In this article we will discuss several applications of vectors, namely Resultant Vectors, Normal Vectors, Orthogonal Projections of Vectors, Distance from point to line, and Area and Volume.

In this article we will discuss several applications of vectors, namely Resultant Vectors, Normal Vectors, Orthogonal Projections of Vectors, Distance from point to line, and Area and Volume.

1. Result of Two Vectors

Look at the image on the side below. Resultan Vector Given two vectors, namely vectors a⃗\vec{a} and b⃗\vec{b} and the angle formed by vector b⃗\vec{b} with respect to vector a⃗\vec{a}, namely α\alpha . The resultant of the vectors a⃗\vec{a} and b⃗\vec{b} is the same as finding the length of OC.

Using the triangle rule, we can find the length of OC in the following way. ∣OCβ†’βˆ£2=∣OAβ†’βˆ£2+∣ACβ†’βˆ£2+2∣OAβ†’βˆ£βˆ£ACβ†’βˆ£cos⁑α\left| \overrightarrow{OC} \right|^{2}={{\left| \overrightarrow{OA} \right|}^{2}}+{{\left| \overrightarrow{AC} \right|}^{2}}+2\left| \overrightarrow{OA} \right|\left| \overrightarrow{AC} \right|\cos \alpha Thus the resultant of two vectors aβƒ—\vec{a} and bβƒ—\vec{b} is: R=∣aβ†’βˆ£2+∣bβ†’βˆ£2+2∣aβ†’βˆ£βˆ£bβ†’βˆ£cos⁑αR=\sqrt{{{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+2\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \alpha } Furthermore, if the resultant of vectors aβƒ—\vec{a} and bβƒ—\vec{b}, namely vector rβƒ—\vec{r} forms an angle ΞΈ with respect to vector aβƒ—\vec{a} then the direction of the resultant vector R can be found with the following formula. sin⁑θ=∣bβƒ—βˆ£sin⁑αR\sin \theta =\frac{\left| \vec{b} \right|\sin \alpha }{R}

Example:

A ship is stuck in the middle of the sea. To bring the ship back to port two tugboats are needed. The force required by the two ships and the angle formed can be seen in the picture on the side. Determine the magnitude of the resultant force produced by the two ships!
Alternative Solutions:
The resultant forces of the two ships are depicted in the force diagram on the side. Ship Result Vector The resultant forces for both ships are given as follows. R=R12+R22+2R1R2cos⁑αR=802+1052+2β‹…80β‹…105cos⁑75∘Rβ‰ˆ6.400+11.025+16.800β‹…0,26Rβ‰ˆ6.400+11.025+4.368Rβ‰ˆ21.793Rβ‰ˆ147,62\begin{align*} & R=\sqrt{R_1^2+R_2^2+2R_1R_2\cos \alpha } \\ & R=\sqrt{{{80}^{2}}+{{105}^{2}}+2\cdot 80\cdot 105\cos 75{}^\circ } \\ & R\approx \sqrt{6.400+11.025+16.800\cdot 0,26} \\ & R\approx \sqrt{6.400+11.025+4.368} \\ & R\approx \sqrt{21.793} \\ & R\approx 147,62 \end{align*} So, the resultant force on both ships is 147.62 N.

2. Straight Line Normal Vector in R2 and Plane Normal in R3

The normal vector of a line is a vector that is perpendicular to that line. Because the condition is that it is perpendicular, the normal vector can be long or short, as long as it is not a zero vector. Usually the normal vector chosen is the simplest normal vector.
Theorem
nβƒ—=(ab) perpendicular to the line ax+by+c=0\vec{n}=\left( \begin{matrix} a \\ b \end{matrix} \right)\text{ perpendicular to the line }ax+by+c=0 in two-dimensional space R2 nβƒ—=(abc) perpendicular to line ax+by+cz+d=0\vec{n}=\left( \begin{matrix} a \\ b \\ c \end{matrix} \right)\text{ perpendicular to line }ax+by+cz+d=0 in three-dimensional space R3

3. Orthogonal Projection of a Vector to Another Vector

Orthogonal projection of a vector onto another vector, the result is a vector. Meanwhile, the length of the projection of an orthogonal vector on another vector is always a positive real number/scalar. vector-projection The orthogonal projection of vector u⃗\vec{u} on vector v⃗\vec{v} can be denoted by u⃗v⃗{\vec{u}}_{\vec{v}} or p⃗\vec{p} and is defined by the following argument.

Proposition:

  1. Scalar Projection orthogonal uβƒ—\vec{u} falls on vβƒ—\vec{v} is ∣∣pβƒ—βˆ£βˆ£=uβƒ—β‹…vβƒ—βˆ£vβƒ—βˆ£\left| \left| \vec{p}\right| \right|=\frac{\vec{u}\cdot\vec{v}}{\left| \vec{v} \right|}
  2. The projection of the vector uβƒ—\vec{u} onto the vector vβƒ—\vec{v} is the vector uβƒ—vβƒ—=(uβƒ—β‹…vβƒ—βˆ£.vβƒ—βˆ£2)vβƒ— atau pβƒ—=uβƒ—βˆ£vβƒ—βˆ£βˆ£βˆ£pβƒ—βˆ£βˆ£{\vec{u}}_{\vec{v}}=\left( \frac{\vec{u } \cdot \vec{v}}{\left|. \vec{v} \right|^2} \right)\vec{v}\text{ atau } \vec{p}=\frac{\vec{u }}{\left| \vec{v} \right|} \left| \left| \vec{p} \right| \right|
  3. Panjang project vector uβƒ—\vec{u} to vector vβƒ—\vec{v} is ∣uβƒ—vβƒ—βˆ£=∣uβƒ—β‹…evβƒ—βˆ£ \lvert \vec{u}_{\vec{v}}\rvert=\left| \vec{u}\cdot e_{\vec{v}} \right| dengan evβƒ—e_{\vec{v}} is vector satuan ke arah vβƒ—\vec{v} atau ∣uβƒ—vβƒ—βˆ£=∣uβƒ—.vβƒ—βˆ£vβƒ—βˆ£βˆ£\left| \vec{u}_{\vec{v}} \right|=\left| \frac{\vec{u}.\vec{v}}{\left| \vec{v} \right|} \right| .

Example 1:

Given aβƒ—=2i^βˆ’6j^βˆ’3k^\vec{a}=2\widehat{i}-6\widehat{j}-3\widehat{k} and bβƒ—=4i^+2j^βˆ’4k^\vec{b}=4\widehat{i}+2\widehat{j} -4\widehat{k}. Determine:

  1. The length of the vector projection a⃗\vec{a} on b⃗\vec{b}
  2. Orthogonal projection of vector a⃗\vec{a} on b⃗\vec{b}
  3. Orthogonal projection of vector b⃗\vec{b} on a⃗\vec{a}

Alternative Solutions:

  1. The length of the vector projection aβƒ—\vec{a} on bβƒ—\vec{b} ∣aβƒ—bβƒ—βˆ£=∣aβƒ—.bβƒ—βˆ£bβƒ—βˆ£βˆ£=∣(2i^βˆ’6j^βˆ’3k^)β‹…(4i^+2j^βˆ’4k^)42+22+(βˆ’4)2∣=∣(2)(4)+(βˆ’6)(2)+(βˆ’3)(βˆ’4)16+4+16∣=∣8βˆ’12+1236∣=∣86∣\begin{align*} \left| \vec{a}_{\vec{b}} \right|=\left| \frac{\vec{a}.\vec{b}}{\left| \vec{b} \right|} \right|&=\left| \frac{(2\widehat{i}-6\widehat{j}-3\widehat{k})\cdot (4\widehat{i}+2\widehat{j}-4\widehat{k})}{\sqrt{{{4}^{2}}+{{2}^{2}}+{{(-4)}^{2}}}} \right| \\ &=\left| \frac{(2)(4)+(-6)(2)+(-3)(-4)}{\sqrt{16+4+16}} \right| \\ &=\left| \frac{8-12+12}{\sqrt{36}} \right|=\left| \frac{8}{6} \right| \end{align*} ∴∣aβƒ—bβƒ—βˆ£=43 \therefore \left| \vec{a}_{\vec{b}} \right|=\frac{4}{3}

  2. Orthogonal projection of vector a⃗\vec{a} on b⃗\vec{b}
    aβƒ—bβƒ—=∣∣aβƒ—bβƒ—βˆ£βˆ£bβƒ—βˆ£bβƒ—βˆ£\vec{a}_{\vec{b}}=\left| \left| \vec{a}_{\vec{b}} \right| \right|\frac{\vec{b}}{\left| \vec{b} \right|}, karena ∣bβƒ—βˆ£=6\left| \vec{b} \right|=6 dan ∣∣aβƒ—bβƒ—βˆ£βˆ£=43\left| \left| {{\vec{a}}_{\vec{b}}} \right| \right|=\frac{4}{3}
    aβƒ—bβƒ—=43.4i^+2j^βˆ’4k^6=89i^+49j^βˆ’89k^\begin{align*} \vec{a}_{\vec{b}}&=\frac{4}{3}.\frac{4\widehat{i}+2\widehat{j}-4\widehat{k}}{6} \\ & =\frac{8}{9}\widehat{i}+\frac{4}{9}\widehat{j}-\frac{8}{9}\widehat{k} \end{align*}

  3. Orthogonal projection of vector b⃗\vec{b} on a⃗\vec{a}
    bβƒ—aβƒ—=(bβƒ—β‹…aβƒ—βˆ£aβƒ—βˆ£2)aβƒ—=(4i^+2j^βˆ’4k^)β‹…(2i^βˆ’6j^βˆ’3k^)(22+(βˆ’6)2+(βˆ’3)2)2(2i^βˆ’6j^βˆ’3k^)=(4)(2)+(2)(βˆ’6)+(βˆ’4)(βˆ’3)22+(βˆ’6)2+(βˆ’3)2(2i^βˆ’6j^βˆ’3k^)=849(2i^βˆ’6j^βˆ’3k^)bβƒ—aβƒ—=1649i^βˆ’4849j^βˆ’2449k^)\begin{align*}\vec{b}_{\vec{a}}&=\left( \frac{\vec{b}\cdot \vec{a}}{{{\left| \vec{a} \right|}^{2}}} \right)\vec{a} \\ &=\frac{(4\widehat{i}+2\widehat{j}-4\widehat{k})\cdot (2\widehat{i}-6\widehat{j}-3\widehat{k})}{{{\left( \sqrt{{{2}^{2}}+{{(-6)}^{2}}+{{(-3)}^{2}}} \right)}^{2}}}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ &=\frac{(4)(2)+(2)(-6)+(-4)(-3)}{{{2}^{2}}+{{(-6)}^{2}}+{{(-3)}^{2}}}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ &=\frac{8}{49}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ \vec{b}_{\vec{a}}&=\frac{16}{49}\widehat{i}-\frac{48}{49}\widehat{j}-\frac{24}{49}\widehat{k})\end{align*}

4. The point-to-line distance in R2 and the point-to-plane distance in R3

Proposition:

In two-dimensional space (R2), the distance of the point P(x1,y1)P(x_1,y_1) to the line ax+by+c=0ax+by+c=0 is d=∣ax1+by1+ca2+b2∣d=\left| \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \right|

On the three-dimensional space (R3) , the distance of the point P(x1,y1,z1)P(x_1,y_1,z_1) to the line ax+by+cz+d=0ax+by+cz+d=0 is d=∣ax1+by1+cz1+da2+b2+c2∣d=\left| \frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \right|

Example

Determine the distance from point (7,1) to the line 4x –3y +10 = 0!
Alternative Solution

  1. Vector way
    Normal garis g : 4x-3y+10 = 0 is nβƒ—=(4βˆ’3)\vec{n}=\left( \begin{matrix}4\\-3 \end{matrix} \right)
    Choose a point on the line 4xβˆ’3y+10=04x-3y+10 = 0 which has round coordinates, for example (–1,2)(–1 ,2). uβƒ—=(71)βˆ’(βˆ’12)=(8βˆ’1)\vec{u}=\left( \begin{matrix}7 \\ 1 \end{matrix} \right)-\left( \begin{matrix} -1 \\ 2\end{matrix} \right)=\left( \begin{matrix} 8 \\ -1 \end{matrix} \right) The distance from the point to the line in question is: d=∣uβƒ—.nβƒ—βˆ£nβƒ—βˆ£βˆ£d=∣(8βˆ’1)(4βˆ’3)42+(βˆ’3)2∣=∣32+35∣=7\begin{align*}& d=\left| \frac{\vec{u}.\vec{n}}{\left| \vec{n} \right|} \right| \\ & d=\left| \frac{\left( \begin{matrix} 8 \\ -1 \end{matrix} \right)\left( \begin{matrix} 4 \\ -3 \end{matrix} \right)}{\sqrt{4^2+(-3)^2}} \right|=\left| \frac{32+3}{5} \right|=7 \end{align*}
  2. Analytical way
    d=∣ax1+by1+ca2+b2∣=d=∣(4)(7)βˆ’3(1)+1042+32∣d=∣28βˆ’3+1025∣=355d=7\begin{align*} & d=\left| \frac{ax_1+by_1+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|= \\ & d=\left| \frac{(4)(7)-3(1)+10}{\sqrt{4^2+3^2}} \right| \\ & d=\left| \frac{28-3+10}{\sqrt{25}} \right|=\frac{35}{5} \\ & d=7\end{align*}

5. Surface area and volume of a geometric shape

The area of ​​the parallelogram formed by vector uβƒ—\vec{u} and vector vβƒ—\vec{v} is L=alas Γ— tinggiL=∣uβƒ—βˆ£βˆ£vβƒ—sin⁑θ∣L=∣uβƒ—βˆ£βˆ£vβƒ—βˆ£sin⁑θ, 0≀θ≀180\begin{align*} & L=\text{alas }\times \text{ tinggi} \\ & L=\left| \vec{u} \right|\left| \vec{v}\sin \theta \right| \\ & L=\left| \vec{u} \right|\left| \vec{v} \right|\sin \theta ,\text{ }0\le \theta \le 180 \end{align*} Because the value sin⁑θ\sin \theta is always positive then L=∣uβƒ—βˆ£βˆ£vβƒ—βˆ£sin⁑θL=\left| \vec{u} \right|\left| \vec{v} \right|\sin \theta can also be written L=∣uβƒ—Γ—vβƒ—βˆ£L=\left| \vec{u}\times \vec{v} \right|.

Since the area of ​​the triangle is exactly 12\dfrac12 of the parallelogram area, the area of ​​the triangle is: Lβ–³=12∣uβƒ—Γ—vβƒ—βˆ£{{L}_{\vartriangle }}=\frac{1}{2}\left| \vec{u}\times \vec{v} \right|

A parallel epipedum is a 6-sided space object whose parallel sides are congruent and each side is a parallelogram. The volume of the parallel epipedum formed by 3 vectors uβƒ—, vβƒ—,\vec{u},\text{ }\vec{v}, and wβƒ—\vec{w} is: V=∣uβƒ—β‹…(vβƒ—Γ—wβƒ—)∣=∣vβƒ—β‹…(uβƒ—Γ—wβƒ—)∣=∣wβƒ—β‹…(uβƒ—Γ—vβƒ—)∣V=\left| \vec{u}\cdot (\vec{v}\times \vec{w}) \right|=\left| \vec{v}\cdot (\vec{u}\times \vec{w}) \right|=\left| \vec{w}\cdot (\vec{u}\times \vec{v}) \right|

Example:

It is known that block ABCD EFGH with AB = 6, BC = 5, and CG = 4 is located in space coordinates R3 as shown in the figure. Determine the area of ​​the ACH field!
Alternative Solution
Parallel Epipedum From the image provided we can state that the coordinates of point A(5, 0, 0) ; B(5, 6, 0) ; C(0, 6, 0) ; H(0, 0, 4) ; F(5, 6, 4) ; and G(0, 6, 4). From this data, the area of ​​the ACH field L=12∣ACβ†’Γ—AHβ†’βˆ£L=12∣(cβƒ—βˆ’aβƒ—)Γ—(hβƒ—βˆ’aβƒ—)∣L=12∣(βˆ’560)Γ—(βˆ’504)∣=12∣(∣6004∣∣04βˆ’5βˆ’5βˆ£βˆ£βˆ’5βˆ’560∣)∣=12∣242030∣L=∣121015∣=122+102+152=469β‰ˆ21,66\begin{align*} & L=\frac{1}{2}\left| \overrightarrow{AC}\times \overrightarrow{AH} \right| \\ & L=\frac{1}{2}\left| (\vec{c}-\vec{a})\times (\vec{h}-\vec{a}) \right| \\ & L=\frac{1}{2}\left| \left( \begin{matrix} -5 \\ 6 \\0 \end{matrix} \right)\times \left( \begin{matrix} -5 \\ 0 \\ 4 \end{matrix} \right) \right|=\frac{1}{2}\left| \left( \begin{matrix} \left| \begin{matrix} 6 & 0 \\ 0 & 4 \end{matrix} \right| \\ \left| \begin{matrix}0 & 4 \\ -5 & -5 \end{matrix} \right| \\ \left| \begin{matrix}-5 & -5 \\ 6 & 0 \end{matrix} \right| \end{matrix} \right) \right|=\frac{1}{2}\left| \begin{matrix} 24 \\ 20 \\ 30 \end{matrix} \right| \\ & L=\left| \begin{matrix} 12 \\ 10 \\ 15 \end{matrix} \right|=\sqrt{{{12}^{2}}+{{10}^{2}}+{{15}^{2}}}=\sqrt{469}\approx 21,66 \end{align*}