Compute Magnitude and Direction of a Scalar Multiple: CCSS.Math.Content.HSN-VM.B.5b

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Common Core: High School - Number and Quantity › Compute Magnitude and Direction of a Scalar Multiple: CCSS.Math.Content.HSN-VM.B.5b

Questions 1 - 10

Calculate $\left | -10v\right |$ , where . Also determine the direction of the resulting vector.

$\left | -10:v\right |=10\sqrt{85}$ , Direction is away from .

$\left | -10:v\right |=-10\sqrt{85}$ , Direction is away from .

$\left | -10:v\right |=10\sqrt{85}$ , Direction is the same as .

$\left | -10:v\right |=\sqrt{85}$ , Direction is the same as .

$\left | -10:v\right |=-\sqrt{85}$ , 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.

$=10\cdot\sqrt{(-6)^2+7^2}=10\cdot\sqrt{36+49}=10\sqrt{85}$

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

See below for a picture.

Screen shot 2016 03 07 at 12.02.14 pm

Calculate $\left | 7v\right |$ , where . Also determine the direction of the resulting vector.

$\left | 7:v\right |=70$ , Direction is away from .

$\left | 7:v\right |=-70$ , Direction is away from .

$\left | 7:v\right |=70$ , Direction is the same as .

$\left | 7:v\right |=7$ , Direction is the same as .

$\left | 7:v\right |=-7$ , 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.

$=7\cdot\sqrt{8^2+6^2}=7\cdot\sqrt{64+36}=7\sqrt{100}=7\cdot10=70$

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

See below for a picture.

Screen shot 2016 03 07 at 12.10.09 pm

Calculate $\left | 11v\right |$ , where . Also determine the direction of the resulting vector.

$\left | 11:v\right |=11\sqrt{26}$ , Direction is away from .

$\left | 11:v\right |=-11\sqrt{26}$ , Direction is away from .

$\left | 11:v\right |=11\sqrt{26}$ , Direction is the same as .

$\left | 11:v\right |=\sqrt{26}$ , Direction is the same as .

$\left | 11:v\right |=-\sqrt{26}$ , 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.

$=11\cdot\sqrt{5^2+(-1)^2}=11\cdot\sqrt{25+1}=11\sqrt{26}$

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

See below for a picture.

Screen shot 2016 03 07 at 11.45.20 am

Calculate $\left | -9v\right |$ , where . Also determine the direction of the resulting vector.

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

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

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

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

$\left | -9:v\right |=9$ , 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 , the resulting vector will be against the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.32.32 am

Calculate $\left | 13v\right |$ , where . Also determine the direction of the resulting vector.

$\left | 13:v\right |=78$ , Direction is away from .

$\left | 13:v\right |=-78$ , Direction is away from .

$\left | 13:v\right |=78$ , Direction is the same as .

$\left | 13:v\right |=13$ , Direction is the same as .

$\left | 13:v\right |=-13$ , 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.

$=13\cdot\sqrt{0^2+6^2}=13\cdot\sqrt{0+36}=13\sqrt{36}=13\cdot6=78$

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

See below for a picture.

Screen shot 2016 03 07 at 11.09.17 am

Calculate $\left | -12v\right |$ , where . Also determine the direction of the resulting vector.

$\left | -12:v\right |=48\sqrt{2}$ , Direction is away from .

$\left | -12:v\right |=-48\sqrt{2}$ , Direction is away from .

$\left | -12:v\right |=48\sqrt{2}$ , Direction is the same as .

$\left | -12:v\right |=\sqrt{2}$ , Direction is the same as .

$\left | -12:v\right |=-\sqrt{2}$ , 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.

$=12\cdot\sqrt{4^2+4^2}=12\cdot\sqrt{16+16}=12\sqrt{32}=48\sqrt{2}$

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

See below for a picture.

Screen shot 2016 03 07 at 11.53.48 am

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

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

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

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

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

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

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

As for the direction of the vector, since , 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

Calculate $\left | -4v\right |$ , where . Also determine the direction of the resulting vector.

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

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

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

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

$\left | -4:v\right |=-4$ , 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 , the resulting vector will be against the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.20.27 am

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

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

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

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

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

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

. Screen shot 2016 03 07 at 10.10.02 am

Calculate $\left | 10v\right |$ , where . Also determine the direction of the resulting vector.

$\left | 10:v\right |=10\sqrt{5}$ , Direction is away from .

$\left | 10:v\right |=-10\sqrt{5}$ , Direction is away from .

$\left | 10:v\right |=10\sqrt{5}$ , Direction is the same as .

$\left | 10:v\right |=5\sqrt{5}$ , Direction is the same as .

$\left | 5:v\right |=-5\sqrt{5}$ , 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.

$=10\cdot\sqrt{2^2+(-1)^2}=10\cdot\sqrt{4+1}=10\sqrt{5}$

As for the direction of the vector, since , the resulting vector in the same direction as the original vector .

See below for a picture.

Screen shot 2016 03 07 at 10.54.13 am

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