Answer
$\lambda=\pm\frac{1}{\sqrt{ 13}}$.
Work Step by Step
The length is given by
$$|\lambda|\sqrt{4+9}=1\Longrightarrow 13\lambda^2=1. $$
Hence, $\lambda=\pm\frac{1}{\sqrt{ 13}}$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 13 - Vector Geometry - 13.1 Vectors in the Plane - Exercises - Page 650: 49
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