Everything About Cross Product of Two Vectors for Physics Class 11

The vector or cross product of two vectors $\vec{A}$  and $\vec{B}$ is denoted as $\vec{A}\times \vec{B}$ . That results in a vector whose magnitude depends on the magnitudes of $\vec{A}$  and $\vec{B}$ . Besides the magnitude, the angle between them also matters. The direction is perpendicular to the plane formed by the two vectors. This mathematically proven operation captures the rotational effect of one vector acting relative to another. 

If you are beginning to practice Class 11 NCERT Solutions for Chapter 6 Physics straightaway, remember these features of Vector Cross Product.

• What is the Vector Result? The cross product is a vector, not a scalar.
• Is it Perpendicular? The resulting vector is normal to the plane of $\vec{A}$ and $\vec{B}$ .
• What's the Magnitude Dependence? The magnitude involves the sine of the angle between the vectors.

The vector product of vectors $\vec{A}$  and $\vec{B}$ is 

$\vec{A}\times \vec{B}=\left|\vec{A}\right|\left|\vec{B}\right|\mathrm{s}\mathrm{i}\mathrm{n}\theta \hat{n}$

You can see that $-\left|\vec{A}\right|$  and $\left|\vec{B}\right|$  are the magnitudes of the vectors.

Following that, we have  $-\theta \left(0^{\circ }◻\theta ◻{180}^{\circ }\right)$   . That's the angle between $\vec{A}$  and $\vec{B}$ . - $\hat{n}$  is a unit vector perpendicular to both $\vec{A}$  and $\vec{B}$ . It's pretty much determined by the right-hand rule, where you curl your fingers from $\vec{A}$  to $\vec{B}$ , while the thumb points along $\hat{n}$ .

Based on all, the magnitude of the cross product is

$|\vec{A}\times \vec{B}|=\left|\vec{A}\right|\left|\vec{B}\right|\mathrm{s}\mathrm{i}\mathrm{n}\theta$

This represents the area of the parallelogram formed by $\vec{A}$ and $\vec{B}$ .
Now, if we see in Cartesian coordinates, we can say that for $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$  and $\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$ , the cross product is computed using the determinant

$\vec{A}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|=\left(A_y{B}_z-A_z{B}_y\right)\hat{i}-\left(A_x{B}_z-A_z{B}_x\right)\hat{j}+\left(A_x{B}_y-A_y{B}_x\right)\hat{k}$

The cross product has some important properties that are applicable in solving problems faster.

• Cross product of a vector is non-commutative. Mathematically, that means  $\vec{A}\times \vec{B}=-\vec{B}\times \vec{A}$ . Simply put, reversing the order of the cross vector product flips the direction.
• The Result is zero for parallel vectors.  If $\vec{A}$ and $\vec{B}$  are parallel ( $\theta =0^{\circ }$  or ${180}^{\circ }$  ), $\mathrm{s}\mathrm{i}\mathrm{n}\theta =0$ , so $\vec{A}\times \vec{B}=\vec{0}$ .
• Cross vector product follow orthogonality. What that means is that $\vec{A}\times \vec{B}$ is perpendicular to both $\vec{A}$  and $\vec{B}$ .
• Cross vector product also has a distributive property. We can show that as $\vec{A}\times (\vec{B}+\vec{C})=\vec{A}\times \vec{B}+\vec{A}\times \vec{C}$ .

Exam Tip on Cross Vector Product

Unit vector cross products are useful shortcuts.

$\hat{i}\times \hat{j}=\hat{k},\mathrm{}\hat{j}\times \hat{k}=\hat{i},\mathrm{}\hat{k}\times \hat{i}=\hat{j},\mathrm{}\hat{i}\times \hat{i}=\vec{0}$

Everything you learn about rotational mechanics is dependent on the cross product of vectors. 

• Torque is the Cause of Rotation: Torque about a point $O$  is defined as $\vec{\tau }=\vec{r}\times \vec{F}$ , where $\vec{r}$  is the position vector from $O$  to the point of application of force $\vec{F}$ The cross product of the vectors shows that torque is perpendicular to both $\vec{r}$  and $\vec{F}$ . That itself is behind driving rotation.
• Angular Momentum that Defines the Quantity of Motion: If you have learned about torque and angular momentum, you will be clear on the fact that angular momentum about a point is $\vec{L}=\vec{r}\times \vec{p}$ . The $\vec{p}$ is linear momentum. This is essential when you are considering the conservation laws. 
• Velocity in Rotational Motion: For a rigid body, the velocity of a point $B$  relative to $A$  is ${\vec{v}}_{BA}=\vec{\omega }\times {\vec{r}}_{BA}$ , where $\vec{\omega }$ is angular velocity. This equation shows us, perfectly, how the rotational velocity vector ω creates a linear velocity v that is perpendicular to both the axis of rotation and the position vector r.

The magnitude $|\vec{A}\times \vec{B}|$  equals the area of the parallelogram formed by $\vec{A}$  and $\vec{B}$ . For a triangle formed by these vectors, the area is:

$\text{Area}=\frac{1}{2}|\vec{A}\times \vec{B}|$

Geometrically, we can represent that as the image below. 

 

 

When preparing for exams or JEE main, be careful to avoid these common errors. 

• Making Direction Errors. The quick fix for that would be to always use the right-hand rule to determine $\hat{n}$ . Else, incorrect direction can lead to wrong signs.  
• Wrongly Assuming Parallel Vectors. If $\vec{A}\times \vec{B}=\vec{0}$ , always check whether the vectors are parallel. It's important to check that before you begin to assume that it has zero magnitude.

Here is an example problem on cross vector of product for you to work on. 
Example: A force $\vec{F}=2\hat{i}-3\hat{j}+\hat{k}\mathrm{N}$ acts at point $P(\mathrm{1,2},3)\mathrm{m}$  relative to the origin $O$ . Find the torque about the origin.

Solution: Position vector $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}$ .

Torque is:

$\vec{\tau }=\vec{r}\times \vec{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 2& -3& 1\end{array}\right|$

$=\hat{i}\left[\right(2\left)\right(1)-(3\left)\right(-3\left)\right]-\hat{j}\left[\right(1\left)\right(1)-(3\left)\right(2\left)\right]+\hat{k}\left[\right(1\left)\right(-3)-(2\left)\right(2\left)\right]=\hat{i}(2+9)+\hat{j}(1-6)+\hat{k}(-3-4)=11\hat{i}-5\hat{j}-7\hat{k}\mathbf{I}$