Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Cross Product
The cross product of 2 vectors is a VECTOR!  

Wize Tip
The cross product is perpendicular to both Vectors. So, it could be used to find a vector perpendicular to two other vectors. 


There are different ways we can find the cross product of two vectors:

If we know the components of two vectors, we can put them in the following template:
a⃗×b⃗=(a2b3−a3b2,a3b1−a1b3,a1b2−a2b1)\vec a\times\vec b= (a_2b_3 - a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)a×b=(a2​b3​−a3​b2​,a3​b1​−a1​b3​,a1​b2​−a2​b1​)

Alternatively, we can use the following determinant to memorize above template and calculate the cross product:
a⃗×b⃗=ı⃗ȷ⃗k⃗a1a2a3b1b2b3=ı⃗(a2b3−a3b2)−ȷ⃗(a1b3−a3b1)+k⃗(a1b2−a2b1)\vec a\times\vec b=\begin{array}{|ccc|}
\vec{\imath} &\vec{\jmath}&\vec k\\
a_1 & a_2 & a_3\\
b_1 & b_2 & b_3
\end{array}=\vec{\imath}(a_2b_3-a_3b_2)-\vec{\jmath}(a_1b_3-a_3b_1)+\vec k(a_1b_2-a_2b_1)a×b=a1​b1​​​a2​b2​​ka3​b3​​=(a2​b3​−a3​b2​)−​(a1​b3​−a3​b1​)+k(a1​b2​−a2​b1​)
Note that both templates produce the same vector with three components.
Alternatively you can remember the result of cross product of unit vectors and find the cross product of a vector by finding the cross product of their components along different unit vectors directly (It usually takes much longer time than using above templates!)


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Sometimes we are only interested to find the magnitude of the resultant vector of a cross product. Then we can use the following formula:
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\boxed{|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta}∣a×b∣=∣a∣∣b∣sinθ​
       where θ\thetaθ is the angle between vectors a⃗\vec aa and b⃗\vec bb

Consequently, vectors a⃗\vec aa and b⃗\vec bb are parallel if:
a⃗×b⃗=0⃗\vec a\times\vec b=\vec0a×b=0

Watch Out!
The order of vectors in a cross product is very important because : a⃗×b⃗=−b⃗×a⃗\vec a\times\vec b = -\vec b\times\vec aa×b=−b×a

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Right Hand Rule

We can find the direction of the cross product result using right hand rule. This is a very useful rule specially for magnetism and fining torques. 

Watch Out!
There are different versions of this rule. The one described in the picture below is using fingers of the right hand to find the direction of the cross product of two vectors shown by blue and red colors. Be careful about the order of vectors and the finger used for each vector!

Alternatively, to find the direction of a⃗×b⃗\vec{a}\times \vec{b}a×b, one can put the fingers of the right hand along the first vector a⃗\vec{a}aand bend fingers toward the second vector direction b⃗\vec{b}b. Then, the thumb should point along a⃗×b⃗\vec{a}\times \vec{b}a×b

For vectors pointing in or out of the page, we use a cross symbol (⊗\otimes⊗) and a dot symbol (⊙\odot⊙), respectively.

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Example: Cross Product of Two Vectors 

If A⃗=−2ı^+3ȷ^\vec A=-2\hat{\imath}+3\hat{\jmath}A=−2^+3^​ and B⃗=−2ı^+2ȷ^\vec B=-2\hat{\imath}+2\hat{\jmath}B=−2^+2^​ and C⃗=−4ı^+1ȷ^\vec C=-4\hat{\imath}+1\hat{\jmath}C=−4^+1^​ , find the cross product of A and B-C? 

Solution

cross-product
{A⃗=−2ı^+3ȷ^B⃗=−2ı^+2ȷ^C⃗=−4ı^+1ȷ^\left\{\begin{aligned}
\vec A &=-2\hat{\imath} + 3\hat{\jmath}\\
\vec B &=-2\hat{\imath} + 2\hat{\jmath}\\
\vec C &=-4\hat{\imath} + 1\hat{\jmath}\\
\end{aligned}\right.⎩⎨⎧​ABC​=−2^+3^​=−2^+2^​=−4^+1^​​
B⃗−C⃗⏟R⃗=(−2ı^+2ȷ^)−(−4ı^+1ȷ^)R⃗=2ı^+ȷ^A⃗×R⃗=(AyRz−AzRy)ı^−(AzRx−AxRz)ȷ^+(AxRy−AyRx)k^=(3(0)−(0)(1))ı^+(0−0)ȷ^+((−2)(1)−(3)(2))k^=−8k^\begin{aligned}
\underbrace{\vec B-\vec C}_{\vec R}&= (-2\hat{\imath} + 2\hat{\jmath})-(-4\hat{\imath} + 1\hat{\jmath})\\
\vec R&=2\hat{\imath} + \hat{\jmath}\\
\vec A\times\vec R &= (A_yR_z-A_zR_y)\hat{\imath}- (A_zR_x-A_xR_z)\hat{\jmath}+ (A_xR_y-A_yR_x)\hat k\\
&= (\cancel{3(0)-(0)}(1))\hat{\imath}+ (\cancel{0-0})\hat{\jmath}+ ((-2)(1)-(3)(2))\hat k\\
&=-8\hat k\\
\end{aligned}RB−C​​RA×R​=(−2^+2^​)−(−4^+1^​)=2^+^​=(Ay​Rz​−Az​Ry​)^−(Az​Rx​−Ax​Rz​)^​+(Ax​Ry​−Ay​Rx​)k^=(3(0)−(0)​(1))^+(0−0​)^​+((−2)(1)−(3)(2))k^=−8k^​

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

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