# Vector (or Cross) Product of Two Vectors

Vectors can be multiplied in two ways, a scalar product where the result is a scalar and cross or vector product where is the result is a vector. In this article, we will look at the cross or vector product of two vectors.

## Explanation

We have already studied the three-dimensional right-handed rectangular coordinate system. As shown in the figure below, when the positive x-axis is rotated counter-clockwise into the positive y-axis, then a right-handed standard screw moves in the direction of the positive z-axis.

As can be seen above, in a three-dimensional right-handed rectangular coordinate system, the thumb of the right-hand points in the direction of the positive z-axis when the fingers are curled from the positive x-axis towards the positive y-axis.

## Definition

The cross or vector product of two non-zero vectors $$\vec{a}$$ and $$\vec{b}$$, is

$$\vec{a}$$ x $$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$| $$\sin{\theta}$$$$\hat{n}$$

Where $$\theta$$ is the angle between $$\vec{a}$$ and $$\vec{b}$$, 0 ≤ $$\theta$$ ≤ $$\pi$$. Also, $$\hat{n}$$ is a unit vector perpendicular to both $$\vec{a}$$ and $$\vec{b}$$ such that $$\vec{a}$$, $$\vec{b}$$, and $$\hat{n}$$ form a right-handed system as shown below.

As can be seen above, when the system is rotated from $$\vec{a}$$ to $$\vec{b}$$, it moves in the direction of $$\hat{n}$$. Also, if either $$\vec{a}$$ = 0 or $$\vec{b}$$ = 0, then $$\theta$$ is not defined and we can say,

$$\vec{a}$$ x $$\vec{b}$$ = $$\vec{0}$$

## Important Observations

• $$\vec{a}$$ x $$\vec{b}$$ is a vector.
• If $$\vec{a}$$ and $$\vec{b}$$ are two non-zero vectors, then $$\vec{a}$$ x $$\vec{b}$$ = 0, if and only if $$\vec{a}$$ and $$\vec{b}$$ are parallel (or collinear) to each other, i.e.

$$\vec{a}$$ x $$\vec{b}$$ = 0 ⇔ $$\vec{a}$$ $$\parallel$$ $$\vec{b}$$

Hence, $$\vec{a}$$ x $$\vec{a}$$ = 0 and $$\vec{a}$$ x $$\vec{(- a)}$$ = 0. This is because in the first case $$\theta$$ = 0. Also, in the second case $$\theta$$ = $$\pi$$, giving the value of $$\sin{\theta}$$ = 0.

• If $$\theta$$ = $$\frac{\pi}{2}$$, then $$\vec{a}$$ x $$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|
• Considering observations 2 and 3 above, for mutually perpendicular vectors $$\vec{i}$$, $$\vec{j}$$, and $$\vec{k}$$, we have

$$\vec{i}$$ x $$\vec{i}$$ = $$\vec{j}$$ x $$\vec{j}$$ = $$\vec{k}$$ x $$\vec{k}$$ = 0

Also,

$$\vec{i}$$ x $$\vec{j}$$ = $$\vec{k}$$
$$\vec{j}$$ x $$\vec{k}$$ = $$\vec{i}$$
$$\vec{k}$$ x $$\vec{i}$$ = $$\vec{j}$$

• The angle between the two vectors $$\vec{a}$$ and $$\vec{b}$$ is,

$$\sin{\theta}$$ = $$\frac{|\vec{a} × \vec{b}|}{|\vec{a}||\vec{b}|}$$

• A cross or vector product is not commutative. We know this because $$\vec{a}$$ x $$\vec{b}$$ = $$\vec{- b}$$ x $$\vec{a}$$. Now, we know that,

$$\vec{a}$$ x $$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$| $$\sin{\theta}$$$$\hat{n}$$.

Where $$\vec{a}$$, $$\vec{b}$$, and $$\hat{n}$$ form a right-handed system. Or, $$\theta$$ is traversed from $$\vec{a}$$ to $$\vec{b}$$. On the other hand,

$$\vec{b}$$ x $$\vec{a}$$ = |$$\vec{b}$$| |$$\vec{a}$$| $$\sin{\theta}$$$$\hat{n_1}$$.

Where $$\vec{b}$$, $$\vec{a}$$, and $$\hat{n_1}$$ form a right-handed system. Or, $$\theta$$ is traversed from $$\vec{b}$$ to $$\vec{a}$$. So, if $$\vec{a}$$ and $$\vec{b}$$ lie on a plane of paper, then $$\hat{n}$$ and $$\hat{n_1}$$ are both perpendicular to the plane of the paper. However, $$\hat{n}$$ is directed above the paper and $$\hat{n_1}$$ is directed below it. Or, $$\hat{n}$$ = – $$\hat{n_1}$$. Hence,

$$\vec{a}$$ x $$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$| $$\sin{\theta}$$$$\hat{n}$$ = – |$$\vec{a}$$||$$\vec{b}$$|$$\sin{\theta}$$$$\hat{n_1}$$

= – $$\vec{b}$$ x $$\vec{a}$$

• From the observations 4 and 6 above, we have

$$\vec{j}$$ x $$\vec{i}$$ = – $$\vec{k}$$
$$\vec{k}$$ x $$\vec{j}$$ = – $$\vec{i}$$
$$\vec{i}$$ x $$\vec{k}$$ = – $$\vec{j}$$

• If $$\vec{a}$$ and $$\vec{b}$$ represent the two sides of a triangle, then its area is $$\frac{1}{2}$$|$$\vec{a}$$ x $$\vec{b}$$|. To understand this, look at the figure given below.

By the definition of the area of a triangle, we have area of ΔABC = $$\frac{1}{2}$$(AB).(CD). We know that, AB = |$$\vec{b}$$| and CD = |$$\vec{a}$$|$$\sin{\theta}$$. Therefore,

Area of ΔABC = $$\frac{1}{2}$$|$$\vec{a}$$||$$\vec{b}$$|$$\sin{\theta}$$ = $$\frac{1}{2}$$|$$\vec{a}$$ x $$\vec{b}$$|

• If $$\vec{a}$$ and $$\vec{b}$$ represent the two adjacent sides of a parallelogram, then its area is |$$\vec{a}$$ x $$\vec{b}$$|. To understand this, look at the figure given below.

By the definition of the area of a parallelogram, we have area of parallelogram ABCD = (AB).(DE). We know that, AB = |$$\vec{b}$$| and DE = |$$\vec{a}$$|$$\sin{\theta}$$. Therefore,

Area of parallelogram ABCD = |$$\vec{a}$$||$$\vec{b}$$|$$\sin{\theta}$$ = |$$\vec{a}$$ x $$\vec{b}$$|

## Property: Distributivity of a cross or vector product over addition

If $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are any three vectors and $$\lambda$$ is a scalar, then

• $$\vec{a}$$ x ($$\vec{b}$$ + $$\vec{c}$$) = $$\vec{a}$$ x $$\vec{b}$$ + $$\vec{a}$$ x $$\vec{c}$$
• $$\lambda$$($$\vec{a}$$ x $$\vec{b}$$) = ($$\lambda$$$$\vec{a}$$) x $$\vec{b}$$ = $$\vec{a}$$ x ($$\lambda$$$$\vec{b}$$)

## Component Form of Vector Product

If $$\vec{a}$$ and $$\vec{b}$$ are two vectors given in the component form as a1$$\hat{i}$$ + a2$$\hat{j}$$ + a3$$\vec{k}$$ and b1$$\hat{i}$$ + b2$$\hat{j}$$ + b3$$\vec{k}$$. Then, their cross or vector product is,

$$\vec{a}$$ x $$\vec{b}$$ = $$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix}$$

### Explanation

$$\vec{a}$$ x $$\vec{b}$$ = (a1$$\hat{i}$$ + a2$$\hat{j}$$ + a3$$\vec{k}$$) x (b1$$\hat{i}$$ + b2$$\hat{j}$$ + b3$$\vec{k}$$)

= a1b1($$\hat{i}$$ x $$\hat{i}$$) + a1b2($$\hat{i}$$ x $$\hat{j}$$) + a1b3($$\hat{i}$$ x $$\hat{k}$$) + a2b1($$\hat{j}$$ x $$\hat{i}$$) + a2b2($$\hat{j}$$ x $$\hat{j}$$) + a2b3($$\hat{j}$$ x $$\hat{k}$$) + a3b1($$\hat{k}$$ x $$\hat{i}$$) + a3b2($$\hat{k}$$ x $$\hat{j}$$) + a3b3($$\hat{k}$$ x $$\hat{k}$$)

We know that, $$\vec{i}$$ x $$\vec{i}$$ = $$\vec{j}$$ x $$\vec{j}$$ = $$\vec{k}$$ x $$\vec{k}$$ = 0. Also,  $$\vec{j}$$ x $$\vec{i}$$ = – $$\vec{k}$$, $$\vec{k}$$ x $$\vec{j}$$ = – $$\vec{i}$$, $$\vec{i}$$ x $$\vec{k}$$ = – $$\vec{j}$$. Therefore,

$$\vec{a}$$ x $$\vec{b}$$ = a1b2($$\hat{i}$$ x $$\hat{j}$$) – a1b3($$\hat{k}$$ x $$\hat{i}$$) – a2b1($$\hat{i}$$ x $$\hat{j}$$) + a2b3($$\hat{j}$$ x $$\hat{k}$$) + a3b1($$\hat{k}$$ x $$\hat{i}$$) – a3b2($$\hat{j}$$ x $$\hat{k}$$)

We also know that, $$\vec{i}$$ x $$\vec{j}$$ = $$\vec{k}$$, $$\vec{j}$$ x $$\vec{k}$$ = $$\vec{i}$$, $$\vec{k}$$ x $$\vec{i}$$ = $$\vec{j}$$. Therefore, we have,

$$\vec{a}$$ x $$\vec{b}$$ = a1b2$$\hat{k}$$ – a1b3$$\hat{j}$$ – a2b1$$\hat{k}$$ + a2b3$$\hat{i}$$ + a3b1$$\hat{j}$$ – a3b2$$\hat{i}$$
= (a2b3 – a3b2)$$\hat{i}$$ – (a1b3 – a3b1)$$\hat{j}$$ + (a1b2 – a2b1)$$\hat{k}$$
= $$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix}$$.

## Solved Examples for You

Question 1: Find the area of the parallelogram whose adjacent sides are determined by the following vectors,

$$\vec{a}$$ = $$\hat{i}$$ – $$\hat{j}$$ + 3$$\vec{k}$$ and

$$\vec{b}$$ = 2$$\hat{i}$$ – 7$$\hat{j}$$ + $$\vec{k}$$.

Answer : We know that if $$\vec{a}$$ and $$\vec{b}$$ represent the two adjacent sides of a parallelogram, then its area is |$$\vec{a}$$ x $$\vec{b}$$|. Also,

$$\vec{a}$$ x $$\vec{b}$$ = $$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix}$$

Substituting the values of a1,a2,a3,b1,b2,and b3, we get

$$\vec{a}$$ x $$\vec{b}$$ = $$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1 \end{vmatrix}$$

Solving the determinant, we get

$$\vec{a}$$ x $$\vec{b}$$ = $$\hat{i}$${[(-1) x 1)] – [(-7) x 3]} – $$\hat{j}$${[1 x 1)] – [2 x 3]} + $$\hat{k}$${[1 x (-7))] – [2 x (-1)]}
= 20$$\hat{i}$$ + 5$$\hat{j}$$ – 5$$\hat{k}$$.

Also, the magnitude of $$\vec{a}$$ x $$\vec{b}$$ is,

|$$\vec{a}$$ x $$\vec{b}$$| = $$\sqrt{20^2 + 5^2 + (-5)^2}$$ = $$\sqrt{450}$$ = $$\sqrt{25 × 9 × 2}$$ = 15$$\sqrt{2}$$.

Therefore, the area of the parallelogram is 15$$\sqrt{2}$$.

Question 2: What is meant by vector product of two vectors?

Answer: The vector product of two vectors refers to a vector that is perpendicular to both of them. One can obtain its magnitude by multiplying their magnitudes by the sine of the angle that exists between them.

Question 3: Explain the characteristics of vector product?

Answer: The characteristics of vector product are as follows:

• Vector product two vectors always happen to be a vector.
• Vector product of two vectors happens to be noncommutative.
• Vector product is in accordance with the distributive law of multiplication.
• If a • b = 0 and a ≠ o, b ≠ o, then the two vectors shall be parallel to each other.

Question 4: When can we say that two vectors are parallel?

Answer: Two vectors A and B will be said to be parallel if and only if they happen to be scalar multiples of one another. A = k B, where k is a constant and not equal to zero.

Question 5: What does the square of a vector mean?

Answer: Square of a vector refers to the Dot Product with itself.

Share with friends

## Customize your course in 30 seconds

##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Biology
Dr. Nazma Shaik
VTU
Chemistry
Gaurav Tiwari
APJAKTU
Physics
Get Started

Subscribe
Notify of