Find the Angle Between the Vectors Using Dot Product

$(1, 0)$ , $(0, 1)$

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.

$a⃗ \cdot b⃗ = 1 \cdot 0 + 0 \cdot 1$

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 $0$ by $1$ .

$a⃗ \cdot b⃗ = 0 + 0 \cdot 1$

Step 2.2.1.2

Multiply $0$ by $1$ .

$a⃗ \cdot b⃗ = 0 + 0$

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.

$| a⃗ | = \sqrt{1^{2} + 0^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

One to any power is one.

$| a⃗ | = \sqrt{1 + 0^{2}}$

Step 3.2.2

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

$| a⃗ | = \sqrt{1 + 0}$

Step 3.2.3

Add $1$ and $0$ .

$| a⃗ | = \sqrt{1}$

Step 3.2.4

Any root of $1$ is $1$ .

$| a⃗ | = 1$

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.

$| b⃗ | = \sqrt{0^{2} + 1^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

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

$| b⃗ | = \sqrt{0 + 1^{2}}$

Step 4.2.2

One to any power is one.

$| b⃗ | = \sqrt{0 + 1}$

Step 4.2.3

Add $0$ and $1$ .

$| b⃗ | = \sqrt{1}$

Step 4.2.4

Any root of $1$ is $1$ .

$| b⃗ | = 1$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{0}{1 \cdot 1})$

Step 6

Simplify.

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Step 6.1

Multiply $1$ by $1$ .

$θ = \arccos (\frac{0}{1})$

Step 6.2

Divide $0$ by $1$ .

$θ = \arccos (0)$

Step 6.3

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

$θ = 90$

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