In this article, we will learn about a unit vector, various components of a vector and the addition, multiplication comparison of vectors using components. Then we will look at some examples to get a better grip on the topic.

### Suggested Videos

## Unit Vector

Let’s take a point each on the x, y, and z-axis as follows:

- A (1, 0, 0) on x-axis
- B (0, 1, 0) on y-axis and
- C (0, 0, 1) on z-axis

So, we have

|\( \vec{OA} \)| = 1 , |\( \vec{OB} \)| = 1, and |\( \vec{OC} \)| = 1

These vectorsÂ \( \vec{OA} \),Â \( \vec{OB} \), andÂ \( \vec{OC} \), each having magnitude 1 are Unit Vectors along the axes OX, OY, and OZ respectively. They are denoted byÂ \( \vec{i} \),Â \( \vec{j} \), andÂ \( \vec{k} \) as shown in Fig. 1 above.

**Browse more Topics under Vector Algebra**

- Basic Concepts of Vectors
- Types of Vectors
- Addition of Vectors
- Scalar (or Dot) Product of Two Vectors
- Vector (or Cross) Product of Two Vectors
- Section Formula
- Projection of a Vector on a Line

**You can download Vector Algebra Cheat Sheet by clicking on the download button below**

## Component Form of a Vector

Let’s consider a position vectorÂ \( \vec{OP} \) of a point P (x, y, z) as shown below

As shown in the figure, Let P_{1} be the foot of the perpendicular from point P on the plane XOY. Observe that P_{1}P is parallel to the z-axis. We know that,Â \( \vec{i} \),Â \( \vec{j} \), andÂ \( \vec{k} \) are unit vectors along the x, y, and z-axes, respectively. Hence, by definition of the coordinates of point P, we have

\( \vec{P_1P} \) =Â \( \vec{OR} \) = z\( \vec{k} \)

Similarly, we have

\( \vec{QP_1} \) =Â \( \vec{OS} \) = y\( \vec{j} \) and

\( \vec{P_1S} \) =Â \( \vec{OQ} \) = x\( \vec{i} \)

Now, by using the triangle law of vector addition, we can write

\( \vec{OP_1} \) =Â \( \vec{OQ} \) + \( \vec{QP_1} \) = x\( \vec{i} \) + y\( \vec{j} \)

And,

\( \vec{OP} \) =Â \( \vec{OP_1} \) + \( \vec{P_1P} \) = x\( \vec{i} \) + y\( \vec{j} \) +Â z\( \vec{k} \)

Therefore, the position vector of P with reference to O is

\( \vec{OP} \) (orÂ \( \vec{r} \)) =Â x\( \vec{i} \) + y\( \vec{j} \) +Â z\( \vec{k} \)

This is the Component Form of a vector. Here, x, y, and z are the *scalar components* ofÂ \( \vec{r} \) andÂ x\( \vec{i} \), y\( \vec{j} \), and z\( \vec{k} \) are the *vector components* ofÂ \( \vec{r} \) along the respective axes. The scalar components are also referred to as *rectangular components* at times.

## Calculation of vectors

### Length of a vector

Applying the Pythagoras theoremtwice can help us calculate the length of any vectorÂ \( \vec{r} \) =Â x\( \vec{i} \) + y\( \vec{j} \) +Â z\( \vec{k} \). In Fig. 2 given above, we note that in the right-angled triangle OQP_{1},

|\( \vec{OP_1} \)| =Â \( \sqrt{|\vec{OQ}|^2 + |\vec{QP_1}|^2} \) =Â =Â \( \sqrt{x^2 + y^2} \)

Also, in the right-angled triangle OP_{1}P, we have

|\( \vec{OP} \)| =Â \( \sqrt{|\vec{OP_1}|^2 + |\vec{P_1P}|^2} \) = \( \sqrt{(x^2 + y^2) + z^2} \)

Therefore, the length of any vectorÂ \( \vec{r} \) =Â x\( \vec{i} \) + y\( \vec{j} \) +Â z\( \vec{k} \) is

|\( \vec{r} \)| = |x\( \vec{i} \) + y\( \vec{j} \) +Â z\( \vec{k} \)| =Â \( \sqrt{x^2 + y^2 + z^2} \)

Now, let’s look at some general calculations of vectors: IfÂ \( \vec{a} \) and \( \vec{b} \) are two vectors given in the component form, that is

- \( \vec{a} \) = a
_{1}\( \vec{i} \) + a_{2}\( \vec{j} \) + a_{3}\( \vec{k} \) - \( \vec{b} \) = b
_{1}\( \vec{i} \) + b_{2}\( \vec{j} \) + b_{3}\( \vec{k} \)

then,

### Sum of Vectors

The sum or resultant of the vectorsÂ \( \vec{a} \) and \( \vec{b} \) is,

\( \vec{a} \) + \( \vec{b} \) = (a_{1Â }+ b_{1})\( \vec{i} \) + (a_{2Â }+ b_{2})\( \vec{j} \) + (a_{3} + b_{3})\( \vec{k} \)

### Difference of Vectors

The difference between the vectorsÂ \( \vec{a} \) and \( \vec{b} \) is,

\( \vec{a} \) – \( \vec{b} \) = (a_{1Â }– b_{1})\( \vec{i} \) + (a_{2Â }– b_{2})\( \vec{j} \) + (a_{3} – b_{3})\( \vec{k} \)

### Equal Vectors

The vectorsÂ \( \vec{a} \) and \( \vec{b} \) are equal if and only if,

- a
_{1}= b_{1} - a
_{2}= b_{2} - a
_{3}= b_{3}

### Multiplication by a Scalar

The multiplication of a vectorÂ \( \vec{a} \) by a scalarÂ \( \lambda \) is,

\( \lambda \)\( \vec{a} \) = (\( \lambda \)a_{1})\( \vec{i} \) +Â (\( \lambda \)a_{2})\( \vec{j} \) +Â (\( \lambda \)a_{3})\( \vec{k} \)

The addition of vectors and multiplication of a vector by a scalar offer the following *distributive laws*: ifÂ \( \vec{a} \) and \( \vec{b} \) are any two vectors and ‘k’ and ‘m’ are any two scalars, then

- k\( \vec{a} \) + m\( \vec{a} \) = (k + m)\( \vec{a} \)
- k(m\( \vec{a} \)) = (km)\( \vec{a} \)
- k(\( \vec{a} \) +Â \( \vec{b} \)) = k\( \vec{a} \) + k\( \vec{b} \)

## Points to remember

### Point 1

Regardless of the value ofÂ \( \lambda \), the vectorÂ \( \lambda \)\( \vec{a} \) is always collinear to the vectorÂ \( \vec{a} \). For that matter, two vectorsÂ \( \vec{a} \) andÂ \( \vec{b} \) are collinear if and only if there exists a non-zero scalarÂ \( \lambda \) such thatÂ \( \vec{b} \) =Â \( \lambda \)\( \vec{a} \). Let’s look at the component forms. If,

\( \vec{a} \) = a_{1}\( \vec{i} \) + a_{2}\( \vec{j} \) + a_{3}\( \vec{k} \) and

\( \vec{b} \) = b_{1}\( \vec{i} \) + b_{2}\( \vec{j} \) + b_{3}\( \vec{k} \)

Then, these two vectorsÂ \( \vec{a} \) andÂ \( \vec{b} \) are collinear if and only if,

b_{1}\( \vec{i} \) + b_{2}\( \vec{j} \) + b_{3}\( \vec{k} \) =Â \( \lambda \)(a_{1}\( \vec{i} \) + a_{2}\( \vec{j} \) + a_{3}\( \vec{k} \))

\( \Longleftrightarrow \)Â b_{1}\( \vec{i} \) + b_{2}\( \vec{j} \) + b_{3}\( \vec{k} \) = (\( \lambda \)a_{1})\( \vec{i} \) +Â (\( \lambda \)a_{2})\( \vec{j} \) +Â (\( \lambda \)a_{3})\( \vec{k} \)

By comparing coefficients ofÂ \( \vec{i} \),Â \( \vec{j} \), andÂ \( \vec{k} \), we get

b_{1} =Â \( \lambda \)a_{1},Â b_{2} =Â \( \lambda \)a_{2}, and b_{3} =Â \( \lambda \)a_{3}

\( \Longleftrightarrow \)Â \( \lambda \) =Â \( \frac{b_1}{a_1} \) =Â \( \frac{b_2}{a_2} \) =Â \( \frac{b_3}{a_3} \)

### Point 2

IfÂ \( \vec{a} \) = a_{1}\( \vec{i} \) + a_{2}\( \vec{j} \) + a_{3}\( \vec{k} \), then a_{1}, a_{2}, and a_{3} are also called direction ratios ofÂ \( \vec{a} \).

### Point 3

Let’s say that l, m, n are the direction cosines of a vector. Then, l\( \vec{i} \) + m\( \vec{j} \) + n\( \vec{k} \) = (\( \cos{\alpha} \))\( \vec{i} \) + (\( \cos{\beta} \))\( \vec{j} \) + (\( \cos{\gamma} \))\( \vec{k} \) is a unit vector in the direction of the original vector. Also,Â \( \alpha \),Â \( \beta \), andÂ \( \gamma \) are angles that the vector makes with the x, y, and z-axes, respectively.

Let’s look at some examples now:

### Example 1

Find the values of x, y, and z so that the vectors \( \vec{a} \) = x\( \vec{i} \) + 2\( \vec{j} \) + z\( \vec{k} \) andÂ \( \vec{b} \) = 2\( \vec{i} \) + y\( \vec{j} \) + \( \vec{k} \) are equal.

Solution: Two vectors are equal only when their corresponding components are the same. Hence, by comparing the coefficients ofÂ \( \vec{i} \),Â \( \vec{j} \), andÂ \( \vec{k} \), we get x = 2, y = 2 and z = 1

### Example 2

Find a unit vector in the direction of the vectorÂ \( \vec{a} \) = 2\( \vec{i} \) + 3\( \vec{j} \) + \( \vec{k} \)

Solution: The unit vector in the direction of a vectorÂ \( \vec{a} \) is, \( \hat{a} \) =Â \( \frac{1}{|\vec{a}|} \)\( \vec{a} \)

Now, |\( \vec{a} \)| =Â \( \sqrt{2^2 + 3^2 + 1^2} \) =Â \( \sqrt{14} \)

Therefore, \( \hat{a} \) =Â \( \frac{1}{\sqrt{14}} \)(2\( \vec{i} \) + 3\( \vec{j} \) + \( \vec{k} \))

=Â \( \frac{2}{\sqrt{14}} \)\( \vec{i} \) +Â \( \frac{3}{\sqrt{14}} \)\( \vec{j} \) +Â \( \frac{1}{\sqrt{14}} \)\( \vec{k} \)

## More Solved Examples for You

**Question 1: For the given vectorsÂ \( \vec{a} \) = 2\( \vec{i} \) – \( \vec{j} \) + 2\( \vec{k} \) andÂ \( \vec{b} \) = – \( \vec{i} \) + \( \vec{j} \) – \( \vec{k} \), find a unit vector in the direction of the vectorÂ \( \vec{a} \) +Â \( \vec{b} \).**

**Answer :** Let \( \vec{c} \) = \( \vec{a} \) +Â \( \vec{b} \). Calculating its value, we have

\( \vec{a} \) +Â \( \vec{b} \) = (2 – 1)\( \vec{i} \) + (- 1 + 1)\( \vec{j} \) + (2 – 1)\( \vec{k} \)

Therefore, \( \vec{c} \) = 1\( \vec{i} \) + 0\( \vec{j} \) + 1\( \vec{k} \)

Hence, |\( \vec{c} \)| =Â \( \sqrt{1^2 + 0^2 + 1^2} \) =Â \( \sqrt{1 + 0 + 1} \) =Â \( \sqrt{2} \)

We know that, the unit vector in the direction ofÂ \( \vec{c} \) is,

\( \hat{c} \) =Â \( \frac{1}{|\vec{c}|} \)\( \vec{c} \)

Hence,

\( \hat{c} \) =Â \( \frac{1}{\sqrt{2}} \)(1\( \vec{i} \) + 0\( \vec{j} \) + 1\( \vec{k} \))

=Â \( \frac{1}{\sqrt{2}} \)\( \vec{i} \) + 0\( \vec{j} \) +Â \( \frac{1}{\sqrt{2}} \)\( \vec{k} \)

Or, the unit vector in the direction ofÂ \( \vec{c} \) is,

\( \hat{c} \) =Â \( \frac{1}{\sqrt{2}} \)\( \vec{i} \) + \( \frac{1}{\sqrt{2}} \)\( \vec{k} \)

**Some Solved Questions for You:**

**Question 1: What is said to be a unit vector?**

**Answer:** A unit vector is of length 1, sometimes we also call it as a direction vector. The unit vector that has the same direction as a nonzero vector is defined by a unique symbol. The symbol denotes the standard of the unit vector in the same direction as the finite vector.

**Question 2: Does the unit vector always remains 1?**

**Answer:** The unit vector is something that is all about the direction that tells us that in what direction any vector of particular magnitude is being directed. So the magnitude of the unit vector is always one (1) so that, on being multiplied by the magnitude of the vector it does not causes any effect on its originality.

**Question 3: Can unit vectors ever be negative (-)?**

**Answer:** A unit vector is a vector that has the magnitude of one (1) with no units. This says that the Fx vector is writable as â€˜And maybe now we can see why the negative sign is essentialâ€™. The vector Fx is lying in the opposite direction as the x-hat vector. In addition, this is the reason why we need a negative (-) sign.

**Question 4: Is acceleration a vector?**

**Answer:** Yes, the acceleration is a vector quantity. The reason behind this is that â€˜it has both magnitudes as well as directionâ€™.

## Leave a Reply