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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 9: Q. 19 (page 612)

In Problems 19鈥�24, decompose v into two vectors $v_{1}$ and $v_{2}$ , where $v_{1}$ is parallel to w and $v_{2}$ is orthogonal to w.
$v = 2 i - 3 j w = i - j$

Short Answer

Expert verified

$v_{1} = 2.5 i - 2.5 j v_{2} = - 0.5 i - 0.5 j$

Step by step solution

01

Step 1. Concept Used

The decomposition of $v$ into $v_{1}$ and localid="1646825977563" $v_{2}$ , where $v_{1}$ is parallel to $w$ and $v_{2}$ is orthogonal to $w$ , is

localid="1646826554773" $v_{1} = \frac{v 路 w}{{|w|}^{2}} w v_{2} = v - v_{1}$

02

Step 2. Find the dot product and w

For the vectors $v = 2 i - 3 j$ and $w = i - j$ the dot product is given as

$v 路 w = (2 i - 3 j) 路 (i - j) v 路 w = 2 路 1 + (- 3) 路 (- 1) v 路 w = 2 + 3 v 路 w = 5$

The magnitude of vector $w = i - j$ is given as

$|w| = \sqrt{1^{2} + {(- 1)}^{2}} |w| = \sqrt{1 + 1} |w| = \sqrt{2}$

03

Step 3. Find v1 and v2

The vector $v_{1}$ which is parallel to $w$ is given by the formula

role="math" $v_{1} = \frac{v 路 w}{{|w|}^{2}} w$

On substituting the calculated values we get

$v_{1} = \frac{5}{{(\sqrt{2})}^{2}} (i - j) v_{1} = \frac{5}{2} (i - j) v_{1} = 2.5 i - 2.5 j$

and the vector $v_{2}$ orthogonal to $w$ is found as

$v_{2} = v - v_{1} v_{2} = 2 i - 3 j - (2.5 i - 2.5 j) v_{2} = 2 i - 3 j - 2.5 i + 2.5 j v_{2} = (2 - 2.5) i + (- 3 + 2.5) j v_{2} = - 0.5 i - 0.5 j$

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