The magnitude of a vector formula helps to know the length of a vector, a quantity that has a definite direction along with magnitude.

There are two types of quantities – Scalar and Vector. The scalar quantities only have a magnitude while vector quantities refer to those which have direction along with magnitude.

The magnitude of a vector, also known as modulus of vector, is the measurement of length between the initial point A and the terminal point B of a vector AB and is expressed as |AB^{→}|

Hence, it can be said that the vectors have the same magnitude although their direction is different (opposite to each other).

If there is a vector AB where coordinates of A and B are (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) respectively, then its magnitude denoted by AB is given by the formula AB = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}

In particular, for A (0, 0, 0) and B (x, y, z), it will be

r^{→} = |AB^{→}| = xi^ + yj^ + zk^

And |AB^{→}| = √x^{2} +y^{2} + z^{2}, where r is the position vector of the point B and xi^, yj^, zk^ are called vector components of r.

## Solved Examples

**Question 1. Find the magnitude of the vector MN joining the points M (1, 2, 3) and N (0, -1, 5).**

**Solution:** Here, x_{1} = 1, y_{1} = 2, z_{1} = 3, x_{2} = 0, y_{2} = -1, z_{2} = 5

Therefore, magnitude =|MN| = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}

Or magnitude = √(0 – 1)^{2} + (-1 – 2)^{2} + (5 – 3)^{2}

Or magnitude = √( – 1)^{2} + (-3)^{2} + (2)^{2}

Or magnitude = √(1 + 9 + 4)

Or magnitude = √14

**Question 2. What will be the value of x for which x(i^+j^+k^) is a unit vector?**

**Solution:** Since it is given to be a unit vector, its magnitude is 1, therefore, | x(i^+j^+k^) | = 1

Applying the magnitude of vector formula,

√x^{2} + x^{2} + x^{2} = 1

Or √3 x^{2} = 1

Or √3x = 1

Or x = 1/√3 or -1/√3