Linear Algebra Examples

(1, - 2)

$(1, - 2)$ ,

(- 2, 1)

$(- 2, 1)$

Step 1

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

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

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

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

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

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

$- 2$ by

1

$1$ .

a⃗ \cdot b⃗ = - 2 - 2 \cdot 1

$a⃗ \cdot b⃗ = - 2 - 2 \cdot 1$

Step 2.2.1.2

Multiply

- 2

$- 2$ by

1

$1$ .

a⃗ \cdot b⃗ = - 2 - 2

$a⃗ \cdot b⃗ = - 2 - 2$

a⃗ \cdot b⃗ = - 2 - 2

$a⃗ \cdot b⃗ = - 2 - 2$

Step 2.2.2

Subtract

2

$2$ from

- 2

$- 2$ .

a⃗ \cdot b⃗ = - 4

$a⃗ \cdot b⃗ = - 4$

a⃗ \cdot b⃗ = - 4

$a⃗ \cdot b⃗ = - 4$

a⃗ \cdot b⃗ = - 4

$a⃗ \cdot b⃗ = - 4$

Step 3

Find the magnitude of

a⃗

$a⃗$ .

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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}}

$| a⃗ | = \sqrt{1^{2} + {(- 2)}^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

One to any power is one.

| a⃗ | = \sqrt{1 + {(- 2)}^{2}}

$| a⃗ | = \sqrt{1 + {(- 2)}^{2}}$

Step 3.2.2

Raise

- 2

$- 2$ to the power of

2

$2$ .

| a⃗ | = \sqrt{1 + 4}

$| a⃗ | = \sqrt{1 + 4}$

Step 3.2.3

Add

1

$1$ and

4

$4$ .

| a⃗ | = \sqrt{5}

$| a⃗ | = \sqrt{5}$

| a⃗ | = \sqrt{5}

$| a⃗ | = \sqrt{5}$

| a⃗ | = \sqrt{5}

$| a⃗ | = \sqrt{5}$

Step 4

Find the magnitude of

b⃗

$b⃗$ .

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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}}

$| b⃗ | = \sqrt{{(- 2)}^{2} + 1^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

| b⃗ | = \sqrt{4 + 1^{2}}

$| b⃗ | = \sqrt{4 + 1^{2}}$

Step 4.2.2

One to any power is one.

| b⃗ | = \sqrt{4 + 1}

$| b⃗ | = \sqrt{4 + 1}$

Step 4.2.3

Add

4

$4$ and

1

$1$ .

| b⃗ | = \sqrt{5}

$| b⃗ | = \sqrt{5}$

| b⃗ | = \sqrt{5}

$| b⃗ | = \sqrt{5}$

| b⃗ | = \sqrt{5}

$| b⃗ | = \sqrt{5}$

Step 5

Substitute the values into the formula.

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

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

Step 6

Simplify.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

Raise

\sqrt{5}

$\sqrt{5}$ to the power of

1

$1$ .

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

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

Step 6.1.2

Raise

\sqrt{5}

$\sqrt{5}$ to the power of

1

$1$ .

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

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

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

Step 6.1.4

Add

1

$1$ and

1

$1$ .

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

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

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

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

Step 6.2

Rewrite

{\sqrt{5}}^{2}

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

5

$5$ .

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Step 6.2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{5}

$\sqrt{5}$ as

5^{\frac{1}{2}}

$5^{\frac{1}{2}}$ .

θ = arccos ⎛ ⎜ ⎜ ⎝ \frac{- 4}{{(5^{\frac{1}{2}})}^{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}

${(a^{m})}^{n} = a^{m n}$ .

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

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

Step 6.2.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

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

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

Step 6.2.4

Cancel the common factor of

2

$2$ .

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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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