## Calculus 8th Edition

Published by Cengage

# Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 853: 39

#### Answer

$4$, $\lt\frac{-20}{13}, \frac{48}{13}\gt$

#### Work Step by Step

Given: $a=\lt-5,12\gt$ , $b=\lt4,6\gt$ Scalar Projection $b$ onto $a$ can be calculated as follows: $\frac{a \times b }{|a|}=\frac{(-5 \times 4)+( 12 \times 6)}{\sqrt {(5)^{2}+(12)^{2}}}$ $=\frac{-20+72}{13}$ $=\frac{52}{13}$ $=4$ Vector Projection $b$ onto $a$ can be calculated as follows: $\frac{a \times b }{|a|^{2}}\times a=\frac{4}{13}\lt-5,12\gt$ $=\lt\frac{-20}{13}, \frac{48}{13}\gt$ Hence, Scalar Projection $b$ onto $a$ = $4$, Vector Projection $b$ onto $a$=$\lt\frac{-20}{13}, \frac{48}{13}\gt$

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