Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 9 - Section 9.1 - Vectors in Two Dimensions - 9.1 Exercises - Page 637: 39

Answer

$|\vec{u}|=\sqrt{101}$ $|\vec{v}|=2\sqrt{2}$ $|2\vec{u}|=2\sqrt{101}$ $|\frac{1}{2}\vec{v}|=\sqrt{2}$ $|\vec{u}+\vec{v}|=\sqrt{73}$ $|\vec{u}-\vec{v}|=\sqrt{145}$ $|\vec{u}|-|\vec{v}|=\sqrt{101}-2\sqrt{2}$

Work Step by Step

$\vec{u}=\langle10,-1\rangle$ $,$ $\vec{v}=\langle-2,-2\rangle$ To obtain the magnitude of a vector $\vec{v}=\langle a_{1},a_{2}\rangle$, apply the formula $\vec{v}=\sqrt{a_{1}^{2}+a_{2}^{2}}$ $|\vec{u}|$ For the vector $\vec{u}$, $a_{1}=10$ and $a_{2}=-1$ $|\vec{u}|=\sqrt{10^{2}+(-1)^{2}}=\sqrt{100+1}=\sqrt{101}$ $|\vec{v}|$ For the vector $\vec{v}$, $a_{1}=-2$ and $a_{2}=-2$ $|\vec{v}|=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$ $|2\vec{u}|$ Multiply both components of $\vec{u}$ by $2$: $2\vec{u}=2\langle10,-1\rangle=\langle20,-2\rangle$ For the vector $2\vec{u}$, $a_{1}=20$ and $a_{2}=-2$ $|2\vec{u}|=\sqrt{20^{2}+(-2)^{2}}=\sqrt{400+4}=\sqrt{404}=2\sqrt{101}$ $|\frac{1}{2}\vec{v}|$ Multiply both components of $\vec{v}$ by $\dfrac{1}{2}$: $\dfrac{1}{2}\vec{v}=\dfrac{1}{2}\langle-2,-2\rangle=\langle-1,-1\rangle$ For the vector $\frac{1}{2}\vec{v}$, $a_{1}=-1$ and $a_{2}=-1$: $|\frac{1}{2}\vec{v}|=\sqrt{(-1)^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2}$ $|\vec{u}+\vec{v}|$ Add the vectors $\vec{u}$ and $\vec{v}$ together and simplify: $\vec{u}+\vec{v}=\langle10,-1\rangle+\langle-2,-2\rangle=\langle8,-3\rangle$ For the vector $\vec{u}+\vec{v}$, $a_{1}=8$ and $a_{2}=-3$: $|\vec{u}+\vec{v}|=\sqrt{8^{2}+(-3)^{2}}=\sqrt{64+9}=\sqrt{73}$ $|\vec{u}-\vec{v}|$ Subtract the vectors $\vec{u}$ and $\vec{v}$ and simplify: $\vec{u}-\vec{v}=\langle10,-1\rangle-\langle-2,-2\rangle=\langle12,1\rangle$ For the vector $\vec{u}-\vec{v}$, $a_{1}=12$ and $a_{2}=1$: $|\vec{u}-\vec{v}|=\sqrt{12^{2}+1^{2}}=\sqrt{144+1}=\sqrt{145}$ $|\vec{u}|-|\vec{v}|$ Subtract the magnitude of $\vec{v}$ from the magnitude of $\vec{u}$. These two magnitudes were found in the first two parts of this problem. $|\vec{u}|=\sqrt{101}$ and $|\vec{v}|=2\sqrt{2}$ $|\vec{u}|-|\vec{v}|=\sqrt{101}-2\sqrt{2}$
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