Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.2 - Vectors - 12.2 Exercises: 21

Answer

See the solution.

Work Step by Step

$ \overline a + \overline b $ $(4+2) \overline i + (-3) \overline j + (2-4) \overline k$ $= 6\overline i -3 \overline j -2 \overline k$ $ 4\overline a + 2\overline b $ $ 4(4\overline i-3\overline j+2\overline k) + 2( 2 \overline i-4 \overline k)$ $=16\overline i-12\overline j+8\overline k + 4 \overline i-8 \overline k$ $=(16+4) \overline i-12\overline j+(8-8)\overline k $ $=20\overline i-12\overline j$ $\| \overline a \|$ $ \sqrt{ \overline a_i^{2}+\overline a_j ^{2}+\overline a_k^{2}}$ $ = \sqrt{ 4^{2}+(-3)^{2}+2^{2}}$ $ = \sqrt{ 29}$ $\| \overline a - \overline b \|$ $a-b = (4-2)\overline i + (-3)\overline j + (2-(-4)) \overline k $ $= 2\overline i -3\overline j + 6 \overline k $ $ =\sqrt{ (\overline a-\overline b)_i^{2}+(\overline a-\overline b)_j ^{2}+(\overline a-\overline b)_k^{2}}$ $= \sqrt{2^2+ (-3)^2+6^2}$ $= \sqrt{49}$ $= 7$
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.