Let $P _ { 1 } = ( 3 , - 2,5 )$ and $P _ { 2 } = ( 10,4,3 )$ If v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle 7 , - 6 , - 2 \rangle$
B) $\langle 7 , - 6,2 \rangle$
C) $\langle 7,6 , - 2 \rangle$
D) $\langle 7,6,2 \rangle$
E) $\langle - 7,6 , - 2 \rangle$

Question

Let $P _ { 1 } = ( 3 , - 2,5 )$ and $P _ { 2 } = ( 10,4,3 )$ If v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle 7 , - 6 , - 2 \rangle$
B) $\langle 7 , - 6,2 \rangle$
C) $\langle 7,6 , - 2 \rangle$
D) $\langle 7,6,2 \rangle$
E) $\langle - 7,6 , - 2 \rangle$

Quizplus · Accepted Answer

To find the vector $\overrightarrow{P_1P_2}$, subtract the coordinates of $P_1$ from $P_2$: $P_2 - P_1 = (10 - 3, 4 - (-2), 3 - 5) = (7, 6, -2)$.

Let P1=(3,−2,5)P _ { 1 } = ( 3 , - 2,5 )P1​=(3,−2,5)

Let $P _ { 1 } = ( 3 , - 2,5 )$