A vector in a geometric shape (three dimensions) is a vector that has 3 axes, namely X, Y and Z, which are perpendicular to each other and the intersection of the three axes as the base

After previously studying vectors in the plane (R2), next we will develop our discussion regarding vectors in geometric shapes (R3). Vectors in geometric shapes (three dimensions) are vectors that have 3 axes, namely X, Y and Z, which are perpendicular to each other and the intersection of the three axes as the base.

1. Vector Writing in R3

Vectors in space are vectors that are located in 3-dimensional space. This space is formed by 3 axes, namely the X axis, Y axis, and Z axis. These three axes intersect perpendicularly. The result of this intersection is O. Next, point O is called the central axis. Look at the picture of the right hand finger rule on the side. right hand finger rule This rule explains several things, namely:

  1. The index finger shows the Y axis. The numbers after O and in the direction of the index finger are positive numbers. The opposite direction and location means a negative number.
  2. The thumb points to the X axis. Numbers that are in the direction of the thumb and located after O are positive numbers. The opposite direction and location are negative numbers.
  3. The middle finger shows the Z axis. Numbers that are in the direction of the middle finger and located after O are positive numbers. The opposite direction and location are negative numbers.

Look at the example of a space vector image on the side. vector ruang The vector OA→\overrightarrow{OA} on the side is a space vector with base O (0, 0, 0) and end A (1, 1, 1). This osition vector OA→\overrightarrow{OA} can be written with a column vector, becoming: OA→=(111)\overrightarrow{OA}=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix}

Space vectors can also be written in units i^,j^\widehat{i},\widehat{j} and k^\widehat{k}. Units i^\widehat{i} correspond to the X axis, units j^\widehat{j} correspond to the Y axis, and units k^\widehat{k} correspond to the Z axis. OB→=(111)\overrightarrow{OB}=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix} can be written as 1i^+1j^+1k^=i^+j^+k^1\widehat{i}+1\widehat{j}+1\widehat{k}=\widehat{i}+\widehat{j}+\widehat{k}.

Notes
Two or more vectors are called coplaner if they lie in the same plane.
Two or more vectors are called collinear if they lie on the same line.

2. Modulus or magnitude of vector

Vector modulus is the magnitude or length of a vector. The vector length OP→=(xyz)\overrightarrow{OP}=\begin{pmatrix} x \\y \\z\end{pmatrix} is formulated as follows. ∣OP→∣=x2+y2+z2\lvert \overrightarrow{OP} \rvert=\sqrt{x^2+y^2+z^2} If we know the points A(x1,y1,z1)A(x_1,y_1,z_1) and B(x2,y2,z2)B(x_2,y_2,z_2), analytically, we obtain the vector components AB→=(x2−x1y2−y1z2−y1)\overrightarrow{AB}=\begin{pmatrix} x_2-x_1 \\ y_2-y_1 \\z_2-y_1 \end{pmatrix}. So the length of the vector AB→\overrightarrow{AB} can be formulated:

∣AB→∣=(x2−x1)2+(y2−y1)2+(z2−z1)2\lvert \overrightarrow{AB} \rvert=\sqrt{\left( x_2-x_1 \right)^2+\left(y_2-y_1\right)^2+\left( z_2-z_1 \right)^2}

If the vector a⃗\vec{a} is presented in linear form a⃗=a1i^+a2j^+a3k^\vec{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}, then the modulus of the vector a⃗\vec{a } is ∣a⃗∣=a12+a22+a32\lvert \vec{a} \rvert=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}

Example:
Determine the modulus/size of the following vector!

  1. AB→\overrightarrow{AB} with points A (1, 4, 6) and B (3, 7, 9)
  2. a⃗=2i^+j^+3k^\vec{a}=2\widehat{i}+\widehat{j}+3\widehat{k}

Solution ✍️

  1. Given a⃗=(146)\vec{a}=\begin{pmatrix}1 \\4 \\6\end{pmatrix} and b⃗=(379)\vec{b}=\begin{pmatrix}3 \\7 \\9 \\ \end{pmatrix} maka AB→=b⃗−a⃗\overrightarrow{AB}=\vec{b}-\vec{ a} AB→=b⃗−a⃗AB→=(379)−(146)AB→=(3−17−49−6)AB→=(233)\begin{align*} \overrightarrow{AB}&=\vec{b}-\vec{a} \\\overrightarrow{AB}&=\begin{pmatrix} 3 \\7 \\9\end{pmatrix}-\begin{pmatrix}1 \\4 \\6\end{pmatrix} \\ \overrightarrow{AB}&=\begin{pmatrix} 3-1 \\7-4 \\9-6\end{pmatrix}\\ \overrightarrow{AB}&=\begin{pmatrix}2 \\3 \\3\end{pmatrix} \end{align*} Up to the length of the vector ∣AB→∣=22+32+32=4+9+9=22\lvert \overrightarrow{AB} \rvert=\sqrt{2^2+3^2+3^2}=\sqrt{4+9+9}=\sqrt{22}
    So, the modulus of the vector AB→\overrightarrow{AB} is 22.\sqrt{22}.
  2. ∣a⃗∣=22+12+32=14\lvert \vec{a} \rvert=\sqrt{2^2+1^2+3^2}=\sqrt{14}
    So, the modulus of the vector a⃗\vec{a} is 14.\sqrt{14}.

3. Unit Vector

A unit vector is a vector that has a length of 1 unit and is denoted as ee. The unit vector of vector a⃗\vec{a} is defined as vector a⃗\vec{a} divided by the magnitude of vector a⃗\vec{a} itself, which is defined as ea⃗=a⃗∣a⃗∣=1∣a⃗∣a⃗{{e}_{\vec{a}} }=\frac{\vec{a}}{\lvert \vec{a} \rvert}=\frac{1}{\lvert \vec{a} \rvert}\vec{a}

Example:

Determine the unit vector of Vector a⃗=(245)\vec{a}=\begin{pmatrix}2 \\4 \\\sqrt{5}\end{pmatrix}

Alternative solution:
First the length of the Vector a⃗\vec{a} is determined
∣a⃗∣=22+42+(5)2=25=5\lvert \vec{a} \rvert=\sqrt{2^2+4^2+(\sqrt{5})^2}=\sqrt{25}=5
ea⃗=15(245)e_{\vec{a}}=\frac{1}{5}\begin{pmatrix} 2 \\4 \\\sqrt{5} \end{pmatrix}
So, the unit vector of a⃗\vec{a} is ea⃗=(2/54/55/5)e_{\vec{a}}=\begin{pmatrix} {2}/{5} \\{4}/{5} \\ {\sqrt{5}}/{5} \end{pmatrix}

Apart from the unit vector, there are unit vectors that are parallel to the coordinate axes, including the following.

  1. The unit vector parallel to the X axis is denoted i^=(100),\widehat{i}=\begin{pmatrix}1 \\0 \\0\end{pmatrix},
  2. The unit vector parallel to the Y axis is denoted j^=(010)\widehat{j}=\begin{pmatrix}0 \\1 \\0\end{pmatrix}
  3. The unit vector parallel to the Z axis is denoted k^=(001)\widehat{k}=\begin{pmatrix}0 \\0 \\1 \end{pmatrix}

4. Position Vector

The position vector of point P is a vector that starts at point O (0, 0, 0) and ends at point P (x, y, z). Algebraically the position vector OP→\overrightarrow{OP} or p⃗\vec{p} can be written as follows. OP→=p⃗=(xyz)=xi^++yj^+zk^\overrightarrow{OP}=\vec{p}=\begin{pmatrix}x \\y \\z\end{pmatrix}=x\widehat{i}++y\widehat{j}+z\widehat{k} The vector AB→\overrightarrow{AB} with starting point A(x1,y1,z1)A(x_1,y_1,z_1) and ending point B(x2,y2,z2)B(x_2,y_2,z_2), has the following position vector.

AB→=OB→−OA→=(x2y2z2)−(x1y1z1)=(x2−x1y2−y1z2−z1)\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix} x_2 \\y_2 \\z_2 \end{pmatrix}-\begin{pmatrix} x_1 \\y_1 \\z_1 \end{pmatrix}=\begin{pmatrix} x_2-x_1 \\y_2-y_1 \\z_2-z_1 \end{pmatrix}

Example:

Given point A(−5,3,4)A(-5, 3, 4) and point B(−2,9,1)B(-2, 9, 1). Line AB intersects the plane XY at point C. Determine the coordinates of point C!

Solution ✍️

Known:
A(−5,3,4)⇒a⃗=(−534)A(-5,3,4)\Rightarrow \vec{a}=\begin{pmatrix}-5 \\3 \\4 \end{pmatrix}, B(−2,9,1)⇒b⃗=(−291)B(-2,9, 1)\Rightarrow \vec{b}=\begin{pmatrix} -2 \\ 9 \\1 \end{pmatrix} C on AB, so that vector AC→\overrightarrow{AC} is in line with Vector AB→ \overrightarrow{AB}. Therefore, AC→=k.AB→c⃗−a⃗=k(b⃗−a⃗)(xyz)−(−534)=k((−291)−(−534))(x+5y−3z−4)=(3k6k−3k)\begin{align*} \overrightarrow{AC}&=k.\overrightarrow{AB} \\ \vec{c}-\vec{a}&=k(\vec{b}-\vec{a}) \\ \begin{pmatrix}x \\ y \\ z \end{pmatrix}-\begin{pmatrix}-5 \\ 3 \\ 4 \end{pmatrix}&=k\left( \begin{pmatrix}-2 \\9 \\1 \end{pmatrix}-\begin{pmatrix}-5 \\3 \\4 \end{pmatrix} \right) \\ \begin{pmatrix}x+5 \\ y-3 \\ z-4 \end{pmatrix}&=\begin{pmatrix} 3k \\ 6k \\ -3k \end{pmatrix} \end{align*} Since AB is in the XY plane then z=0z=0 so z−4=−3k0−4=−3kk=43\begin{align*} z-4&=-3k \\ 0-4&=-3k \\ k&=\frac{4}{3} \end{align*} x+5=3kx+5=3.43x=−1\begin{align*} x+5&=3k \\ x+5&=3.\frac{4}{3} \\ x&=-1 \end{align*} y−3=6ky−3=6.43y=11\begin{align*} y-3&=6k \\ y-3&=6.\frac{4}{3} \\ y&=11 \end{align*} So, the position vector c⃗=(−1110)\vec{c}=\begin{pmatrix}-1 \\11 \\0 \end{pmatrix} so that the coordinates of point C are C(−1,11,0)C(-1,11,0)  

Exercise 4

  1. Determine the modulus of the following vectors:

    1. a⃗=(4−5−3)\vec{a} = \begin{pmatrix}4 \\-5 \\-3 \end{pmatrix}
    2. AB⃗\vec{AB} with point A(−2,3,−1)A (-2 , 3 , -1) and point B(2,1,−4)B (2 , 1 , -4)

  2. Given the vector PQ⃗\vec{PQ} with points P (2,5,−4)(2 , 5 , -4) and Q(1,0,−3)Q (1 , 0 , -3). Determine:

    1. The coordinates of point R if SR⃗\vec{SR} are the same as the vector PQ⃗\vec{PQ} if point S(2,−2,4)S (2 , -2 , 4)
    2. Coordinate of point N if MN⃗\vec{MN} is negative vector PQ⃗\vec{PQ} if point M(−1,3,2)M (-1 , 3 , 2)

  3. Determine the unit vector of the following vectors:

    1. u⃗=(00−1)\vec{u} = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix}
    2. v⃗=(111)\vec{v} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
    3. KL⃗\vec{KL} with K(3,−2,1)K (3 , -2 , 1) and L(2,−2,1)L (2 , -2 , 1)
    4. MN⃗\vec{MN} with M(2,1,2)M (2 , 1 , 2) and N(2,0,3)N (2 , 0 , 3)

  4. Draw a vector with points P(2,−3,1)P (2 , -3 , 1) and Q(1,3,−2)Q (1 , 3 , -2)

    1. Calculate the modulus of the vector PQ⃗\vec{PQ}
    2. Create a negative vector from PQ⃗\vec{PQ}, then calculate its modulus/magnitude!
    3. What can you conclude from the above work?

  5. If point P(1,1,1)P (1 , 1 , 1) and point Q(−1,4,−6)Q (-1 , 4 , -6), determine:

    1. position vector of point P and point Q
    2. vector components PQ⃗\vec{PQ}
    3. negative vector PQ⃗\vec{PQ}
    4. unit vector PQ⃗\vec{PQ}

  6. Determine the magnitude of the following vectors and their unit vectors!

    1. u⃗=(241)\vec{u} = \begin{pmatrix}2 \\4 \\1 \end{pmatrix}
    2. w⃗=−i^+5j^+k^\vec{w} = -\widehat{i} + 5\widehat{j} + \widehat{k}
    3. PQ⃗=(−305)\vec{PQ} = \begin{pmatrix} -3 \\0 \\5 \end{pmatrix}