# 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.

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## Definition

TheÂ cross or vector product of two non-zero vectorsÂ $$\vec{a}$$ andÂ $$\vec{b}$$, is

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

WhereÂ $$\theta$$ is the angle betweenÂ $$\vec{a}$$ andÂ $$\vec{b}$$, 0 â‰¤ $$\theta$$ â‰¤ $$\pi$$. Also,Â $$\hat{n}$$ is a unit vector perpendicular to bothÂ $$\vec{a}$$ andÂ $$\vec{b}$$ such thatÂ $$\vec{a}$$, $$\vec{b}$$, andÂ $$\hat{n}$$ form a right-handed system as shown below.

As can be seen above, when the system is rotated fromÂ $$\vec{a}$$ toÂ $$\vec{b}$$, it moves in the direction ofÂ $$\hat{n}$$. Also, if eitherÂ $$\vec{a}$$ = 0 orÂ $$\vec{b}$$ = 0, thenÂ $$\theta$$ is not defined and we can say,

$$\vec{a}$$ xÂ $$\vec{b}$$ =Â $$\vec{0}$$

## Important Observations

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

$$\vec{a}$$ xÂ $$\vec{b}$$ = 0Â â‡”Â $$\vec{a}$$ $$\parallel$$ $$\vec{b}$$

Hence,Â $$\vec{a}$$ xÂ $$\vec{a}$$ = 0 andÂ $$\vec{a}$$ xÂ $$\vec{(- a)}$$ = 0. This is because in the first caseÂ $$\theta$$ = 0. Also, in the second caseÂ $$\theta$$ =Â $$\pi$$, giving the value ofÂ $$\sin{\theta}$$ = 0.

• IfÂ $$\theta$$ =Â $$\frac{\pi}{2}$$, thenÂ $$\vec{a}$$ xÂ $$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|
• Considering observations 2 and 3 above, for mutually perpendicular vectorsÂ $$\vec{i}$$, $$\vec{j}$$, andÂ $$\vec{k}$$, we have

$$\vec{i}$$ xÂ $$\vec{i}$$ =Â $$\vec{j}$$ xÂ $$\vec{j}$$ =Â $$\vec{k}$$ xÂ $$\vec{k}$$ = 0

Also,

$$\vec{i}$$ xÂ $$\vec{j}$$ =Â $$\vec{k}$$
$$\vec{j}$$ xÂ $$\vec{k}$$ =Â $$\vec{i}$$
$$\vec{k}$$ xÂ $$\vec{i}$$ =Â $$\vec{j}$$

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

$$\sin{\theta}$$ =Â $$\frac{|\vec{a} Ã— \vec{b}|}{|\vec{a}||\vec{b}|}$$

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

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

Where $$\vec{a}$$, $$\vec{b}$$, andÂ $$\hat{n}$$ form a right-handed system. Or,Â $$\theta$$ is traversed fromÂ $$\vec{a}$$ toÂ $$\vec{b}$$. On the other hand,

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

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

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

= –Â $$\vec{b}$$ xÂ $$\vec{a}$$

• From the observations 4 and 6 above, we have

$$\vec{j}$$ xÂ $$\vec{i}$$ = –Â $$\vec{k}$$
$$\vec{k}$$ xÂ $$\vec{j}$$ = –Â $$\vec{i}$$
$$\vec{i}$$ xÂ $$\vec{k}$$ = –Â $$\vec{j}$$

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

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

Area ofÂ Î”ABC =Â $$\frac{1}{2}$$|$$\vec{a}$$||$$\vec{b}$$|$$\sin{\theta}$$ =Â $$\frac{1}{2}$$|$$\vec{a}$$ x $$\vec{b}$$|

• IfÂ $$\vec{a}$$ 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 ofÂ parallelogram ABCD = (AB).(DE). We know that, AB = |$$\vec{b}$$| and DE = |$$\vec{a}$$|$$\sin{\theta}$$. Therefore,

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

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

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

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

## Component Form of Vector Product

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

### 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($$\hat{i}$$ xÂ $$\hat{i}$$) +Â a1b2($$\hat{i}$$ xÂ $$\hat{j}$$) +Â a1b3($$\hat{i}$$ xÂ $$\hat{k}$$) +Â a2b1($$\hat{j}$$ xÂ $$\hat{i}$$) +Â a2b2($$\hat{j}$$ xÂ $$\hat{j}$$) +Â a2b3($$\hat{j}$$ xÂ $$\hat{k}$$) +Â a3b1($$\hat{k}$$ xÂ $$\hat{i}$$) +Â a3b2($$\hat{k}$$ xÂ $$\hat{j}$$) +Â a3b3($$\hat{k}$$ xÂ $$\hat{k}$$)

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

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

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

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

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

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

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

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

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

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

Solving the determinant, we get

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

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

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

Therefore, the area of the parallelogram isÂ 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 â€¢ b = 0 and a â‰  o, b â‰  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.

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