Unit Vector: Practice Problems, Formula

A unit vector is defined with the following properties
   (i) Its magnitude is always 1.
   (ii) It has no unit.
   (iii) It is parallel to a coordinate axis or any arbitrary vector. In the other words, this vector describes a direction.

Therefore, any vector with these properties is called a unit vector.

If each component of an arbitrary vector is divided by its magnitude, the resulting vector is a unit vector.

Recall that a vector in physics is defined as a quantity that has both a magnitude and a direction.

Therefore, the unit vector in the direction of vector $\vec{w}$ with magnitude of $|\vec{w}|$ is found by the following formula \[\hat{w}=\frac{\vec{w}}{|\vec{w}|}\] To distinguish a unit vector from an ordinary vector, a caret or $\hat{}$ is used.

If you are getting ready for AP physics exam 1, check this:
Vectors in the AP physics exam.

In the following, a number of simple problems about unit vectors are answered that are useful for high school students.

Unit Vector Practice Problems

Problem (1): Find the unit vector in the direction $\vec{w}=(5,2)$.

Solution: A unit vector in physics is defined as a dimensionless vector whose magnitude is exactly 1.

A unit vector that points in the direction of $\vec{A}$ is determined by formula \[\hat{A}=\frac{\vec{A}}{|\vec{A}|}\] Where $|A|$ is the magnitude of the vector $\vec{A}$ with the following formula \[|\vec{A}|=\sqrt{A_x^2+A_y^2}\]

To find the unit vector points in the direction of the given vector, $\vec{W}$, first find its magnitude as \[|\vec{w}|=\sqrt{5^2+2^2}=\sqrt{29}\] Next divide the given vector by its magnitude as below \[\hat{w}=\frac{(5,2)}{\sqrt{29}}=\left(\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}\right)\] Therefore, the unit vector in the direction of $\vec{w}=(5,2)$ is $\left(\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}\right)$.

Problem (2): What is the unit vector point in the direction of $\vec{v}=(-4,-8)$.

Solution: The unit vector along an arbitrary vector is computed as the vector divided by its magnitude.

The magnitude of the given vector is found by the formula below \[|\vec{v}|=\sqrt{(-4)^2+(-8)^2}=\sqrt{80}\] Therefore, applying unit vector definition, we have \[\hat{v}=\frac{(-4,-8)}{\sqrt{80}}=\left(\frac{-1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\right)\] where in the last step, we used the simplification $\sqrt{80}=\sqrt{16\times 5}=4\sqrt{5}$.

Problem (3): Find the unit vector in the direction of sum of two vectors $\vec{v}=(2,-4)$ and $\vec{w}=(-3,2)$.

Solution: the sum of two given vectors, we call it $\vec{c}$, is calculated as below \begin{align*} \vec{c}&=\vec{v}+\vec{w}\\&=(2,-4)+(-3,2)\\&=(2+(-3),-4+2)\\&=(-1,-2)\end{align*} The magnitude of this vector is also found as \[|\vec{c}|=\sqrt{(-1)^2+(-2)^2}=\sqrt{5}\] Now, according to unit vector definition, divide the obtained vector by its magnitude to find the unit vector points in the direction of sum of the two vectors given as below \[\hat{c}=\frac{(-1,-2)}{\sqrt{5}}=\left(\frac{-1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\right)\]

Problem (4): Find a vector in the direction of unit vector $\hat{v}=\left(1/3,\sqrt{8}/3\right)$ and a magnitude of 11.

Solution: By definition, a unit vector has a length (magnitude) of 1. To check this condition for the given vector, we use the Pythagorean theorem to find its magnitude as \[|\hat{v}|=\sqrt{(1/3)^2+(\sqrt{8}/3)^2}=1\] As expected for a unit vector. With the help of unit vector definition, we can construct a vector with an arbitrary magnitude.

If you scale up or down a unit vector by a constant coefficient, then you construct a vector with the magnitude of that coefficient.

Consequently, the vector $\vec{w}=k\hat{v}$, where $k$ is a constant coefficient, has a magnitude of $k$.

To find a vector with magnitude of 11 and in the direction of the unit vector given above, $\hat{v}$, we simply scale up that unit vector by 11. Hence, the desired vector is \[\vec{u}=11\hat{v}=\left(\frac{11}{3},\frac{11\sqrt{8}}{3}\right)\]
where in the last step, we used the definition of multiplication of a vector by a constant coefficient (scalar) as \[k(a,b)=(ka,kb)\]

Problem (5): Find the unit vector in the direction of the resultant vectors $\vec{A}=2\hat{i}-3\hat{j}+4\hat{k}$ and $\vec{B}=-\hat{i}+\hat{j}+2\hat{k}$.

Solution: The vectors are in space and their components were given. The sum of two vectors is called resultant.

By applying vector addition formula, the components of the resultant vector, $\vec{C}=\vec{A}+\vec{B}$ are found as below \begin{align*}\vec{C}&=\vec{A}+\vec{B}\\&=(2,-3,4)+(-1,1,2)\\&=(2+(-1),-3+1,4+2)\\&=(1,-2,6)\end{align*} Thus, the resultant vector is $\vec{C}=(1,-2,6)$.

Using unit vector definition, the components of vector divided by its magnitude, the unit vector in the direction of the resultant vector is found as follows \begin{align*}\hat{c}&=\frac{\vec{C}}{|\vec{C}|}\\\\&=\frac{(1,-2,6)}{\sqrt{1^2+(-2)^2+6^2}}\\\\&=\left(\frac {1}{\sqrt{41}} ,\frac {-2}{\sqrt{41}} ,\frac {6}{\sqrt{41}}\right)\end{align*}

More related:
Practice problems on vectors

Problem (6): In the following figure, (a) write each vector in terms of unit vectors $\hat{i}$ and $\hat{j}$, (b) Use unit vector definition to express the vector $\vec{C}=3\vec{A}-2\vec{B}$.

Solution: The notation $\hat{i}$ and $\hat{j}$ are the unit vectors (magnitude of 1) in the direction of x and y axes. Here, the magnitude and direction (angle) of the vectors are given.

(a) First, resolve the vectors into their components.

The components of a vector $\vec{A}$ with magnitude of $|\vec{A}|$ are determined by the following formula \begin{gather*} \vec{A}_x=|\vec{A}|\cos\theta \\ \vec{A}_y=|\vec{A}|\sin\theta \end{gather*} We cal also vector sum the above components to write the vector in the usual form as below \[\vec{A}=\vec{A}_x\hat{i}+\vec{A}_y\hat{j}\] Therefore, the decomposition of the given vectors are \begin{align*} \vec{A}_x&=6\cos 63^\circ =2.72\,{\rm m}\\ \vec{A}_y&=6\sin 63^\circ=5.34 \\\\ \vec{B}_x&=5\cos 27^\circ=4.45 \\ \vec{B}_y&=5\sin 27^\circ=2.7\end{align*} The relationship between components of a ordinary vector and unit vectors in the $x$ and $y$ directions, are as below \begin{gather*} \vec{A}_x=|\vec{A}|\hat{i}\\\vec{A}_y=|\vec{A}|\hat{j}\end{gather*} Combining all things together gives the following results for the given vectors in terms of units vectors \begin{gather*} \vec{A}=2.72\hat{i}+5.34\hat{j} \\ \vec{B}=4.45\hat{i}+2.7\hat{j}\end{gather*}
(b) We are to multiply the vector $\vec{A}$ by 2 and subtract 2 times of vector $\vec{B}$ from the result.

According to the rules of multiplication of a vector by a scalar (a constant coefficient), to multiply $\vec{A}$ by 2, we must multiply each of its components by 2 as below \[2\vec{A}=2(2.72,5.34)=(5.44,10.68)\] Similarly, \[2\vec{B}=2(4.45,2.7)=(8.9,5.4)\] To subtract two vectors from each other, simply subtract its components as below \begin{align*}\vec{C}&=3\vec{A}-2\vec{B}\\&=(5.44,10.68)-(8.9,5.4)\\&=(-3.46,5.28)\end{align*} Using the Pythagorean theorem the magnitude of this vector is found as \[|\vec{C}|=\sqrt{(-3.46)^2+(5.28)^2}=6.31\] To find the unit vector in the direction of the vector $\vec{C}$, we use the definition of unit vector \begin{align*} \hat{c}&=\frac{\vec{C}}{|\vec{C}|}\\\\&=\frac{(-3.46,5.28)}{6.31}\\\\&=(0.55,0.84)\end{align*} We leave this to you to check that the magnitude of this vector is 1.

Author: Ali Nemati
Date Published: 6/19/2021