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

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Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 13:v\right |=\left | 13\right |\left | v\right |$

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

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

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Question

Calculate $\left | -3v\right |$ , where . Also determine the direction of the resulting vector.

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -3:v\right |=\left | -3\right |\left | v\right |$

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

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

See below for a picture.

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Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 2:v\right |=\left | 2\right |\left | v\right |$

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -4:v\right |=\left | -4\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 1:v\right |=\left | 1\right |\left | v\right |$

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

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

.

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Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -9:v\right |=\left | -9\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -2:v\right |=\left | -2\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 10:v\right |=\left | 10\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 11:v\right |=\left | 11\right |\left | v\right |$

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

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.

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Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -12:v\right |=\left | -12\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -10:v\right |=\left | -10\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 7:v\right |=\left | 7\right |\left | v\right |$

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

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.

Compare your answer with the correct one above

Question

Calculate $\left | -3v\right |$ , where . Also determine the direction of the resulting vector.

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -3:v\right |=\left | -3\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 2:v\right |=\left | 2\right |\left | v\right |$

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -4:v\right |=\left | -4\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 1:v\right |=\left | 1\right |\left | v\right |$

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

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

.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -9:v\right |=\left | -9\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | -2:v\right |=\left | -2\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 10:v\right |=\left | 10\right |\left | v\right |$

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

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

See below for a picture.

Compare your answer with the correct one above

Question

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

Answer

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

$\left | c:v\right |=\left | c\right |:\left | v\right |$ , where is a scalar.

Now lets calculate this.

$\left | 13:v\right |=\left | 13\right |\left | v\right |$

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

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.

Compare your answer with the correct one above

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