Vector Point Multiplication Theory, Vector Cross Product, Vector Problems and Solutions

Physics Questions and Solutions - In vectors there are several types of multiplication. We know vector multiplication by scalar, dot multiplication and cross multiplication between vectors.

In this article we will discuss dot multiplication which produces scalar and cross multiplication which produces vector magnitude , along with Questions and Solutions dot multiplication and vector cross multiplication.

A. Multiplication of Dot (.)

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Example vector $\vec{A} = (A_i+ A_j+ A_k )$ and vektor $\vec{B} = (B_i+ B_j+B_k )$ wherein the two vectors forming an angle of θ, so:
$\vec{A} . \vec{B} = |\vec{A}||\vec{B}| \cos \theta$
with:
$\small \begin{align*} \left |A \right | &= \sqrt{A_i^2+A_j^2+A_k^2}\\\left |B \right | &= \sqrt{B_i^2+B_j^2+B_k^2}\end{align*}$
Dot Multiplication (dot multiplication) Algebraically
$\vec{A} . \vec{B} = A_i.B_i + A_j.B_j + A_k.B_k$
dot multiplication result:
$\begin{align*}i . i &= j . j = k . k = 1\\i . j &= j . k = j . k = 0\end{align*}$

B. Multiplication Cross ($\times$)

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In cross multiplication, the symbol 'x' is used. Let vector $\vec {A} = (A_i+ A_j+A_k)$ and vektor $\vec {B} = (B_i+ B_j+ B_k)$. so:
$\begin{align*}A \times B = \left | A \right |\left | B \right |\sin\theta \end{align*}$
with:
$\small \begin{align*} \left |A \right | &= \sqrt{A_i^2+A_j^2+A_k^2}\\\left |B \right | &= \sqrt{B_i^2+B_j^2+B_k^2}\end{align*}$

To make it easier for friends to understand the Cross Operation ($\times$), see the picture below, in the direction of the positive arrow, opposite the negative arrow.

Thus we can write:
$$\begin{align*}i \times j &= +k\textrm{, } j \times i =-k\\k \times i &= +j\textrm{, }i \times k =-j\\j \times k&=+ i \textrm{, }k \times j = -i\\&\textrm{with:}\\i \times i &= j \times j = k\times k = 0\end{align*}$$

C. Dot and Cross Multiplication Questions and Solutions

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Problem: 1. Know the vector $\vec{A} = (3, -5, 4)$ and $\vec{B} = (-2,1,2).$ . Find the sine of the angle between the vectors  $\vec{A}$ and $\vec{B}$! !

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Vector solution:

angle value:
$\small \begin{align} \cos \theta & = \frac{ \vec{A} . \vec{B} }{|\vec{A}||\vec{B}| } \\ & = \frac{ -6 - 5 + 8 }{\sqrt{3^2 + (-5)^2 + 4^2} . \sqrt{(-2)^2 + 1^2 + 2^2} } \\ & = \frac{ -3 }{\sqrt{50} . \sqrt{9} } = \frac{ -3 }{5\sqrt{2} .3 } \\ & = \frac{ -1 }{5\sqrt{2} } \end{align}$
With the formula of trigonometric identity $\sin ^2 \theta + \cos ^2 \theta = 1$ , so:$\begin{align*} \sin ^2 \theta &+ \cos ^2 \theta = 1 \\ \sin ^2 \theta & = 1 - \cos ^2 \theta \\ \sin \theta & = \sqrt{ 1 - \cos ^2 \theta } \\ & = \sqrt{ 1 - ( \frac{ -1 }{5\sqrt{2} } )^2 } \\ & = \sqrt{ 1 - \frac{1 }{50} } \\ & = \sqrt{ \frac{49 }{50} } \\ & = \frac{7}{5\sqrt{2}} =0,7\sqrt{2} \end{align*}$

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Problem: 2. Given a vector:

$\begin{align*} \vec{A} &= (2i + 3j-4k) \\\vec{B}&= (i-2j + 3k) \\ \vec{C} &= (i-10j-7k)\end{align*}$

so  $\vec{A} \times \vec{B}$  dan $(\vec{A} \times \vec{B}) \times \vec{C}$ is...

Vector solution:

$\small \begin{align*} \vec{A} \times \vec{B} &= (2i + 3j-4k) \times (i-2j + 3k) \\=&-4k-6j-3k + 9i-4j-8i\\=& i-10j-7k\\ (\vec{A} \times \vec{B}) &\times \vec{C} = (i-10j-7k)\times (i-j+k)\\ =&-k-j + 10k-10i-7j-7i \\=&-17i-8j + 9k \end{align*}$  
A, B, C = besar vektor A, B, C
i, j, k = unit vector for the x, y, z axes

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Problem: 3. If vectors $\vec{A}$ and $\vec{B}$ form an angle of 60 , and vector $\vec{A}$ is perpendicular to vector $\vec{C}$ , with$|\vec{A}| = 4$ , $|\vec{B}| = 3$ , dan $|\vec{C}| = 6$, then specify:

• $\vec{A} (\vec{B} + \vec{C})$ 
• $\vec{A}(2\vec{C} - 3\vec{B})$ 

Vector solution:
Since $\vec{A}$ is perpendicular to $\vec{C}$ , then $\vec{A}.\vec{C} = 0$.
a. Value $\vec{A} (\vec{B} + \vec{C})$
$\small \begin{align} \vec{A} (\vec{B} + \vec{C}) & = \vec{A} . \vec{B} + \vec{A}. \vec{C} \\ & = |\vec{A}| |\vec{B} | \cos 60^\circ + 0 \\ & = 4. 3. \frac{1}{2} \\ & = 6 \end{align}$
b. Value $\vec{A}(2\vec{C} - 3\vec{B})$
$\small \begin{align} \vec{A}(2\vec{C} - 3\vec{B}) & = 2(\vec{A}.\vec{C}) - 3(\vec{ A}.\vec{B}) \\ & = 2.0 - 3|\vec{A}| |\vec{B} | \cos 60^\circ \\ & = 0 - 3 . 4. 3. \frac{1}{2} \\ & = - 18 \end{align}$

Hope it is useful. regards..!