Calculus 8th Edition

Published by Cengage

Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises: 44

Answer

$\frac{2}{\sqrt {14}}$, $\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$

Work Step by Step

Given: $a=i+2j+3k$ , $b=5i-k$ Change the form of $a$ and $b$. $a=\lt1,2,3\gt$ , $b=\lt5,0,-1\gt$ Scalar Projection $b$ onto $a$ can be calculated as follows: $\frac{a \times b }{|a|}=\frac{(1 \times 5)+( 2 \times 0)+(3 \times -1)}{\sqrt {{(1)^{2}+(2)^{2}}+(3)^{2}}}$ $=\frac{5+0-3}{\sqrt {14}}$ $=\frac{2}{\sqrt {14}}$ Vector Projection $b$ onto $a$ can be calculated as follows: $\frac{a \times b }{|a|^{2}}\times a=\frac{2}{14}\lt1,2,3\gt$ $=\lt\frac{2}{14}, \frac{4}{14},\frac{6}{14}\gt$ Change vector projection back into $i+j+k$ form. $=\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$ Hence, Scalar Projection $b$ onto $a$ = $\frac{2}{\sqrt {14}}$, Vector Projection $b$ onto $a$=$\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$

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