Pre-Algebra Examples

$(5, 0)$ , $(0, 9)$

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.

$a⃗ \cdot b⃗ = 5 \cdot 0 + 0 \cdot 9$

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 $5$ by $0$ .

$a⃗ \cdot b⃗ = 0 + 0 \cdot 9$

Step 2.2.1.2

Multiply $0$ by $9$ .

$a⃗ \cdot b⃗ = 0 + 0$

Step 2.2.2

Add $0$ and $0$ .

$a⃗ \cdot b⃗ = 0$

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.

$| a⃗ | = \sqrt{5^{2} + 0^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

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

$| a⃗ | = \sqrt{25 + 0^{2}}$

Step 3.2.2

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

$| a⃗ | = \sqrt{25 + 0}$

Step 3.2.3

Add $25$ and $0$ .

$| a⃗ | = \sqrt{25}$

Step 3.2.4

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

$| a⃗ | = \sqrt{5^{2}}$

Step 3.2.5

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

$| a⃗ | = 5$

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.

$| b⃗ | = \sqrt{0^{2} + 9^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

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

$| b⃗ | = \sqrt{0 + 9^{2}}$

Step 4.2.2

Raise $9$ to the power of $2$ .

$| b⃗ | = \sqrt{0 + 81}$

Step 4.2.3

Add $0$ and $81$ .

$| b⃗ | = \sqrt{81}$

Step 4.2.4

Rewrite $81$ as $9^{2}$ .

$| b⃗ | = \sqrt{9^{2}}$

Step 4.2.5

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

$| b⃗ | = 9$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{0}{5 \cdot 9})$

Step 6

Simplify.

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Step 6.1

Cancel the common factor of $0$ and $5$ .

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Step 6.1.1

Factor $5$ out of $0$ .

$θ = \arccos (\frac{5 (0)}{5 \cdot 9})$

Step 6.1.2

Cancel the common factors.

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Step 6.1.2.1

Factor $5$ out of $5 \cdot 9$ .

$θ = \arccos (\frac{5 (0)}{5 (9)})$

Step 6.1.2.2

Cancel the common factor.

$θ = \arccos (\frac{5 \cdot 0}{5 \cdot 9})$

Step 6.1.2.3

Rewrite the expression.

$θ = \arccos (\frac{0}{9})$

Step 6.2

Cancel the common factor of $0$ and $9$ .

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Step 6.2.1

Factor $9$ out of $0$ .

$θ = \arccos (\frac{9 (0)}{9})$

Step 6.2.2

Cancel the common factors.

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Step 6.2.2.1

Factor $9$ out of $9$ .

$θ = \arccos (\frac{9 \cdot 0}{9 \cdot 1})$

Step 6.2.2.2

Cancel the common factor.

$θ = \arccos (\frac{9 \cdot 0}{9 \cdot 1})$

Step 6.2.2.3

Rewrite the expression.

$θ = \arccos (\frac{0}{1})$

Step 6.2.2.4

Divide $0$ by $1$ .

$θ = \arccos (0)$

Step 6.3

The exact value of $\arccos (0)$ is $90$ .

$θ = 90$