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Physics Formulas

Cross Product Formula

This article will help in increasing our knowledge on the topic of the Cross Product Formula. Similarly, we will here get to know about what it actually is? And also will find out how to solve math problems by this method with examples. The cross product or we can say the vector product (occasionally directed area product for emphasizing the significance of geometry) is a binary operation that occurs on two vectors in 3D space $$R^3$$ and we denote it by the symbol \(\times\). We have two linearly independent vectors a and b, the cross product, a \(\times\) b, is a vector which is perpendicular to both a and to b as well, moreover, normal to the plane containing them. It has a lot of applications in mathematics, physics, engineering, and computer programming, etc. we should not confuse it with the dot product (projection product).

Cross Product Formula and Derivation

cross product formula

The cross product for two vectors can find a third vector that is perpendicular to the original two vectors that we are having.

We can assume the given vectors to be perpendicular (orthogonal) to the vector that would result from the cross product.  This means that the dot product of all of the original vectors with the new vector will be 0.

So, we are having 2 vectors:

$$a = \begin{vmatrix}
a_1\\ a_2
\\ a_3
\end{vmatrix}$$

And,

$$b = \begin{vmatrix}
b_1\\ b_2
\\ b_3
\end{vmatrix}$$

Now, we have to find the third vector i.e. n:

$$n = \begin{vmatrix}
n_1\\ n_2
\\ n_3
\end{vmatrix}$$

So that the n is perpendicular to both a and to b as well:

As we had mentioned above this means that we want the dot product of n with each of the two original vectors to be 0.

$$ a\cdot n =  a_{1}n_{1} + a_{2}n_{2} + a_{3}n_{3} = 0$$

And,
$$ b\cdot n =  b_{1}n_{1} + b_{2}n_{2} + b_{3}n_{3} = 0$$

This provides us 2 equations to work with.  Since we have 3 variables to solve for $$\left ( n_{1}, n_{2}, n_{3}\right )$$ we will need another equation also to work with.

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It introduces us to a third constraint that the modulus of the cross product vector n will be equal to = 1.

This forms a third equation \(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}\) = 1 and allows us to solve for \(n_{1},n_{2}\) and \(n_{3}\) in terms of \(a_{1},a_{2},a_{3},b_{1},b_{2}\) and \(b_{3}\).

As mentioned above the derivation we are having of the values for \(n_{1},n_{2}\) and \(n_{3}\) and ends up with the formula given below:

$$n = \begin{vmatrix}
a_{2} b_{3} – a_{3} b_{2}\\ a_{3} b_{1} – a_{1} b_{3}
\\ a_{1} b_{2} – a_{2} b_{1}
\end{vmatrix}$$

 

We are not able to see it in the derivation, but it is visible that the \(\sqrt{z}\) term that is factored out and defined to make the derivation work more smoothly is the modulus in actual of the cross product vector n.

Solved Examples on Cross product Formula

Question:

Find out the cross product between a = (3, −3, 1) and b = (4, 9, 2).

Solution:

The cross product is:

$$\begin{vmatrix}
i& j & k\\
3& -3 & 1\\
4& 9 & 2
\end{vmatrix}$$

 

$$= i\left ( -3\cdot 2 – 1\cdot 9 \right ) – j\left ( 3\cdot 2 – 1\cdot 4 \right ) + k\left ( 3\cdot 9 + 3\cdot 4 \right ) = -15i -2j + 39k$$

Question:

Find out the area of the parallelogram that is spanned by the vectors a = (3, −3, 1) and b = (4, 9, 2).

Solution:

The area is ∥a×b∥. Using the expression that is provided above for the cross product, we find that the area is \(\sqrt{15^{2}+2^{2}+39^{2}}\) = 5\(\sqrt{70}\).

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