## Thomas' Calculus 13th Edition

$\lt 6,-8 \gt$ and $10$
The magnitude of a vector can be calculated as: $|v|=\sqrt{v_1^2+v_2^2}$ Now,$-2u=-2\lt -3,4 \gt =\lt 6,-8 \gt$ and $|\lt 6,-8\gt|=\sqrt{(6)^2+(-8)^2}=\sqrt {100}=10$