Let $P _ { 1 } = ( 0,1 , - 4 )$ and $P _ { 2 } = ( - 2,8,0 )$ If v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle - 2,7 , - 4 \rangle$
B) $\langle - 2,7,4 \rangle$
C) $\langle - 2,7 , - 4 \rangle$.
D) $\langle 2 , - 7,4 \rangle$
E) $\langle 2,7 , - 4 \rangle$.

Question

Let $P _ { 1 } = ( 0,1 , - 4 )$ and $P _ { 2 } = ( - 2,8,0 )$ If v is $\overrightarrow { P _ { 1 } P _ { 2 } }$ in standard position, then v is

A) $\langle - 2,7 , - 4 \rangle$
B) $\langle - 2,7,4 \rangle$
C) $\langle - 2,7 , - 4 \rangle$.
D) $\langle 2 , - 7,4 \rangle$
E) $\langle 2,7 , - 4 \rangle$.

Quizplus · Accepted Answer

To find the vector $\overrightarrow{P_1P_2}$ in standard position, subtract the coordinates of $P_1$ from $P_2$: $(-2 - 0, 8 - 1, 0 - (-4)) = (-2, 7, 4)$, which matches choice B.

Let P1=(0,1,−4)P _ { 1 } = ( 0,1 , - 4 )P1​=(0,1,−4)

Let $P _ { 1 } = ( 0,1 , - 4 )$