Find the Angle Between the Vectors Using Dot Product

$(1, - 2)$ , $(- 2, 1)$

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.

$a⃗ \cdot b⃗ = 1 \cdot - 2 - 2 \cdot 1$

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 $- 2$ by $1$ .

$a⃗ \cdot b⃗ = - 2 - 2 \cdot 1$

Step 2.2.1.2

Multiply $- 2$ by $1$ .

$a⃗ \cdot b⃗ = - 2 - 2$

Step 2.2.2

Subtract $2$ from $- 2$ .

$a⃗ \cdot b⃗ = - 4$

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.

$| a⃗ | = \sqrt{1^{2} + {(- 2)}^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

One to any power is one.

$| a⃗ | = \sqrt{1 + {(- 2)}^{2}}$

Step 3.2.2

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

$| a⃗ | = \sqrt{1 + 4}$

Step 3.2.3

Add $1$ and $4$ .

$| a⃗ | = \sqrt{5}$

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.

$| b⃗ | = \sqrt{{(- 2)}^{2} + 1^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

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

$| b⃗ | = \sqrt{4 + 1^{2}}$

Step 4.2.2

One to any power is one.

$| b⃗ | = \sqrt{4 + 1}$

Step 4.2.3

Add $4$ and $1$ .

$| b⃗ | = \sqrt{5}$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{- 4}{\sqrt{5} \sqrt{5}})$

Step 6

Simplify.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

Raise $\sqrt{5}$ to the power of $1$ .

$θ = \arccos (\frac{- 4}{{\sqrt{5}}^{1} \sqrt{5}})$

Step 6.1.2

Raise $\sqrt{5}$ to the power of $1$ .

$θ = \arccos (\frac{- 4}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}})$

Step 6.1.3

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

$θ = \arccos (\frac{- 4}{{\sqrt{5}}^{1 + 1}})$

Step 6.1.4

Add $1$ and $1$ .

$θ = \arccos (\frac{- 4}{{\sqrt{5}}^{2}})$

Step 6.2

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Step 6.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$θ = \arccos (\frac{- 4}{{(5^{\frac{1}{2}})}^{2}})$

Step 6.2.2

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

$θ = \arccos (\frac{- 4}{5^{\frac{1}{2} \cdot 2}})$

Step 6.2.3

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

$θ = \arccos (\frac{- 4}{5^{\frac{2}{2}}})$

Step 6.2.4

Cancel the common factor of $2$ .

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Step 6.2.4.1

Cancel the common factor.

$θ = \arccos (\frac{- 4}{5^{\frac{2}{2}}})$

Step 6.2.4.2

Rewrite the expression.

$θ = \arccos (\frac{- 4}{5^{1}})$

Step 6.2.5

Evaluate the exponent.

$θ = \arccos (\frac{- 4}{5})$

Step 6.3

Move the negative in front of the fraction.

$θ = \arccos (- \frac{4}{5})$

Step 6.4

Evaluate $\arccos (- \frac{4}{5})$ .

$θ = 143.13010235$

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