Mathematics: If d subscript 1 baseline overscript right-arrow endscripts plus d subscript 2 baseline overscript right-arrow endscripts equals 7.0 d subscript 3 baseline overscript right-arrow endscripts, d subscript 1 baseline overscript right-arrow endscripts minus d subscript 2 baseline overscript right-arrow endscripts equals 1.0 d subscript 3 baseline overscript right-arrow endscripts and d overscript right-arrow endscripts subscript 3 baseline equals 2.1 i overscript ̂ endscripts plus 5.0 j overscript ̂

If d subscript 1 baseline overscript right-arrow

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nonispn606

11.03.2022

Given that

$\overrightarrow{d_1}+\overrightarrow{d_2}=5.0\overrightarrow{d_3}, \ and \\ \\ \overrightarrow{d_1}-\overrightarrow{d_2}=1.0\overrightarrow{d_3}$

Also, given that
$\overrightarrow{d_3}=2.6i+5.6j$

Let
$\overrightarrow{d_1}=x_1i+y_1j$ , and
$\overrightarrow{d_2}=x_2i+y_2j$ , then

$x_1i+y_1j+x_2i+y_2j=5.0(2.6i+5.6j) \\ \\ \Rightarrow(x_1+x_2)i+(y_1+y_2)j=13i+28j \\ \\ \Rightarrow x_1+x_2=13 \ . \ . \ . \ . \ . \ (1) \\ \\ \Rightarrow y_1+y_2=28 \ . \ . \ . \ . \ . \ (2)$

Also,

$(x_1i+y_1j)-(x_2i+y_2j)=1.0(2.6i+5.6j) \\ \\ \Rightarrow(x_1-x_2)i+(y_1-y_2)j=2.6i+5.6j \\ \\ \Rightarrow x_1-x_2=2.6 \ . \ . \ . \ . \ . \ (3) \\ \\ \Rightarrow y_1-y_2=5.6 \ . \ . \ . \ . \ . \ (4)$

Part A:

Solving for (1) and (3), we have

$x_1+x_2=13 \\ x_1-x_2=2.6 \\ \\ \Rightarrow2x_2=10.4 \\ \\ \Rightarrow x_2=5.2\\ \\ \Rightarrow x_1=7.8$

Therefore, the x-component of d_1

is 7.8

Part B

Solving for (2) and (4), we have

$y_1+y_2=28 \\ y_1-y_2=5.6 \\ \\ \Rightarrow2y_2=22.4 \\ \\ \Rightarrow y_2=11.2 \\ \\ \Rightarrow y_1=16.8$

Therefore, the y-component of d_1

is 16.8

Part C

Solving for (1) and (3), we have

$x_1+x_2=13 \\ x_1-x_2=2.6 \\ \\ \Rightarrow2x_2=10.4 \\ \\ \Rightarrow x_2=5.2\\ \\ \Rightarrow x_1=7.8$

Therefore, the x-component of d_2

is 5.2

Part D

Solving for (2) and (4), we have

$y_1+y_2=28 \\ y_1-y_2=5.6 \\ \\ \Rightarrow2y_2=22.4 \\ \\ \Rightarrow y_2=11.2 \\ \\ \Rightarrow y_1=16.8$

Therefore, the y-component of d_2

is 11.2

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