One thing that only applies to three-dimensional vector spaces R3 is cross vectors (vector multiplication between 2 vectors), namely the multiplication of 2 vectors which produces a single vector.

One thing that only applies to the three-dimensional vector space R3 is cross vector (vector multiplication between 2 vectors), namely the multiplication between 2 vectors which produces a single vector.

Cross product or cross product is the product of two vectors in three-dimensional space (R3) which produces a vector perpendicular to the two vectors being multiplied. Or it can also be said that the cross product between two vectors will produce a new vector whose direction is perpendicular to each vector.

Determining the direction of the vector in cross multiplication can use the right hand rule which involves the palm, four fingers and the thumb. Where, the palm of the hand is in the direction of the first vector to be multiplied and the four fingers are in the direction of the second vector. Then, the direction of the unit vector resulting from the product is indicated by the thumb. Right hand rule

1. Definition

If uβƒ—β‰ 0\vec{u} \ne 0 and vβƒ—β‰ 0\vec{v}\ne 0 in space can be rotated without changing their respective magnitudes or directions so that their starting points coincide, with the right-hand rule (right-hand thread) it is defined that: uβƒ—Γ—vβƒ—=e^∣uβƒ—βˆ£βˆ£vβƒ—βˆ£sin⁑θ, 0≀θ≀π\vec{u}\times \vec{v}=\widehat{e}\left| \vec{u} \right|\left| \vec{v} \right|\sin \theta ,\text{ 0}\le \theta \le \pi

e^\widehat{e}= perpendicular unit vector u⃗\vec{u} and v⃗\vec{v}
uβƒ—Γ—vβƒ—\vec{u}\times \vec{v} reads β€œu vector across vector v” or simply β€œu across v”.

Cross vector determinant formula

The vector product of two vectors written as u⃗×v⃗\vec{u}\times \vec{v} is formulated with the matrix determinant as follows.

If uβƒ—=a1i^+a2j^+a3k^\vec{u}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k} and vβƒ—=b1i^+b2j^+b3k^\vec{v}=b_1\widehat{i}+b_2\widehat{j} +b_3\widehat{k} then uβƒ—Γ—vβƒ—=∣i^j^k^a1a2a3b1b2bβˆ’3∣\vec{u}\times \vec{v}=\left| \begin{matrix}\widehat{i} & \widehat{j} & \widehat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b-3 \end{matrix} \right| Using Sarrus' rule, the following multiplication results will be obtained. sarrus rules uβƒ—Γ—vβƒ—=(a2b3βˆ’a3b2)i^+(a3b1βˆ’a1b3)j^+(a1b2βˆ’a2b1)k^\vec{u}\times \vec{v}=(a_2b_3-a_3b_2)\widehat{i}+(a_3b_1-a_1b_3)\widehat{j}+(a_1b_2-a_2b_1)\widehat{k}

2. Characteristic

  1. Property 1: u⃗×v⃗\vec{u}\times \vec{v} is a vector that is perpendicular to vector u⃗\vec{u} and perpendicular to vector v⃗\vec{v}.
  2. Sifat 2: uβƒ—Γ—vβƒ—\vec{u}\times \vec{v} zapatas arah dengan vβƒ—Γ—uβƒ—\vec{v}\times \vec{u} Ε‘eho uβƒ—Γ—vβƒ—=βˆ’vβƒ—Γ—uβƒ—\vec{u}\times \vec{v} =-\vec{v}\times \vec{u}

Dalil: ∣uβƒ—Γ—vβƒ—βˆ£=∣uβƒ—βˆ£βˆ£vβƒ—βˆ£sin⁑θ\left| \vec{u}\times \vec{v} \right|=\left| \vec{u} \right|\left| \vec{v} \right|\sin \theta

Prove the cross vector properties above as practice!

Example of Cross Vector Questions:

Given the vectors aβƒ—=2i^βˆ’j^+3k^\vec{a}=2\widehat{i}-\widehat{j}+3\widehat{k} and bβƒ—=3i^βˆ’2j^+k^.\vec{b}=3\widehat{i}-2\widehat{j} +\widehat{k}. Determine the result of the operation

  1. a⃗×b⃗\vec{a}\times \vec{b}
  2. b⃗×a⃗\vec{b}\times \vec{a}
  3. ∣bβƒ—Γ—aβƒ—βˆ£\left| \vec{b}\times \vec{a} \right|

Alternative Solutions:

  1. Hasil operasi aβƒ—Γ—bβƒ—\vec{a}\times \vec{b} aβƒ—Γ—bβƒ—=∣i^j^k^2βˆ’133βˆ’21∣bβƒ—Γ—aβƒ—=∣i^j^k^2βˆ’133βˆ’21∣ i^j^2βˆ’13βˆ’2∣bβƒ—Γ—aβƒ—=βˆ’i^+9j^βˆ’4k^βˆ’(βˆ’3)k^βˆ’(βˆ’6)i^βˆ’2j^aβƒ—Γ—bβƒ—=5i^+7j^βˆ’k^\begin{align*} & \vec{a}\times \vec{b}=\left| \begin{matrix}\widehat{i} & \widehat{j} & \widehat{k}\\ 2 & -1 & 3\\ 3 & -2 & 1\end{matrix} \right| \\ & \vec{b}\times \vec{a}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 2 & -1 & 3\\ 3 & -2 & 1\end{matrix} \right|\left. \text{ }\begin{matrix}\widehat{i} &\widehat{j}\\2 & -1\\3 & -2\end{matrix} \right| \\ & \vec{b}\times \vec{a}=-\widehat{i}+9\widehat{j}-4\widehat{k}-(-3)\widehat{k}-(-6)\widehat{i}-2\widehat{j} \\ & \vec{a}\times \vec{b}=5\widehat{i}+7\widehat{j}-\widehat{k}\end{align*}
  2. Hasil operasi bβƒ—Γ—aβƒ—\vec{b}\times \vec{a} bβƒ—Γ—aβƒ—=∣i^j^k^3βˆ’212βˆ’13∣bβƒ—Γ—aβƒ—=∣i^j^k^3βˆ’212βˆ’13∣ i^j^3βˆ’22βˆ’1∣bβƒ—Γ—aβƒ—=βˆ’6i^+2j^βˆ’3k^βˆ’(βˆ’4)k^βˆ’(βˆ’1)i^βˆ’9j^bβƒ—Γ—aβƒ—=βˆ’5i^βˆ’7j^+k^\begin{align*} & \vec{b}\times \vec{a}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 3 & -2 & 1\\ 2 & -1 & 3\end{matrix} \right| \\ & \vec{b}\times \vec{a}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 3 & -2 & 1\\ 2 & -1 & 3\end{matrix} \right|\left. \text{ }\begin{matrix}\widehat{i} & \widehat{j}\\ 3 & -2\\ 2 & -1\end{matrix} \right| \\ & \vec{b}\times \vec{a}=-6\widehat{i}+2\widehat{j}-3\widehat{k}-(-4)\widehat{k}-(-1)\widehat{i}-9\widehat{j}\\ & \vec{b}\times \vec{a}=-5\widehat{i}-7\widehat{j}+\widehat{k} \\ \end{align*}
  3. Hasil operasi ∣bβƒ—Γ—aβƒ—βˆ£\left| \vec{b}\times \vec{a} \right| ∣bβƒ—Γ—aβƒ—βˆ£=(βˆ’5)2+(βˆ’7)2+12∣bβƒ—Γ—aβƒ—βˆ£=25+49+1∣bβƒ—Γ—aβƒ—βˆ£=75∣bβƒ—Γ—aβƒ—βˆ£=53\begin{align*} & \left| \vec{b}\times \vec{a} \right|=\sqrt{{{(-5)}^{2}}+{{(-7)}^{2}}+{{1}^{2}}} \\ & \left| \vec{b}\times \vec{a} \right|=\sqrt{25+49+1} \\ & \left| \vec{b}\times \vec{a} \right|=\sqrt{75} \\ & \left| \vec{b}\times \vec{a} \right|=5\sqrt{3}\end{align*}

Example 2 Cross Vector Questions

It is known that the angle between the vectors pβƒ— \vec{p} and qβƒ— \vec{q} is 30∘ 30^\circ . If ∣pβƒ—βˆ£=4 |\vec{p}| = 4 and ∣qβƒ—βˆ£=5 |\vec{q}| = 5 , then determine ∣pβƒ—Γ—qβƒ—βˆ£ |\vec{p} \times \vec{q}| !
Solution:

Determines the result ∣pβƒ—Γ—qβƒ—βˆ£ |\vec{p} \times \vec{q}| : ∣pβƒ—Γ—qβƒ—βˆ£=∣pβƒ—βˆ£βˆ£qβƒ—βˆ£sin⁑θ=4Γ—5sin⁑30∘=20Γ—12=10\begin{align*}|\vec{p} \times \vec{q}|& = |\vec{p} | |\vec{q}| \sin \theta \\& = 4 \times 5 \sin 30^\circ \\& = 20 \times \frac{1}{2} \\& = 10 \end{align*}
So, the result of ∣pβƒ—Γ—qβƒ—βˆ£=10 |\vec{p} \times \vec{q}| = 10 .