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

$(4,-3,5)$
Since, we have $\overrightarrow{AB}=i+4j-2k$ and suppose $A=(p,q,r)$ $\overrightarrow{AB}=(5-p)i+(1-q)j+(3-r)k=i+4j-2k$ Now, $5-p=1, 1-q=4, 3-r=-2$ Thus, $p=4,q=-3,r=5$ Hence, $A=(4,-3,5)$