The magnitude and the direction of D → = A → + B → + C → .

Question

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Answer 1

Question

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

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Answer 2

Question

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

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Answer 3

Question

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

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Answer 4

Question

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

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Answer 5

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

Answer 6

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Answer 7

Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of D→=A→+B→+C→ .

Answer 8

(a)

Expert Solution

Answer to Problem 31P

The magnitude of D→=A→+B→+C→ is 2.83 m in the direction of 315° counterclockwise from the positive x -axis.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Section 1:

To determine: The magnitude of D→=A→+B→+C→ .

Answer: The magnitude of D→=A→+B→+C→ is 2.83 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of D→=A→+B→+C→ is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

D=(3 m+1 m+(−2 m))2+(−3 m+(−4 m)+5 m)2=2.83 m

Conclusion:

Therefore, the magnitude of D→=A→+B→+C→ is 2.83 m .

Section 2:

To determine: The direction of D→=A→+B→+C→ .

Answer: The direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the direction of D→=A→+B→+C→ is,

θ=tan−1(Ay+By+CyAx+Bx+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((−3 m+(−4 m)+5 m)(3 m+1 m+(−2 m)))=tan−1(−2 m2 m)=−45°

Thus, the angle measured from counterclockwise is 360°−45°=315°

Conclusion:

Therefore, the direction of D→=A→+B→+C→ is 315° counterclockwise from positive x -axis.

Answer 13

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

Answer 14

(b)

To determine

The magnitude and the direction of E→=−A→−B→+C→ .

Answer 15

(b)

Expert Solution

Answer to Problem 31P

The magnitude of E→=−A→−B→+C→ is 13.42 m in the direction of 296.57° counterclockwise from the positive x -axis.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Section 1:

To determine: The magnitude of E→=−A→−B→+C→ .

Answer: The magnitude of E→=−A→−B→+C→ is 13.42 m .

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to calculate the magnitude of E→=−A→−B→+C→ is,

E=((−Ax)+(−Bx)+Cx)2+((−Ay)+(−By)+Cy)2

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

E=((−3 m)+(−1 m)+(−2 m))2+(3 m+4 m+5 m)2=180 m=13.42 m

Conclusion:

Therefore, the magnitude of E→=−A→−B→+C→ is 13.42 m .

Section 2:

To determine: The direction of E→=−A→−B→+C→ .

Answer: The direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A→=(3i^−3j^) m , B→=(i^−4j^) m and C→=(−2i^+5j^) m .

Formula to determine the direction of E→=−A→−B→+C→ is,

θ=tan−1((−Ay)+(−By)+Cy(−Ax)+(−Bx)+Cx)

Substitute 3 m for Ax , −3 m for Ay , 1 m for Bx , −4 m for By , −2 m for Cx and 5 m for Cy in above expression.

θ=tan−1((3 m+(4 m)+5 m)(−3 m+(−1 m)+(−2 m)))=tan−1(12 m−6 m)=−63.43°

Thus, the angle measured from counterclockwise is 360°−63.43°=296.57°

Conclusion:

Therefore, the direction of E→=−A→−B→+C→ is 296.57° counterclockwise from the positive x -axis.

Answer 20

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The magnitude and the direction of D → = A → + B → + C → .

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The magnitude and the direction of D→=A→+B→+C→ .

Answer to Problem 31P

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The magnitude and the direction of E→=−A→−B→+C→ .

Answer to Problem 31P

Explanation of Solution

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