Pre-Algebra Examples

$(\frac{1}{5}, 3)$ , $(\frac{1}{5}, 0)$

Step 1

Use the dot product formula to find the angle between two vectors.

$θ = \arccos (\frac{a⃗ \cdot b⃗}{| a⃗ | | b⃗ |})$

Step 2

Find the dot product.

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Step 2.1

The dot product of two vectors is the sum of the products of the their components.

$a⃗ \cdot b⃗ = \frac{1}{5} \cdot \frac{1}{5} + 3 \cdot 0$

Step 2.2

Simplify.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $\frac{1}{5} \cdot \frac{1}{5}$ .

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Step 2.2.1.1.1

Multiply $\frac{1}{5}$ by $\frac{1}{5}$ .

$a⃗ \cdot b⃗ = \frac{1}{5 \cdot 5} + 3 \cdot 0$

Step 2.2.1.1.2

Multiply $5$ by $5$ .

$a⃗ \cdot b⃗ = \frac{1}{25} + 3 \cdot 0$

Step 2.2.1.2

Multiply $3$ by $0$ .

$a⃗ \cdot b⃗ = \frac{1}{25} + 0$

Step 2.2.2

Add $\frac{1}{25}$ and $0$ .

$a⃗ \cdot b⃗ = \frac{1}{25}$

Step 3

Find the magnitude of $a⃗$ .

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Step 3.1

The norm is the square root of the sum of squares of each element in the vector.

$| a⃗ | = \sqrt{{(\frac{1}{5})}^{2} + 3^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

Apply the product rule to $\frac{1}{5}$ .

$| a⃗ | = \sqrt{\frac{1^{2}}{5^{2}} + 3^{2}}$

Step 3.2.2

One to any power is one.

$| a⃗ | = \sqrt{\frac{1}{5^{2}} + 3^{2}}$

Step 3.2.3

Raise $5$ to the power of $2$ .

$| a⃗ | = \sqrt{\frac{1}{25} + 3^{2}}$

Step 3.2.4

Raise $3$ to the power of $2$ .

$| a⃗ | = \sqrt{\frac{1}{25} + 9}$

Step 3.2.5

To write $9$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$| a⃗ | = \sqrt{\frac{1}{25} + 9 \cdot \frac{25}{25}}$

Step 3.2.6

Combine $9$ and $\frac{25}{25}$ .

$| a⃗ | = \sqrt{\frac{1}{25} + \frac{9 \cdot 25}{25}}$

Step 3.2.7

Combine the numerators over the common denominator.

$| a⃗ | = \sqrt{\frac{1 + 9 \cdot 25}{25}}$

Step 3.2.8

Simplify the numerator.

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Step 3.2.8.1

Multiply $9$ by $25$ .

$| a⃗ | = \sqrt{\frac{1 + 225}{25}}$

Step 3.2.8.2

Add $1$ and $225$ .

$| a⃗ | = \sqrt{\frac{226}{25}}$

Step 3.2.9

Rewrite $\sqrt{\frac{226}{25}}$ as $\frac{\sqrt{226}}{\sqrt{25}}$ .

$| a⃗ | = \frac{\sqrt{226}}{\sqrt{25}}$

Step 3.2.10

Simplify the denominator.

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Step 3.2.10.1

Rewrite $25$ as $5^{2}$ .

$| a⃗ | = \frac{\sqrt{226}}{\sqrt{5^{2}}}$

Step 3.2.10.2

Pull terms out from under the radical, assuming positive real numbers.

$| a⃗ | = \frac{\sqrt{226}}{5}$

Step 4

Find the magnitude of $b⃗$ .

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Step 4.1

The norm is the square root of the sum of squares of each element in the vector.

$| b⃗ | = \sqrt{{(\frac{1}{5})}^{2} + 0^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

Apply the product rule to $\frac{1}{5}$ .

$| b⃗ | = \sqrt{\frac{1^{2}}{5^{2}} + 0^{2}}$

Step 4.2.2

One to any power is one.

$| b⃗ | = \sqrt{\frac{1}{5^{2}} + 0^{2}}$

Step 4.2.3

Raise $5$ to the power of $2$ .

$| b⃗ | = \sqrt{\frac{1}{25} + 0^{2}}$

Step 4.2.4

Raising $0$ to any positive power yields $0$ .

$| b⃗ | = \sqrt{\frac{1}{25} + 0}$

Step 4.2.5

Add $\frac{1}{25}$ and $0$ .

$| b⃗ | = \sqrt{\frac{1}{25}}$

Step 4.2.6

Rewrite $\sqrt{\frac{1}{25}}$ as $\frac{\sqrt{1}}{\sqrt{25}}$ .

$| b⃗ | = \frac{\sqrt{1}}{\sqrt{25}}$

Step 4.2.7

Any root of $1$ is $1$ .

$| b⃗ | = \frac{1}{\sqrt{25}}$

Step 4.2.8

Simplify the denominator.

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Step 4.2.8.1

Rewrite $25$ as $5^{2}$ .

$| b⃗ | = \frac{1}{\sqrt{5^{2}}}$

Step 4.2.8.2

Pull terms out from under the radical, assuming positive real numbers.

$| b⃗ | = \frac{1}{5}$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{\frac{1}{25}}{\frac{\sqrt{226}}{5} \cdot \frac{1}{5}})$

Step 6

Simplify.

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Step 6.1

Multiply the numerator by the reciprocal of the denominator.

$θ = \arccos (\frac{1}{25} \cdot \frac{1}{\frac{\sqrt{226}}{5} \cdot \frac{1}{5}})$

Step 6.2

Multiply $\frac{\sqrt{226}}{5}$ by $\frac{1}{5}$ .

$θ = \arccos (\frac{1}{25} \cdot \frac{1}{\frac{\sqrt{226}}{5 \cdot 5}})$

Step 6.3

Multiply $5$ by $5$ .

$θ = \arccos (\frac{1}{25} \cdot \frac{1}{\frac{\sqrt{226}}{25}})$

Step 6.4

Multiply the numerator by the reciprocal of the denominator.

$θ = \arccos (\frac{1}{25} (1 \frac{25}{\sqrt{226}}))$

Step 6.5

Multiply $\frac{25}{\sqrt{226}}$ by $1$ .

$θ = \arccos (\frac{1}{25} \cdot \frac{25}{\sqrt{226}})$

Step 6.6

Cancel the common factor of $25$ .

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Step 6.6.1

Cancel the common factor.

$θ = \arccos (\frac{1}{25} \cdot \frac{25}{\sqrt{226}})$

Step 6.6.2

Rewrite the expression.

$θ = \arccos (\frac{1}{\sqrt{226}})$

Step 6.7

Multiply $\frac{1}{\sqrt{226}}$ by $\frac{\sqrt{226}}{\sqrt{226}}$ .

$θ = \arccos (\frac{1}{\sqrt{226}} \cdot \frac{\sqrt{226}}{\sqrt{226}})$

Step 6.8

Combine and simplify the denominator.

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Step 6.8.1

Multiply $\frac{1}{\sqrt{226}}$ by $\frac{\sqrt{226}}{\sqrt{226}}$ .

$θ = \arccos (\frac{\sqrt{226}}{\sqrt{226} \sqrt{226}})$

Step 6.8.2

Raise $\sqrt{226}$ to the power of $1$ .

$θ = \arccos (\frac{\sqrt{226}}{{\sqrt{226}}^{1} \sqrt{226}})$

Step 6.8.3

Raise $\sqrt{226}$ to the power of $1$ .

$θ = \arccos (\frac{\sqrt{226}}{{\sqrt{226}}^{1} {\sqrt{226}}^{1}})$

Step 6.8.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$θ = \arccos (\frac{\sqrt{226}}{{\sqrt{226}}^{1 + 1}})$

Step 6.8.5

Add $1$ and $1$ .

$θ = \arccos (\frac{\sqrt{226}}{{\sqrt{226}}^{2}})$

Step 6.8.6

Rewrite ${\sqrt{226}}^{2}$ as $226$ .

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Step 6.8.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{226}$ as $226^{\frac{1}{2}}$ .

$θ = \arccos (\frac{\sqrt{226}}{{(226^{\frac{1}{2}})}^{2}})$

Step 6.8.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$θ = \arccos (\frac{\sqrt{226}}{226^{\frac{1}{2} \cdot 2}})$

Step 6.8.6.3

Combine $\frac{1}{2}$ and $2$ .

$θ = \arccos (\frac{\sqrt{226}}{226^{\frac{2}{2}}})$

Step 6.8.6.4

Cancel the common factor of $2$ .

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Step 6.8.6.4.1

Cancel the common factor.

$θ = \arccos (\frac{\sqrt{226}}{226^{\frac{2}{2}}})$

Step 6.8.6.4.2

Rewrite the expression.

$θ = \arccos (\frac{\sqrt{226}}{226^{1}})$

Step 6.8.6.5

Evaluate the exponent.

$θ = \arccos (\frac{\sqrt{226}}{226})$

Step 6.9

Evaluate $\arccos (\frac{\sqrt{226}}{226})$ .

$θ = 86.18592516$