## Precalculus (10th Edition)

$-3i-5j+4k.$
If a vector $v$ initiates at point $P_1(x_1,y_1,z_1)$ and terminates at $P_2(x_2,y_2,z_2)$ then $v$ is equal to the position vector $v=(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k.$ Hence here $v=(-3-0)i+(-5-0)j+(4-0)k\\v-3i-5j+4k.$