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Common Core: High School - Number and Quantity : Magnitude and Direction of a Vector Sum: CCSS.Math.Content.HSN-VM.B.4b

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Example Questions

Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=10$ , and $\left \| w\right \|=5$ , and the angle in between them is $60^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=\sqrt{75}$

$\left \| v+w\right \|=\sqrt{85}$

$\left \| v+w\right \|=7$

$\left \| v+w\right \|=\sqrt{65}$

$\left \| v+w\right \|=15$

Correct answer:

$\left \| v+w\right \|=\sqrt{75}$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=10 , b=5 , and $C=60^{\circ}$ , then

$c^2=10^2+5^2-2\cdot10\cdot5\cos(60^{\circ})$

$c^2=100+25-20\cdot5\cos(60^{\circ})$

$c^2=125-100\cos(60^{\circ})$

$c^2=125-100\cdot\frac{1}{2}$

c^2=125-50

c^2=75

$\sqrt{c^2}=\sqrt{75}$

$c=\sqrt{75}\approx8.66$

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Example Question #145 : High School: Number And Quantity

If $\left \| v\right \|=11$ , and $\left \| w\right \|=21$ , and the angle in between them is $21^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=11.43$

$\left \| v+w\right \|=12$

$\left \| v+w\right \|=11.6$

$\left \| v+w\right \|=11.5$

$\left \| v+w\right \|=\sqrt{85}$

Correct answer:

$\left \| v+w\right \|=11.43$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=11 , b=21 , and $C=21^{\circ}$ , then

$c^2=11^2+21^2-2\cdot11\cdot21\cos(21^{\circ})$

$c^2=121+441-462\cos(21^{\circ})$

$c^2=562-462\cos(21^{\circ})$

$c^2=562-462\cdot0.9336$

c^2=562-431.3142

c^2=130.6858

$\sqrt{c^2}=\sqrt{130.6858}$

$c=\sqrt{130.6858}\approx11.43$

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Example Question #146 : High School: Number And Quantity

If $\left \| v\right \|=8$ , and $\left \| w\right \|=22$ , and the angle in between them is $103^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=21.04$

$\left \| v+w\right \|=24.04$

$\left \| v+w\right \|=22.04$

$\left \| v+w\right \|=25.04$

$\left \| v+w\right \|=23.04$

Correct answer:

$\left \| v+w\right \|=25.04$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=8 , b=22 , and $C=103^{\circ}$ , then

$c^2=8^2+22^2-2\cdot8\cdot22\cos(103^{\circ})$

$c^2=64+484-16\cdot22\cos(103^{\circ})$

$c^2=548-352\cos(103^{\circ})$

$c^2=548-352\cdot-0.2249511$

c^2=548+79.18277

c^2=627.1828

$\sqrt{c^2}=\sqrt{627.1828}$

$c=\sqrt{627.1828}\approx25.04$

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Example Question #147 : High School: Number And Quantity

If $\left \| v\right \|=1$ , and $\left \| w\right \|=38$ , and the angle in between them is $116^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=38.45$

$\left \| v+w\right \|=38.5$

$\left \| v+w\right \|=40.8$

$\left \| v+w\right \|=38.6$

$\left \| v+w\right \|=41.34$

Correct answer:

$\left \| v+w\right \|=38.45$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=1 , b=38 , and $C=116^{\circ}$ , then

$c^2=1^2+38^2-2\cdot1\cdot38\cos(116^{\circ})$

$c^2=1+1444-2\cdot38\cos(116^{\circ})$

$c^2=1445-76\cos(116^{\circ})$

$c^2=1445-76\cdot-.4383711$

c^2=1445+33.31621

c^2=1478.316

$\sqrt{c^2}=\sqrt{1478.316}$

$c=\sqrt{1478.316}\approx38.45$

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Example Question #2 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=18$ , and $\left \| w\right \|=16$ , and the angle in between them is $26^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=20.27$

$\left \| v+w\right \|=9.62$

$\left \| v+w\right \|=11.43$

$\left \| v+w\right \|=4.67$

$\left \| v+w\right \|=7.89$

Correct answer:

$\left \| v+w\right \|=7.89$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=10 , b=5 , and $C=60^{\circ}$ , then

$c^2=18^2+16^2-2\cdot18\cdot16\cos(26^{\circ})$

$c^2=324+256-36\cdot16\cos(26^{\circ})$

$c^2=580-576\cos(26^{\circ})$

c^2=580-517.7054

c^2=62.2946

$\sqrt{c^2}=\sqrt{62.2946}$

$c=\sqrt{62.2946}\approx7.89$

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Example Question #149 : High School: Number And Quantity

If $\left \| v\right \|=16$ , and $\left \| w\right \|=19$ , and the angle in between them is $150^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=31.42$

$\left \| v+w\right \|=36.87$

$\left \| v+w\right \|=40.58$

$\left \| v+w\right \|=33.82$

$\left \| v+w\right \|=30.8$

Correct answer:

$\left \| v+w\right \|=33.82$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=16 , b=19 , and $C=150^{\circ}$ , then

$c^2=16^2+19^2-2\cdot16\cdot19\cos(150^{\circ})$

$c^2=256+361-32\cdot19\cos(150^{\circ})$

$c^2=617-608\cos(150^{\circ})$

$c^2=617-608\cdot-0.8660254$

c^2=1143.543

$\sqrt{c^2}=\sqrt{1143.543}$

$c=\sqrt{1143.543}\approx33.82$

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Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=34$ , and $\left \| w\right \|=17$ , and the angle in between them is $10^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=13.56$

$\left \| v+w\right \|=17.51$

$\left \| v+w\right \|=18.09$

$\left \| v+w\right \|=14.69$

$\left \| v+w\right \|=17.84$

Correct answer:

$\left \| v+w\right \|=17.51$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=17 , b=10 , and $C=10^{\circ}$ , then

$c^2=17^2+34^2-2\cdot34\cdot17\cos(10^{\circ})$

$c^2=289+1156-34\cdot34\cos(10^{\circ})$

$c^2=1445-1156\cos(10^{\circ})$

c^2=1445-1138.438

c^2=306.562

$\sqrt{c^2}=\sqrt{306.562}$

$c=\sqrt{306.562}\approx17.51$

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Example Question #2 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=18$ , and $\left \| w\right \|=19$ , and the angle in between them is $131^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=33.67$

$\left \| v+w\right \|=32.89$

$\left \| v+w\right \|=34.90$

$\left \| v+w\right \|=37.92$

$\left \| v+w\right \|=31.56$

Correct answer:

$\left \| v+w\right \|=33.67$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=18 , b=2 , and $C=106^{\circ}$ , then

$c^2=18^2+19^2-2\cdot18\cdot19\cos(131^{\circ})$

$c^2=324+361-36\cdot19\cos(131^{\circ})$

$c^2=685-684\cos(131^{\circ})$

$c^2=685-684\cdot-0.656059$

c^2=685+448.7444

c^2=1133.744

$\sqrt{c^2}=\sqrt{1133.744}$

$c=\sqrt{1133.744}\approx33.67$

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Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=4$ , and $\left \| w\right \|=23$ , and the angle in between them is $149^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=28.13$

$\left \| v+w\right \|=27.76$

$\left \| v+w\right \|=27.23$

$\left \| v+w\right \|=24.57$

$\left \| v+w\right \|=26.42$

Correct answer:

$\left \| v+w\right \|=27.23$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=4 , b=23 , and $C=149^{\circ}$ , then

$c^2=4^2+23^2-2\cdot4\cdot23\cos(149^{\circ})$

$c^2=16+529-8\cdot23\cos(149^{\circ})$

$c^2=545-184\cos(149^{\circ})$

$c^2=545-184\cdot-0.8571673$

c^2=584+157.7188

c^2=741.7188

$\sqrt{c^2}=\sqrt{741.7188}$

$c=\sqrt{741.7188}\approx27.23$

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Example Question #4 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

If $\left \| v\right \|=14$ , and $\left \| w\right \|=1$ , and the angle in between them is $76^{\circ}$ , find the magnitude of the resultant vector.

Possible Answers:

$\left \| v+w\right \|=14.09$

$\left \| v+w\right \|=13.79$

$\left \| v+w\right \|=17.94$

$\left \| v+w\right \|=14.25$

$\left \| v+w\right \|=13.62$

Correct answer:

$\left \| v+w\right \|=13.79$

Explanation:

In order to find the resultants magnitude, we need to use the law of cosines. The law of cosines is $c^2=a^2+b^2-2ab\cos(C)$ . Suppose that a=14 , b=1 , and $C=76^{\circ}$ , then

$c^2=14^2+1^2-2\cdot14\cdot1\cos(76^{\circ})$

$c^2=196+1-28\cdot1\cos(76^{\circ})$

$c^2=197-28\cos(76^{\circ})$

$c^2=197-28\cdot0.2419219$

c^2=197-6.773813

c^2=190.2262

$\sqrt{c^2}=\sqrt{190.2262}$

$c=\sqrt{190.2262}\approx13.79$

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Common Core: High School - Number and Quantity : Magnitude and Direction of a Vector Sum: CCSS.Math.Content.HSN-VM.B.4b

Study concepts, example questions & explanations for Common Core: High School - Number and Quantity

All Common Core: High School - Number and Quantity Resources

Example Questions

Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #145 : High School: Number And Quantity

Example Question #146 : High School: Number And Quantity

Example Question #147 : High School: Number And Quantity

Example Question #2 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #149 : High School: Number And Quantity

Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #2 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #1 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #4 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

All Common Core: High School - Number and Quantity Resources

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