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Common Core: High School - Number and Quantity : Vector & Matrix Quantities

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

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All Common Core: High School - Number and Quantity Resources

6 Diagnostic Tests 49 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #8 : Represent Scalar Multiplication Graphically: Ccss.Math.Content.Hsn Vm.B.5a

If v=<10,2> , what is -19v ?

Possible Answers:

<190,-38>

<190,38>

<-38,-190>

<-190,-38>

<-190,38>

Correct answer:

<-190,-38>

Explanation:

Multiplying a vector by a scalar is as simple as multiplying the scalar by each vector component. In equation form it looks like this.

$x=k\cdot(x\: \mbox{component})$

$y=k\cdot(y\: \mbox{component})$

where is the scalar.

In this case it is

$x=10\cdot-19=-190$

$y=2\cdot-19=-38$

The final answer is then <-190,-38> .

There is a visual representation below.

The solid red arrow is , and the dashed blue line is -19v .

Screen shot 2016 03 07 at 3.54.13 pm

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Example Question #2 : Represent Scalar Multiplication Graphically: Ccss.Math.Content.Hsn Vm.B.5a

If v=<1,0> , what is 10v ?

Possible Answers:

<0,10>

<1,10>

<0,0>

<10,0>

<10,10>

Correct answer:

<10,0>

Explanation:

Multiplying a vector by a scalar is as simple as multiplying the scalar by each vector component. In equation form it looks like this.

$x=k\cdot(x\: \mbox{component})$

$y=k\cdot(y\: \mbox{component})$

where is the scalar.

In this case it is

$x=1\cdot10=10$

$y=0\cdot10=0$

The final answer is then <10,0> .

There is a visual representation below.

The solid red arrow is , and the dashed blue line is 10v .

Screen shot 2016 03 07 at 3.59.16 pm

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Example Question #81 : Vector & Matrix Quantities

If v=<-3,8> , what is 17v ?

Possible Answers:

<51,-136>

<-51,136>

<51,136>

<136,-51>

<-51,-136>

Correct answer:

<-51,136>

Explanation:

Multiplying a vector by a scalar is as simple as multiplying the scalar by each vector component. In equation form it looks like this.

$x=k\cdot(x\: \mbox{component})$

$y=k\cdot(y\: \mbox{component})$

where is the scalar.

In this case it is

$x=-3\cdot17=-51$

$y=8\cdot17=136$

The final answer is then <-51,136> .

There is a visual representation below.

The solid red arrow is , and the dashed blue line is 17v .

Screen shot 2016 03 07 at 4.07.14 pm

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Example Question #82 : Vector & Matrix Quantities

If v=<-9,2> , what is -15v ?

Possible Answers:

<-135,-30>

<135,-30>

<135,30>

<-135,30>

<-30,135>

Correct answer:

<135,-30>

Explanation:

Multiplying a vector by a scalar is as simple as multiplying the scalar by each vector component. In equation form it looks like this.

$x=k\cdot(x\: \mbox{component})$

$y=k\cdot(y\: \mbox{component})$

where is the scalar.

In this case it is

$x=-9\cdot-15=135$

$y=2\cdot-15=-30$

The final answer is then <135,-30> .

There is a visual representation below.

The solid red arrow is , and the dashed blue line is -15v .

Screen shot 2016 03 07 at 4.20.14 pm

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Example Question #1 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Calculate $\left \| -3v\right \|$ , where v=<0,2> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| -3\:v\right \|=6$ , Direction is the same as .

$\left \| -3\:v\right \|=6$ , Direction is away from .

$\left \| -3\:v\right \|=-6$ , Direction is away from .

$\left \| -3\:v\right \|=3$ , Direction is the same as .

$\left \| -3\:v\right \|=-3$ , Direction is away from .

Correct answer:

$\left \| -3\:v\right \|=6$ , Direction is away from .

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=3\cdot\sqrt{0^2+2^2}=3\cdot\sqrt{0+4}=3\sqrt{4}=3\cdot2=6$

As for the direction of the vector, since c<0 , the resulting vector will be against the original vector .

See below for a picture.

Vecscale

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Example Question #2 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Calculate $\left \| 1v\right \|$ , where v=<-2,-4> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| 1\:v\right \|=2\sqrt{5}$ , Direction is the same as .

$\left \| 1\:v\right \|=2$ , Direction is the same as .

$\left \| 1\:v\right \|=2\sqrt{5}$ , Direction is away from .

$\left \| 1\:v\right \|=\sqrt{5}$ , Direction is the same as .

$\left \| 1\:v\right \|=-2\sqrt{5}$ , Direction is away from .

Correct answer:

$\left \| 1\:v\right \|=2\sqrt{5}$ , Direction is the same as .

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=1\cdot\sqrt{(-2)^2+(-4)^2}=1\cdot\sqrt{4+16}=1\sqrt{20}=2\sqrt{5}$

As for the direction of the vector, since c>0 , the resulting vector will be in the same direction as the original vector

. Screen shot 2016 03 07 at 10.10.02 am

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Example Question #2 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Calculate $\left \| 2v\right \|$ , where v=<1,2> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| 2\:v\right \|=2$ , Direction is the same as .

$\left \| 2\:v\right \|=2\sqrt{5}$ , Direction is away from .

$\left \| 2\:v\right \|=4\sqrt{5}$ , Direction is away from .

$\left \| 2\:v\right \|=2\sqrt{5}$ , Direction is the same as

$\left \| 2\:v\right \|=\sqrt{5}$ , Direction is the same as .

Correct answer:

$\left \| 2\:v\right \|=2\sqrt{5}$ , Direction is the same as

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=2\cdot\sqrt{1^2+2^2}=2\cdot\sqrt{1+4}=2\sqrt{5}$

As for the direction of the vector, since c>0 , the resulting vector will be in the same direction as the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.16.42 am

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Example Question #83 : Vector & Matrix Quantities

Calculate $\left \| -4v\right \|$ , where v=<-3,-1> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| -4\:v\right \|=4\sqrt{10}$ , Direction is away from .

$\left \| -4\:v\right \|=4\sqrt{10}$ , Direction is the same as .

$\left \| -4\:v\right \|=-4$ , Direction is away from .

$\left \| -4\:v\right \|=-4$ , Direction is the same as .

$\left \| -4\:v\right \|=-4\sqrt{10}$ , Direction is away from .

Correct answer:

$\left \| -4\:v\right \|=4\sqrt{10}$ , Direction is away from .

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=4\cdot\sqrt{(-3)^2+(-1)^2}=4\cdot\sqrt{9+1}=4\sqrt{10}$

As for the direction of the vector, since c<0 , the resulting vector will be against the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.20.27 am

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Example Question #5 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Calculate $\left \| -9v\right \|$ , where v=<1,4> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| -9\:v\right \|=9\sqrt{17}$ , Direction is the same as .

$\left \| -9\:v\right \|=9\sqrt{17}$ , Direction is away from .

$\left \| -9\:v\right \|=9$ , Direction is away from .

$\left \| -9\:v\right \|=\sqrt{17}$ , Direction is the same as .

$\left \| -9\:v\right \|=-9\sqrt{17}$ , Direction is away from .

Correct answer:

$\left \| -9\:v\right \|=9\sqrt{17}$ , Direction is away from .

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=9\cdot\sqrt{1^2+4^2}=9\cdot\sqrt{1+16}=9\sqrt{17}$

As for the direction of the vector, since c<0 , the resulting vector will be against the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.32.32 am

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Example Question #84 : Vector & Matrix Quantities

Calculate $\left \| -2v\right \|$ , where v=<4,10> . Also determine the direction of the resulting vector.

Possible Answers:

$\left \| -2\:v\right \|=-4\sqrt{29}$ , Direction is away from .

$\left \| -2\:v\right \|=2\sqrt{29}$ , Direction is the same as .

$\left \| -2\:v\right \|=4\sqrt{29}$ , Direction is away from .

$\left \| -2\:v\right \|=-2\sqrt{29}$ , Direction is away from .

$\left \| -2\:v\right \|=4\sqrt{29}$ , Direction is the same as .

Correct answer:

$\left \| -2\:v\right \|=4\sqrt{29}$ , Direction is away from .

Explanation:

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

Now lets calculate this.

$=2\cdot\sqrt{4^2+10^2}=2\cdot\sqrt{16+100}=2\sqrt{116}=4\sqrt{29}$

As for the direction of the vector, since c<0 , the resulting vector will be against the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.42.39 am

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All Common Core: High School - Number and Quantity Resources

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Common Core: High School - Number and Quantity : Vector & Matrix Quantities

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 #8 : Represent Scalar Multiplication Graphically: Ccss.Math.Content.Hsn Vm.B.5a

Example Question #2 : Represent Scalar Multiplication Graphically: Ccss.Math.Content.Hsn Vm.B.5a

Example Question #81 : Vector & Matrix Quantities

Example Question #82 : Vector & Matrix Quantities

Example Question #1 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Example Question #2 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Example Question #2 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Example Question #83 : Vector & Matrix Quantities

Example Question #5 : Compute Magnitude And Direction Of A Scalar Multiple: Ccss.Math.Content.Hsn Vm.B.5b

Example Question #84 : Vector & Matrix Quantities

All Common Core: High School - Number and Quantity Resources

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