,5,8\right),R\left(-1,-5,3\right)$

(a) A nonzero vector N perpendicular to the plane determined by the points are $\left(20,-9,30\right)$.

(b) Two unit vectors perpendicular to the plane determined by the points are $±\frac{1}{\sqrt{1381}}\left(20,-9,30\right)$.

(c) The area of the triangle determined by the points is $\frac{\sqrt{1381}}{2}$.

See the step by step solution

## Step 1. Given Information

In the given exercises use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

The given points are $P\left(2,-5,1\right),Q\left(-4,5,8\right),R\left(-1,-5,3\right)$

## Part (a) Step 1. firstly finding a nonzero vector N perpendicular to the plane determined by the points.

We have $P\left(2,-5,1\right),Q\left(-4,5,8\right),R\left(-1,-5,3\right)$

Now

$\stackrel{\to }{PQ}=\left(-4-2,5-\left(-5\right),8-1\right)=\left(-6,10,7\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{PR}=\left(-1-2,-5-\left(-5\right),3-1\right)=\left(-3,0,2\right)$

## Part (a) Step 2. Now finding PQ→×PR→

$\stackrel{\to }{PQ}×\stackrel{\to }{PR}=\left|\begin{array}{ccc}i& j& k\\ -6& 10& 7\\ -3& 0& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}\stackrel{\to }{PQ}×\stackrel{\to }{PR}=i\left|\begin{array}{cc}10& 7\\ 0& 2\end{array}\right|-j\left|\begin{array}{cc}-6& 7\\ -3& 2\end{array}\right|+k\left|\begin{array}{cc}-6& 10\\ -3& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}\stackrel{\to }{PQ}×\stackrel{\to }{PR}=i\left(10×2-7×0\right)-j\left\{\left(-6\right)×2-7×\left(-3\right)\right\}+k\left\{\left(-6\right)×0-10×\left(-3\right)\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{PQ}×\stackrel{\to }{PR}=i\left(20-0\right)-j\left(-12+21\right)+k\left(0+30\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{PQ}×\stackrel{\to }{PR}=20i-9j+30k$

The points are $\left(20,-9,30\right)$.

## Part (b) Step 1. Now finding two unit vectors perpendicular to the plane determined by the points.

So,

$||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||=\sqrt{{\left(20\right)}^{2}+{\left(-9\right)}^{2}+{\left(30\right)}^{2}}\phantom{\rule{0ex}{0ex}}||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||=\sqrt{400+81+900}\phantom{\rule{0ex}{0ex}}||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||=±\sqrt{1381}$

Required vector

$\frac{\stackrel{\to }{PQ}×\stackrel{\to }{PR}}{||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||}=\frac{\left(20,-9,30\right)}{±\sqrt{1381}}\phantom{\rule{0ex}{0ex}}\frac{\stackrel{\to }{PQ}×\stackrel{\to }{PR}}{||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||}=±\frac{1}{\sqrt{1381}}\left(20,-9,30\right)$

## Part (c) Step 1. Now finding the area of the triangle determined by the points.

$\mathrm{Area}∆ABC=\frac{1}{2}||\stackrel{\to }{PQ}×\stackrel{\to }{PR}||\phantom{\rule{0ex}{0ex}}\mathrm{Area}∆ABC=\frac{1}{2}||\sqrt{1381}||\phantom{\rule{0ex}{0ex}}\mathrm{Area}∆ABC=\frac{\sqrt{1381}}{2}$ ### Want to see more solutions like these? 