$\text { Given } \mathbf { v } = \{ 1,6,6 \} \text {, find } \frac { 3 } { 2 } \mathbf { v } \text {. }$
A) $\mathbf { v } = \left\langle \frac { 2 } { 3 } , 9,9 \right)$
B) $\mathbf { v } = \left( \frac { 2 } { 3 } , 3,3 \right\}$
C) $\mathbf { v } = \left( \frac { 3 } { 2 } , 9,9 \right)$
D) $\mathbf { v } = \left( \frac { 3 } { 2 } , 3,3 \right)$
E) $\mathbf { v } = \left( \frac { 3 } { 2 } , 6,6 \right\}$

Question

$\text { Given } \mathbf { v } = \{ 1,6,6 \} \text {, find } \frac { 3 } { 2 } \mathbf { v } \text {. }$ 
A) $\mathbf { v } = \left\langle \frac { 2 } { 3 } , 9,9 \right)$
B) $\mathbf { v } = \left( \frac { 2 } { 3 } , 3,3 \right\}$
C) $\mathbf { v } = \left( \frac { 3 } { 2 } , 9,9 \right)$
D) $\mathbf { v } = \left( \frac { 3 } { 2 } , 3,3 \right)$
E) $\mathbf { v } = \left( \frac { 3 } { 2 } , 6,6 \right\}$

Quizplus · Accepted Answer

Multiplying a vector by a scalar multiplies all its components by that scalar. Thus, $\frac{3}{2}\mathbf{v} = \left( \frac{3}{2}, 9, 9\right)$. Therefore, the answer is choice C.

Given v={1,6,6}, find 32v. \text { Given } \mathbf { v } = \{ 1,6,6 \} \text {, find } \frac { 3 } { 2 } \mathbf { v } \text {. } Given v={1,6,6}, find 23​v.

$\text { Given } \mathbf { v } = \{ 1,6,6 \} \text {, find } \frac { 3 } { 2 } \mathbf { v } \text {. }$