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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=125-50$

c^2=75 $\displaystyle c^2=75$

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

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=562-431.3142$

$\displaystyle c^2=130.6858$

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

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

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Example Question #3 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=548+79.18277$

$\displaystyle c^2=627.1828$

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

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=1445+33.31621$

$\displaystyle c^2=1478.316$

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

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

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Example Question #5 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

$\displaystyle c^2=580-517.7054$

$\displaystyle c^2=62.2946$

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

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

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Example Question #6 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=1143.543$

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

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

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Example Question #7 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

$\displaystyle c^2=1445-1138.438$

$\displaystyle c^2=306.562$

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

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

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Example Question #8 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=685+448.7444$

$\displaystyle c^2=1133.744$

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

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=584+157.7188$

$\displaystyle c^2=741.7188$

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

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

$\displaystyle c^2=197-6.773813$

$\displaystyle c^2=190.2262$

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

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

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

Example Question #3 : 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

Example Question #5 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #6 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #7 : Magnitude And Direction Of A Vector Sum: Ccss.Math.Content.Hsn Vm.B.4b

Example Question #8 : 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 #2 : 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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