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: 38

Answer

$|\vec{u}|=\sqrt{13}$ $|\vec{v}|=\sqrt{5}$ $|2\vec{u}|=2\sqrt{13}$ $|\frac{1}{2}\vec{v}|=\dfrac{\sqrt{5}}{2}$ $|\vec{u}+\vec{v}|=\sqrt{2}$ $|\vec{u}-\vec{v}|=\sqrt{34}$ $|\vec{u}|-|\vec{v}|=\sqrt{13}-\sqrt{5}$

Work Step by Step

$\vec{u}=-2\vec{i}+3\vec{j}$ $;$ $\vec{v}=\vec{i}-2\vec{j}$ 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 $\vec{u}$, $a_{1}=-2$ and $a_{2}=3$ $|\vec{u}|=\sqrt{(-2)^{2}+3^{2}}=\sqrt{4+9}=\sqrt{13}$ $|\vec{v}|$ For the vector $\vec{v}$, $a_{1}=1$ and $a_{2}=-2$ $|\vec{v}|=\sqrt{1^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5}$ $|2\vec{u}|$ Multiply both components of $\vec{u}$ by $2$: $2\vec{u}=2(-2\vec{i}+3\vec{j})=-4\vec{i}+6\vec{j}$ For the vector $2\vec{u}$, $a_{1}=-4$ and $a_{2}=6$: $|2\vec{u}|=\sqrt{(-4)^{2}+6^{2}}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$ $|\frac{1}{2}\vec{v}|$ Multiply both components of $\vec{v}$ by $\dfrac{1}{2}$: $\dfrac{1}{2}\vec{v}=\dfrac{1}{2}(\vec{i}-2\vec{j})=\dfrac{1}{2}\vec{i}-\vec{j}$ For the vector $|\frac{1}{2}\vec{v}|$, $a_{1}=\dfrac{1}{2}$ and $a_{2}=-1$: $|\frac{1}{2}\vec{v}|=\sqrt{\Big(\dfrac{1}{2}\Big)^{2}+(-1)^{2}}=\sqrt{\dfrac{1}{4}+1}=\sqrt{\dfrac{5}{4}}=\dfrac{\sqrt{5}}{2}$ $|\vec{u}+\vec{v}|$ Add the vectors $\vec{u}$ and $\vec{v}$ together and simplify: $\vec{u}+\vec{v}=-2\vec{i}+3\vec{j}+\vec{i}-2\vec{j}=(-2+1)\vec{i}+(3-2)\vec{j}=-\vec{i}+\vec{j}$ For the vector $\vec{u}+\vec{j}$, $a_{1}=-1$ and $a_{2}=1$: $|\vec{u}+\vec{v}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}$ $|\vec{u}-\vec{v}|$ Subtract vectors $\vec{u}$ and $\vec{v}$ and simplify: $\vec{u}-\vec{v}=(-2\vec{i}+3\vec{j})-(\vec{i}-2\vec{j})=-2\vec{i}+3\vec{j}-\vec{i}+2\vec{j}=...$ $...=(-2-1)\vec{i}+(3+2)\vec{j}=-3\vec{i}+5\vec{j}$ For the vector $\vec{u}-\vec{v}$, $a_{1}=-3$ and $a_{2}=5$: $|\vec{u}-\vec{v}|=\sqrt{(-3)^{2}+5^{2}}=\sqrt{9+25}=\sqrt{34}$ $|\vec{u}|-|\vec{v}|$ Subtract the magnitude of $\vec{v}$ from the magnitude of $\vec{u}$. These two magnitudes where found in the first two parts of this problem. $|\vec{u}|=\sqrt{13}$ and $|\vec{v}|=\sqrt{5}$ $|\vec{u}|-|\vec{v}|=\sqrt{13}-\sqrt{5}$
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