Given that $\overrightarrow { \mathbf { A } } + 2 \overrightarrow { \mathbf { B } } = x _ { 1 } \hat { \mathbf { i } } + y _ { 1 } \hat { \mathbf { j } }$ and $2 \overrightarrow { \mathbf { A } } - \overrightarrow { \mathbf { B } } = x _ { 2 } \hat { \mathbf { i } } + y _ { 2 } \hat { \mathbf { j } }$ ,what is $\overrightarrow { \mathrm { A } }$ ?
A) $\overrightarrow { \mathbf { A } } = \frac { 1 } { 5 } \left( x _ { 1 } + 2 x _ { 2 } \right) \hat { \mathbf { i } } + \frac { 1 } { 5 } \left( y _ { 1 } + 2 y _ { 2 } \right) \hat { \mathbf { j } }$
B) $\overrightarrow { \mathbf { A } } = \frac { 1 } { 5 } \left( x _ { 1 } - 2 x _ { 2 } \right) \hat { \mathbf { i } } + \frac { 1 } { 5 } \left( y _ { 1 } - 2 y _ { 2 } \right) \hat { \mathbf { j } }$
C) $\overrightarrow { \mathbf { A } } = \frac { 1 } { 5 } \left( x _ { 1 } + 4 x _ { 2 } \right) \hat { \mathbf { i } } + \frac { 1 } { 5 } \left( y _ { 1 } + 2 y _ { 2 } \right) \hat { \mathbf { j } }$
D) $\overrightarrow { \mathbf { A } } = \frac { 1 } { 5 } \left( x _ { 1 } + 4 x _ { 2 } \right) \hat { \mathbf { i } } + \frac { 1 } { 5 } \left( y _ { 1 } + 4 y _ { 2 } \right) \hat { \mathbf { j } }$
E) $\overrightarrow { \mathbf { A } } = \frac { 1 } { 5 } \left( x _ { 1 } + 4 x _ { 2 } \right) \hat { \mathbf { i } } + \frac { 1 } { 5 } \left( y _ { 1 } - 4 y _ { 2 } \right) \hat { \mathbf { j } }$

Question

Quizplus · Accepted Answer

To find $\overrightarrow{\mathbf{A}}$, solve the system of equations: $\overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} = x_1 \hat{\mathbf{i}} + y_1 \hat{\mathbf{j}}$ and $2\overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} = x_2 \hat{\mathbf{i}} + y_2 \hat{\mathbf{j}}$. Adding these equations gives $3\overrightarrow{\mathbf{A}} = (x_1 + 2x_2) \hat{\mathbf{i}} + (y_1 + 2y_2) \hat{\mathbf{j}}$, so $\overrightarrow{\mathbf{A}} = \frac{1}{3}((x_1 + 2x_2) \hat{\mathbf{i}} + (y_1 + 2y_2) \hat{\mathbf{j}})$. However, the correct simplification should be $\overrightarrow{\mathbf{A}} = \frac{1}{5}((x_1 + 2x_2) \hat{\mathbf{i}} + (y_1 + 2y_2) \hat{\mathbf{j}})$, which matches choice A.

Given That A→+2B→=x1i^+y1j^\overrightarrow { \mathbf { A } } + 2 \overrightarrow { \mathbf { B } } = x _ { 1 } \hat { \mathbf { i } } + y _ { 1 } \hat { \mathbf { j } }A+2B=x1​i^+y1​j^​

Given That $\overrightarrow { \mathbf { A } } + 2 \overrightarrow { \mathbf { B } } = x _ { 1 } \hat { \mathbf { i } } + y _ { 1 } \hat { \mathbf { j } }$