Find a unit vector in the direction of the vector described below. Initial point: (-9,3,-2)
Terminal point: (4,-7,3)
A) $\sqrt { 294 } ( 13 , - 10,5 )$
B) $\frac { 1 } { \sqrt { 28 } } ( 13 , - 10,5 )$
C) $\sqrt { 28 } ( 13 , - 10,5 )$
D) $( 13 , - 10,5 )$
E) $\frac { 1 } { \sqrt { 294 } } ( 13 , - 10,5 )$

Question

Quizplus · Accepted Answer

The vector from the initial point to the terminal point is (13, -10, 5). Its magnitude is $\sqrt{13^2 + (-10)^2 + 5^2} = \sqrt{294}$. The unit vector is obtained by dividing the vector by its magnitude, resulting in $\frac{1}{\sqrt{294}}(13, -10, 5)$.

Find a Unit Vector in the Direction of the Vector 294(13,−10,5)\sqrt { 294 } ( 13 , - 10,5 )294​(13,−10,5)

Find a Unit Vector in the Direction of the Vector $\sqrt { 294 } ( 13 , - 10,5 )$