Pre-Algebra Examples

$(- 7, 5)$ , $(- 8, - 3)$

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.

$a⃗ \cdot b⃗ = - 7 \cdot - 8 + 5 \cdot - 3$

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 $- 7$ by $- 8$ .

$a⃗ \cdot b⃗ = 56 + 5 \cdot - 3$

Step 2.2.1.2

Multiply $5$ by $- 3$ .

$a⃗ \cdot b⃗ = 56 - 15$

Step 2.2.2

Subtract $15$ from $56$ .

$a⃗ \cdot b⃗ = 41$

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.

$| a⃗ | = \sqrt{{(- 7)}^{2} + 5^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

Raise $- 7$ to the power of $2$ .

$| a⃗ | = \sqrt{49 + 5^{2}}$

Step 3.2.2

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

$| a⃗ | = \sqrt{49 + 25}$

Step 3.2.3

Add $49$ and $25$ .

$| a⃗ | = \sqrt{74}$

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.

$| b⃗ | = \sqrt{{(- 8)}^{2} + {(- 3)}^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

Raise $- 8$ to the power of $2$ .

$| b⃗ | = \sqrt{64 + {(- 3)}^{2}}$

Step 4.2.2

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

$| b⃗ | = \sqrt{64 + 9}$

Step 4.2.3

Add $64$ and $9$ .

$| b⃗ | = \sqrt{73}$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{41}{\sqrt{74} \sqrt{73}})$

Step 6

Simplify.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

Combine using the product rule for radicals.

$θ = \arccos (\frac{41}{\sqrt{74 \cdot 73}})$

Step 6.1.2

Multiply $74$ by $73$ .

$θ = \arccos (\frac{41}{\sqrt{5402}})$

Step 6.2

Multiply $\frac{41}{\sqrt{5402}}$ by $\frac{\sqrt{5402}}{\sqrt{5402}}$ .

$θ = \arccos (\frac{41}{\sqrt{5402}} \cdot \frac{\sqrt{5402}}{\sqrt{5402}})$

Step 6.3

Combine and simplify the denominator.

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Step 6.3.1

Multiply $\frac{41}{\sqrt{5402}}$ by $\frac{\sqrt{5402}}{\sqrt{5402}}$ .

$θ = \arccos (\frac{41 \sqrt{5402}}{\sqrt{5402} \sqrt{5402}})$

Step 6.3.2

Raise $\sqrt{5402}$ to the power of $1$ .

$θ = \arccos (\frac{41 \sqrt{5402}}{{\sqrt{5402}}^{1} \sqrt{5402}})$

Step 6.3.3

Raise $\sqrt{5402}$ to the power of $1$ .

$θ = \arccos (\frac{41 \sqrt{5402}}{{\sqrt{5402}}^{1} {\sqrt{5402}}^{1}})$

Step 6.3.4

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

$θ = \arccos (\frac{41 \sqrt{5402}}{{\sqrt{5402}}^{1 + 1}})$

Step 6.3.5

Add $1$ and $1$ .

$θ = \arccos (\frac{41 \sqrt{5402}}{{\sqrt{5402}}^{2}})$

Step 6.3.6

Rewrite ${\sqrt{5402}}^{2}$ as $5402$ .

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Step 6.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5402}$ as $5402^{\frac{1}{2}}$ .

$θ = \arccos (\frac{41 \sqrt{5402}}{{(5402^{\frac{1}{2}})}^{2}})$

Step 6.3.6.2

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

$θ = \arccos (\frac{41 \sqrt{5402}}{5402^{\frac{1}{2} \cdot 2}})$

Step 6.3.6.3

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

$θ = \arccos (\frac{41 \sqrt{5402}}{5402^{\frac{2}{2}}})$

Step 6.3.6.4

Cancel the common factor of $2$ .

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Step 6.3.6.4.1

Cancel the common factor.

$θ = \arccos (\frac{41 \sqrt{5402}}{5402^{\frac{2}{2}}})$

Step 6.3.6.4.2

Rewrite the expression.

$θ = \arccos (\frac{41 \sqrt{5402}}{5402^{1}})$

Step 6.3.6.5

Evaluate the exponent.

$θ = \arccos (\frac{41 \sqrt{5402}}{5402})$

Step 6.4

Evaluate $\arccos (\frac{41 \sqrt{5402}}{5402})$ .

$θ = 56.09372301$