A vector is a quantity which has both magnitudes, as well as direction. A vector which has a magnitude of 1 is a unit vector. It is also known as Direction Vector.Â
For example, vector v = (1,3) is not a unit vector, because its magnitude is not equal to 1, i.e., |v| =Â âˆš(1^{2}+3^{2})Â â‰ 1.Â Any vector can become a unit vector by dividing it by the magnitude of the given vector.
Unit Vector Symbol:
Unit Vector is represented by the symbol ‘^’, which is called as cap or hat, such as: \(\hat{a}\). It is given by \(\hat{a}= \frac{a}{|a|}\)
Where |a| is for norm or magnitude of vector a.
It can be calculated using a Unit vector formula or by using a calculator.
Unit vectors are usually determined to form the base of a vector space. Every vector in the space can be expressed as a linear combination of unit vectors. The dot products of two unit vectors is a scalar quantity whereas the cross product of two arbitrary unit vectors results in third vector orthogonal to both of them.
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What is the unit normal vector?
The normal vector is a vector which is perpendicular to the surface at a given point. It is also called “normal,” to a surface is a vector. When normals are estimated on closed surfaces, the normal pointing towards the interior of the surface and outward-pointing normal are usually discovered.Â The unit vector acquired by normalizing the normal vector is the unit normal vector, also known as the “unit normal.”Here, weÂ divide a nonzero normal vector by its vector norm.
Unit Vector Formula
As explained above vectors have both magnitude (Value) and a direction. They are shown with an arrow \(\vec{a}\). \(\hat{a}\) denotes a unit vector. If we want to change any vector in unit vector, divide it by the vectorâ€™s magnitude. Usually, xyz coordinates are used to write any vector.
It can be done in two ways:
- \(\vec{a}\) = (x, y, z) using the brackets.
- \(\vec{a}\) = x\(\hat{i}\) + y \(\hat{j}\) +z \(\hat{k}\)
Formula for magnitude of a vector is:
\(\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}+z^{2}}\) |
Unit Vector = \(\frac{Vector}{Vectorâ€™s magnitude}\) |
The above is a unit vector formula.
How to find the unit vector?
To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude. For example, consider a vector v = (1, 4) which has a magnitude of |v|. If we divide each component of vector v by |v| we will get the unit vector u_{v} which is in the same direction as v.
How to represent Vector in a bracket format?
\(\hat{a} \hat{\equiv }\frac{a}{\left|a \right |} =\frac{(x,y,z)}{\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\)
How to represent Vector in a unit vector component format?
\(\hat{a}\hat{\equiv }\frac{a}{\left |a \right | } =\frac{x\hat{i}+ y\hat{j} +z \hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} =(\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{i},\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{j},\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{k})\)Where x, y, z are the value of the vector in the x, y, z axis respectively and
\(\hat{a}\) is a unit vector, \(\vec{a}\) is a vector, \(\left | \vec{a} \right |\) is the magnitude of the vector \(\vec{a}, \hat{i}, \hat{j}, \hat{k}\) are the directed unit vectors along the x , y , z axis.Unit Vector Example
Here is an example based on the unit vector. Observe and follow each step and solve problems based on it.
Question 1:
Find the unit vector \(\vec{p}\) for the given vector, 12\(\hat{i}\) – 3\(\hat{j}\) – 4 \(\hat{k}\). Show it in both the formats – Bracket and Unit vector component.
Solution: Letâ€™s find the magnitude of the given vector first, \(\vec{p}\) is :
\(\left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{144 + 9 + 16} \left |p \right | = \sqrt{169} \left |p \right | = 13\)Letâ€™s use this magnitude to find the unit vector now:
\(\hat{p} = \frac{p}{|p|} = \frac{x\hat{i}+y \hat{j} +z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\)=\(\hat{p}=\frac{12\hat{i}-3 \hat{j} – 4\hat{j}}{13}\)
= \(\hat{p} = \frac{12}{13}\hat{i} -\frac{3}{13}\hat{j}-\frac{4}{13}\hat{k}\)
The unit vector in Bracket form is:
\(\hat{p} = (\frac{(12, -3, -4)}{13} \hat{p} = (\frac{(12)}{13}, -\frac{(3)}{13},\frac{(-4)}{13})\)Unit Vector Problem
Question 2:
Find the unit vector \(\vec{q}\) for the given vector, \(-2\hat{i} + 4\hat{j} – 4 \hat{k}.\). Show it in both the formats – Bracket and Unit vector component.
Solution: Letâ€™s find the magnitude of the given vector first, \(\vec{q}\) is :
\(\left |q \right |\) = \(\sqrt{x^{2}+y^{2}+z^{2}}\) \(\left |q \right | = \sqrt{-2^{2}+(4)^{2}+(-4)^{2}}\) \(\left |q \right |\) = \(\sqrt{4 + 16 + 16}\) \(\left |q \right |\) = \(\sqrt{36}\) \(\left |q \right |\) = 6Letâ€™s use this magnitude to find the unit vector now:
\(\hat{q}= \frac{q}{|q|} ={\frac{x\hat{i} + y\hat{j} +z \hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} \hat{q} = \frac{-2\hat{i} +4 \hat{j} – 4\hat{q}}{6}}= \frac{-2}{6}\hat{i} + \frac{4}{6} \hat{j} -\frac{4}{6}\hat{k}\) \(\hat{q}= \frac{-2\hat{i} +4 \hat{j}- 4 \hat{k}} 6\) \(\hat{p} = \frac{-2}{6}\hat{i} + \frac{4}{6} \hat{j} -\frac{4}{6}\hat{k}\)The unit vector in Bracket form is:
\(\hat{p} = \frac{(-2, 4, -4)}{6} \hat{p} = \frac{(-2)}{6}, \frac{(4)}{6}, \frac{(-4)}{6}= \frac{(-1)}{3}, \frac{(2)}{3}, \frac{(-2)}{3}\)To study more on Vectors and related Mathematical Topics, Visit BYJUâ€™S.