# Vector Product of Two Vectors

We know that a vector has magnitude as well as a direction. But do we know how any two vectors multiply? Let us now study about the cross product of these vectors in detail.

## Vector Product of Two Vectors

Vector product also means that it is the cross product of two vectors.

If you have two vectors a and b then the vector product of a and b is c.Â

c = aÂ Ã— b

So thisÂ aÂ Ã— b actuallyÂ means that the magnitude of c = ab sinÎ¸ whereÂ Î¸ is the angle between a and b and the direction of c is perpendicular toÂ a well as b.Â Â Now, what should be the direction of this cross product? So to find out the direction,Â we use the rule which we call it as the ”right-hand thumb rule”.

Suppose we want to find out the direction ofÂ aÂ Ã— bÂ here we curl our fingers from the direction ofÂ aÂ toÂ b. So if we curl our fingers in a direction as shown in the above figure, your thumb points in the direction of cÂ that is in an upward direction. This thumb denotes the direction of the cross product.

While applying rules to direction, the rotation should be taken to smaller angles that isÂ <180Â° between a and b.Â So the fingers should always be curled in acute angle between a and b.

## Properties of Vector Cross Product

1] Vector product is not commutative. That meansÂ aÂ Ã— bÂ â‰  bÂ Ã— aÂ

We saw thatÂ aÂ Ã— b = cÂ here the thumb is pointing in an upward direction. Whereas in bÂ Ã— aÂ the thumb will point in the downward direction. So,Â bÂ Ã— a = – c.Â So it is not commutative.

2] There is no change in the reflection.

What happens toÂ aÂ Ã— bÂ in the reflection? Suppose vector a goes and strikes the mirror, so the direction of a will become – a.Â So under reflection, a will become – a and b will become – b. NowÂ aÂ Ã— bÂ will become -aÂ Ã— -b =Â Â aÂ Ã— bÂ

3] It is distributive with respect to vector addition.

This means that if aÂ Ã— ( b + c ) = a Ã— b + aÂ Ã— c. This is true in case of addition.

## Vector Product of Unit Vectors

The three unit vectors areÂ $$\hat{i}$$ ,Â $$\hat{j}$$ andÂ $$\hat{k}$$. So,

• $$\hat{i}$$Â Ã—Â $$\hat{i}$$ = 0
• $$\hat{i}$$Â Ã—Â $$\hat{j}$$ = 1Â $$\hat{k}$$
• $$\hat{i}$$Â Ã—$$\hat{k}$$ =1 –Â $$\hat{j}$$
• $$\hat{j}$$ Ã—$$\hat{i}$$ =Â –Â $$\hat{k}$$
• $$\hat{j}$$Â Ã—$$\hat{j}$$ = 0
• $$\hat{j}$$ Ã—$$\hat{k}$$ = 1Â $$\hat{i}$$
• $$\hat{k}$$Ã—Â $$\hat{i}$$=Â $$\hat{j}$$
• $$\hat{k}$$Ã—Â $$\hat{j}$$= -$$\hat{i}$$
• $$\hat{k}$$Ã—Â $$\hat{k}$$= 0

This is how we determine the vector product of unit vectrors.

## Mathematical Form of Vector Product

a = axÂ $$\hat{i}$$ + ayÂ $$\hat{j}$$ +azÂ $$\hat{k}$$

b = bxÂ $$\hat{i}$$ + byÂ $$\hat{j}$$ +bzÂ $$\hat{k}$$

aÃ—bÂ = (Â  axÂ $$\hat{i}$$ + ayÂ $$\hat{j}$$ +azÂ $$\hat{k}$$ )Â Ã— (Â bxÂ $$\hat{i}$$ + byÂ $$\hat{j}$$ +bzÂ $$\hat{k}$$ )

= axÂ $$\hat{i}$$Â Ã— (Â bxÂ $$\hat{i}$$ + byÂ $$\hat{j}$$ +bzÂ $$\hat{k}$$ ) +Â  ayÂ $$\hat{j}$$Â Ã—Â (Â bxÂ $$\hat{i}$$ + byÂ $$\hat{j}$$ +bzÂ $$\hat{k}$$ ) +Â azÂ $$\hat{k}$$Â Ã—Â (Â bxÂ $$\hat{i}$$ + byÂ $$\hat{j}$$ +bzÂ $$\hat{k}$$ )

=Â axÂ byÂ $$\hat{k}$$ –Â  axÂ bzÂ $$\hat{j}$$ + ayÂ bzÂ $$\hat{i}$$ +Â azÂ bxÂ $$\hat{j}$$ – azÂ byÂ $$\hat{i}$$

aÃ—bÂ = (ayÂ bz Â – azÂ by)$$\hat{i}$$Â + (azÂ bxÂ –Â  axÂ bz)$$\hat{j}$$ + (axÂ byÂ Â – ayÂ bx)$$\hat{k}$$

So the determinantÂ  form of the vectors will be,Â aÃ—b =Â

 $$\hat{i}$$ $$\hat{j}$$ $$\hat{k}$$ ax ay Â az bx by bz

## Solved Question

Q1.Â The magnitude of the vector product of two vectors $$\vec{P}$$ and $$\vec{Q}$$ may be:

1. Equal to PQ
2. Less than PQ
3. Equal to zero
4. All of the above.

Answer: The correct option is “D”. |Â $$\vec{P}$$Ã— $$\vec{Q}$$ =Â $$\vec{P}$$ $$\vec{Q}$$ sinÎ¸, whereÂ Î¸ is the angle between P and Q.

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