University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 37


$\dfrac{9 \sqrt {42}}{7}$

Work Step by Step

The formula to calculate the distance for two vectors is given by: $d=\dfrac{|u \times v|}{|v|}$ Thus, we have $u \times v=\lt 30,6,6\gt $ and $|u \times v|=\sqrt{(30)^2+(6)^2+(6)^2}=18 \sqrt {3}$ Now, $d=\dfrac{|u \times v|}{|v|}=\dfrac{18 \sqrt {3}}{ \sqrt {14}}=\dfrac{9 \sqrt {42}}{7}$
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.