Here you will learn what is unit vector, representation of unit vector, formula for unit vector and how to find the unit vector with examples.
Let’s begin –
What is a Vector ?
A quantity that has both magnitude & direction is called a vector. A vector that has magnitude of 1 is called a unit vector. Any vector can become a unit vector by dividing it with the magnitude of the given vector.
In other words, A vector of unit magnitude in direction of a vector \(\vec{a}\) is called unit vector along \(\vec{a}\) and it is denoted by \(\hat{a}\).
symbolically \(\hat{a}\) = \(\vec{a}\over {|\vec{a}|}\) (provided |\(\vec{a}\)| \(\ne\) 0).
Formula for Unit Vector :
Usually, Vector are represented in Two Dimension and Three Dimension :
i) In Two Dimension, any vector can be written as \(x\hat{i}\) + \(y\hat{j}\).
Let \(\vec{a}\) = \(x\hat{i}\) + \(y\hat{j}\)
Then unit vector of \(\vec{a}\) can be calculated as,
\(\hat{a}\) = \(\vec{a}\over {|\vec{a}|}\) = \({x\hat{i} + y\hat{j}}\over {\sqrt{x^2 + y^2}}\)
Example : Find the unit vector for the vector \(-3\hat{i} + 4\hat{j}\).
Solution : \(\vec{a}\) = \(-3\hat{i} + 4\hat{j}\)
Then |\(\vec{a}\)| = \(\sqrt{(-3)^2 + 4^2}\) = 5
Unit Vector is \(\hat{a}\) = \(\vec{a}\over {|\vec{a}|}\)
\(\hat{a}\) = \(-3\hat{i} + 4\hat{j}\over 5\)
ii) In Three Dimension, any vector can be written as \(x\hat{i}\) + \(y\hat{j}\) + \(z\hat{k}\).
Let \(\vec{a}\) = \(x\hat{i}\) + \(y\hat{j}\) + \(z\hat{k}\),
Then unit vector of \(\vec{a}\) can be calculated as,
\(\hat{a}\) = \(\vec{a}\over {|\vec{a}|}\) = \({x\hat{i} + y\hat{j} + z\hat{k}}\over {\sqrt{x^2 + y^2 + z^2}}\)
Example : Find the unit vector for the vector \(3\hat{i} – 6\hat{j} + 2\hat{k}\).
Solution : \(\vec{a}\) = \(3\hat{i} – 6\hat{j} + 2\hat{k}\)
Then |\(\vec{a}\)| = \(\sqrt{3^2 + (-6)^2 + 2^2}\) = 7
and Unit Vector is \(\hat{a}\) = \(\vec{a}\over {|\vec{a}|}\)
\(\hat{a}\) = \(3\hat{i} – 6\hat{j} + 2\hat{k}\over 7\)
Hope you learnt what is unit vector and how to find unit vector, learn more concepts of scalar and vector and practice more questions to get ahead in the competition. Good luck!
