## University Calculus: Early Transcendentals (3rd Edition)

$\lt9, -6\gt$ and $3 \sqrt {13}$
Formula to find the magnitude of a vector can be found as: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $3u=3\lt 3, -2 \gt =\lt9, -6\gt$ and $|\lt9, -6\gt|=\sqrt{9^2+(-6)^2}=\sqrt {117}=3 \sqrt {13}$ Hence, our final answers are: $\lt9, -6\gt$ and $3 \sqrt {13}$