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# Vector calculator

This calculator performs all vector operations in two- and three-dimensional space. You can add, subtract, find length, find vector projections, and find the dot and cross product of two vectors. For each operation, the calculator writes a step-by-step, easy-to-understand explanation of how the work has been done.

## Find the magnitude of the vector $\| \vec{v} \| = \left(\dfrac{ 2 }{ 3 },~\sqrt{ 3 },~2\right)$ .

solution

The magnitude of vector $\vec{v}$ is $\| \vec{v} \| = \dfrac{\sqrt{ 67 }}{ 3 }$ .

explanation

To find magnitude of a vector $v=(a,b,c)$ we use formula $||v|| = \sqrt{a^2+b^2+c^2}$

In this example $a = \frac{ 2 }{ 3 }$ , $b = \sqrt{ 3 }$ and $c = 2$ so:

$$||v|| = \sqrt{ \left(\frac{ 2 }{ 3 }\right)^2 + \left( \sqrt{ 3 } \right)^{ 2 } + 2^2} = \sqrt{ \frac{ 4 }{ 9 } + 3 + 4} = \sqrt{ \frac{ 67 }{ 9 } } = \frac{\sqrt{ 67 }}{ 3 }$$

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Vectors in two dimensions
seven operations on two-dimensional vectors + steps
help ↓↓ examples ↓↓ tutorial ↓↓
V1(1,-3) V2(5,1/2)
V1(√2,-1/3) V2(√5,0)
 Magnitude (length) of V1 V1 · V2 ( dot product ) V1 + V2 V1 - V2 Angle between V1 and V2 Check if V1 and V2 are linearly dependent Find projection of V1 onto V2
Find approximate solution
Hide steps
Vectors in three dimensions
seven operations on three-dimensional vectors + steps
help ↓↓ examples ↓↓ tutorial ↓↓
V1=(1,-√3,3/2) V2=(√2, 1, 2/3)
 Magnitude (length) of V1 V1 · V2 ( dot product ) V1 ✕ V2 ( cross product ) V1 + V2 V1 - V2 Angle between V1 and V2 Check if V1, V2 and V3 are linearly dependent Find projection of V1 onto V2
Find approximate solution
Hide steps
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Examples
ex 1:
Given vector v1=(8,-4), calculate the magnitude.
ex 2:
Calculate the difference of vectors v1=(3/4,2) and v2=(3,-2).
ex 3:
Calculate the dot product of vectors v1=(-1/4,2/5) and v2=(-5,-5/4)
ex 4:
Find the angle between the vectors v1=(3,5,-7) and v2=(-3,4,-2)
ex 5:
Find the cross product of v1=(-2,2/3,-3) and v2=(4,0,-1/2)
ex 6:
Determine if the following set of vectors is linearly independent: v1=(3,-2,4), v2=(1,-2,3) and v3=(3,2,-1)
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TUTORIAL

## Vector operations

In this tutorial, we'll learn how to find: magnitude, dot product, angle between two vectors and cross product of two vectors.

### 1 : Magnitude

Magnitude is the vector length. The formula for the magnitude of a vector $\vec{v} = (v_1, v_2)$ is:

$$\| \vec{v} \| = \sqrt{v_1^2 + v_2^2 }$$

Example 01: Find the magnitude of the vector $\vec{v} = (4, 2)$.

In this example we have $v_1 = 4$ and $v_2 = 2$ so the magnitude is:

$$\| \vec{v} \| = \sqrt{4^2 + 2 ^2} = \sqrt{20} = 2\sqrt{5}$$

Example 02: Find the magnitude of the vector $\vec{v} = \left(\dfrac{2}{3}, \sqrt{3}, 2\right)$.

### 2 : Dot product

The formula for the dot product of vectors $\vec{v} = (v_1, v_2)$ and $\vec{w} = (w_1, w_2)$ is

$$\vec{v} \cdot \vec{w}= v_1 \cdot w_1 + v_2 \cdot w_2$$

Two vectors are orthogonal to each other if their dot product is equal zero.

Example 03: Calculate the dot product of $\vec{v} = \left(4, 1 \right)$ and $\vec{w} = \left(-1, 5 \right)$. Check if the vectors are mutually orthogonal.

To find the dot product we use the component formula:

\begin{aligned} \vec{v} \cdot \vec{w} &= \left(4, 1 \right) \cdot \left(-1, 5 \right) = \\[1 em] &= 4 \cdot (-1) + 1 \cdot 5 = -4 + 5 = 1 \end{aligned}

Since the dot product is not equal to zero, we can conclude that vectors ARE NOT orthogonal.

Example 04: Find the dot product of the vectors $\vec{v_1} = \left(\dfrac{1}{2}, \sqrt{3}, 5 \right)$ and $\vec{v_2} = \left( 4, -\sqrt{3}, 10 \right)$.

### 3 : Angle between two vectors

To find the angle $\alpha$ between vectors $\vec{a}$ and $\vec{b}$, we use the following formula:

$$\cos \alpha = \dfrac{\vec{a} \cdot \vec{b}}{ \|\vec{a}\| \, \|\vec{b}\|}$$

Note that $\vec{a} \cdot \vec{b}$ is a dot product while $\|\vec{a}\|$ and $\|\vec{b}\|$ are magnitudes of vectors $\vec{a}$ and $\vec{b}$.

Example 05: Find the angle between vectors $\vec{a} = ( 4, 3)$ and $\vec{b} = (-2, 2)$.

First we will find the dot product and magnitudes:

\begin{aligned} \vec{a} \cdot \vec{b} &= ( 4, 3) \cdot (-2, 2) = 4 \cdot (-2) + 3 \cdot 2 = -2 \\[1 em] \| \vec{a} \| &= \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \\[1 em] \| \vec{b} \| &= \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \end{aligned}

Now we'll find the $\cos \alpha$

$$\cos \alpha = \dfrac{\vec{a} \cdot \vec{b}}{ \|\vec{a}\| \, \|\vec{b}\|} = \dfrac{-2}{5 \sqrt{8}} \approx -0.1414$$

The angle $\alpha$ is:

$$\alpha = \cos^{-1}(-0.1414) \approx 98^{o}$$

Example 06: Find the angle between vectors $\vec{v_1} = \left(2, 1, -4 \right)$ and $\vec{v_2} = \left( 3, -5, 2 \right)$.

### 4 : Cross product

The cross product of vectors $\vec{v} = (v_1,v_2,v_3)$ and $\vec{w} = (w_1,w_2,w_3)$ is given by the formula:

$$\vec{v} \times \vec{w} = \left( v_2 w_3 - v_3 w_2, v_3 w_1 - v_1 w_3, v_1 w_2 - v_2 w_1 \right)$$

Note that the cross product requires both vectors to be three-dimensional.

If the two vectors are parallel than the cross product is equal zero.

Example 07: Find the cross products of the vectors $\vec{v} = ( -2, 3 , 1)$ and $\vec{w} = (4, -6, -2)$. Check if the vectors are parallel.

We'll find cross product using above formula

v • w = (3×(-2)-1×(-6), 1×4-(-2)×(-2), -2×(-6) - 3×4) = (0, 0, 0)

Since the cross product is zero, we conclude that the vectors are parallel.

Example 08: Find the cross products of the vectors $\vec{v_1} = \left(4, 2, -\dfrac{3}{2} \right)$ and $\vec{v_2} = \left(\dfrac{1}{2}, 0, 2 \right)$.

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