Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - Chapter Review Exercises - Page 701: 12

Answer

(a) $ Q=(5,3,2)$. (b) the unit vector in the direction of $\overrightarrow{ PQ}$ is given by $$ \left\langle \frac{3}{\sqrt{35}},-\frac{1}{\sqrt{35}},\frac{5}{\sqrt{35}}\right \rangle.$$

Work Step by Step

(a) Let $ Q=(a,b,c) $; then we have $$\overrightarrow{ PQ}=Q-P=(a,b,c)- (1,4,-3)=\langle a-1,b-4,c-3 \rangle $$ now since $\langle a-1,b-4,c-3 \rangle $ and $\langle 3,-1,5 \rangle $ are equivalent, then we have $\langle a-1,b-4,c-3 \rangle=\langle 3,-1,5 \rangle $ and hence $$ a-1=4, \quad b-4=-1, \quad c+3=5.$$ That is, $ Q=(5,3,2)$. (b) The unit vector in the direction of $\overrightarrow{ PQ}$ is given by $$\frac{\langle 3,-1,5 \rangle}{\sqrt{9+1+25}}=\left\langle \frac{3}{\sqrt{35}},-\frac{1}{\sqrt{35}},\frac{5}{\sqrt{35}}\right \rangle.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.