Let $P _ { 1 } = ( 6,2,1 )$ and $P _ { 2 } = ( 3,0,2 )$ v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle - 3 , - 2,1 \rangle$
B) $\langle 3,2 , - 1 \rangle$
C) $\langle 3 , - 2,1 \rangle$
D) $\langle - 3 , - 2 , - 1 \rangle$
E) $\langle - 3,2,1 \rangle$

Question

Let $P _ { 1 } = ( 6,2,1 )$ and $P _ { 2 } = ( 3,0,2 )$ v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle - 3 , - 2,1 \rangle$
B) $\langle 3,2 , - 1 \rangle$
C) $\langle 3 , - 2,1 \rangle$
D) $\langle - 3 , - 2 , - 1 \rangle$
E) $\langle - 3,2,1 \rangle$

Quizplus · Accepted Answer

To find the vector $\overrightarrow{P_1P_2}$, subtract the coordinates of $P_1$ from $P_2$: $P_2 - P_1 = (3-6, 0-2, 2-1) = \langle -3, -2, 1 \rangle$.

Let P1=(6,2,1)P _ { 1 } = ( 6,2,1 )P1​=(6,2,1) And P2=(3,0,2)P _ { 2 } = ( 3,0,2 )P2​=(3,0,2)

Let $P _ { 1 } = ( 6,2,1 )$ And $P _ { 2 } = ( 3,0,2 )$