# Chapter 13 - Vector Geometry - 13.1 Vectors in the Plane - Exercises - Page 650: 44

The required vector is $\frac{12}{5}i+ \frac{9}{5}j$.

#### Work Step by Step

Since the vector is in the direction of $v$ then it will be in the form $av= 4ai+3aj$. Now, since the length is $3$, then we have $$\sqrt{16a^2+9a^2}=\sqrt{25a^2}=3\Longrightarrow 5a=3\Longrightarrow a= \frac{3}{5}.$$ The required vector is $\frac{12}{5}i+ \frac{9}{5}j$.

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