The vector v and its initial point are given.Find the terminal point. $\mathbf { v } = \left( 4 , \frac { 3 } { 2 } , - \frac { 1 } { 4 } \right)$
Initial point: $\left( 2,1 , - \frac { 3 } { 2 } \right)$
A)Terminal point is ( $\frac { 5 } { 2 }$ ,6, $- \frac { 7 } { 4 }$ ).
B)Terminal point is ( $- \frac { 7 } { 4 }$ , $\frac { 5 } { 2 }$ ,6).
C)Terminal point is (0, $\frac { 5 } { 2 }$ , $- \frac { 7 } { 4 }$ ).
D)Terminal point is (6, $\frac { 5 } { 2 }$ , $- \frac { 7 } { 4 }$ ).
E)Terminal point is (6,6,6).

Question

The vector v and its initial point are given.Find the terminal point.  $\mathbf { v } = \left( 4 , \frac { 3 } { 2 } , - \frac { 1 } { 4 } \right)$
Initial point: $\left( 2,1 , - \frac { 3 } { 2 } \right)$ 
A)Terminal point is ( $\frac { 5 } { 2 }$ ,6, $- \frac { 7 } { 4 }$ ).
B)Terminal point is ( $- \frac { 7 } { 4 }$ , $\frac { 5 } { 2 }$ ,6).
C)Terminal point is (0, $\frac { 5 } { 2 }$ , $- \frac { 7 } { 4 }$ ).
D)Terminal point is (6, $\frac { 5 } { 2 }$ , $- \frac { 7 } { 4 }$ ).
E)Terminal point is (6,6,6).

Quizplus · Accepted Answer

To find the terminal point, add the vector components to the initial point's coordinates: $2 + 4 = 6$, $1 + \frac{3}{2} = \frac{5}{2}$, $-\frac{3}{2} - \frac{1}{4} = -\frac{7}{4}$.

The Vector V and Its Initial Point Are Given v=(4,32,−14)\mathbf { v } = \left( 4 , \frac { 3 } { 2 } , - \frac { 1 } { 4 } \right)v=(4,23​,−41​)

The Vector V and Its Initial Point Are Given $\mathbf { v } = \left( 4 , \frac { 3 } { 2 } , - \frac { 1 } { 4 } \right)$