## Calculus 8th Edition

Published by Cengage

# Chapter 12 - Vectors and the Geometry of Space - 12.4 The Cross Product - 12.4 Exercises - Page 861: 7

#### Answer

$(1-t)i+(t^3-t^2)k$ Yes, $a \times b$ is orthogonal to both $a$ and $b$.

#### Work Step by Step

$a= \lt t,1,1/t \gt$ and $b= \lt t^2,t^2,1 \gt$ $a \times b= \lt t,1,1/t \gt \times \lt t^2,t^2,1 \gt$ $a \times b=(1-t)i+(t^3-t^2)k$ To verify that it is orthogonal to $a$; we will compute: $(a\times b).a=(1-t,0,t^3-t^2) \cdot (t,1,1/t)=0$ To verify that it is orthogonal to $b$; we will compute: $(a\times b).b=(1-t,0,t^3-t^2) \cdot (t^2,t^2,1)=0$ Yes, $a \times b$ is orthogonal to both $a$ and $b$.

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