Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 684: 18

Answer

$$ y=1.$$

Work Step by Step

We have $$\overrightarrow{PQ}=Q-P=(1,1,2)-(5,1,1)=\langle -4,0,1\rangle,$$ $$\overrightarrow{PR}=R-P=(2,1,1)-(5,1,1)=\langle -3,0,0\rangle .$$ Now, the normal vector $ n $ is given by $$\overrightarrow{PQ} \times \overrightarrow{PR}= \left|\begin{array}{rrr}{i} & {j} & {k} \\ {-4} & {0} & {1} \\ {-3} & {0} & {0}\end{array}\right|=i\left|\begin{array}{rrr} {0} & {1} \\ {0} & {0}\end{array}\right|-j\left|\begin{array}{rrr} {-4} & {1} \\ {-3} & {0}\end{array}\right|+k\left|\begin{array}{rrr} {-4} & {0} \\ {-3} & {0} \end{array}\right|\\ =i(0-0)-j(0+3)+k(0-0)\\ =-3j. $$ Now, using any of the given points, the equation of the plane is given by $$0(x-1)-3(y-1)+0(z-1)=0\Longrightarrow -3y+3=0$$ Hence the equation is given by $$ y=1.$$
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