Calculus III 11.02 Coordinates and Vectors in Three Dimensions

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11.2 Coordinates and Vectors in Three Dimensions

• Understand the 3D rectangular coordinate system.
• Analyze vectors in 3D space.

Coordinates in 3D Space[1]

The 3D Coordinate System Figure 11.2.1 Points in 3D Coordinates. Figure 11.2.2

Take the traditional 2D Cartesian Coordinate System[2] and add a z-axis that passes through the origin and is perpendicular to both the x and y axes as shown in Figure 11.2.1. This adds the \(xz\) and \(yz\) planes to the traditional \(xy\)-plane. This divides the 3D Space into eight zones. The coordinate for any point becomes an ordered triple \((x,y,z)\). For example, in Figure 11.2.2 the point \((2,-5,3)\) breaks down to: \(x=2\) \(y = -5\) and \(z= 3\). A 3D coordinate system can have either a right-handed or left-handed orientation. Stand with your arms pointing in the direction of the positive \(x\) and \(y\)- axes, and with the \(z\)-axis pointing up, as shown in Figure 11.2.3. The system is right-handed or left-handed depending on which hand points along the \(x\)-axis. All 3D coordinate systems used here will be right-handed system. Points in 3D Coordinates. Figure 11.2.3

The distance between two points in 3D space. Figure 11.2.4

To find the distance between two points in 3D Space use the Pythagorean Theorem twice, as shown in Figure 11.2.4. The formula for the distance between the points \((x_{1},y_{1},z_{1})\) and \((x_{2},y_{2},z_{2})\) is the same as in 2D Space, just with the \(z\)-axis points added. \( d=\sqrt{ (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2} }\:\:\:\: \)3D Space Distance Formula

Example 11.2.1 Finding the Distance Between Two Points in 3D Space

Find the distance between the points \((2,-1,3)\) and \((1,0,-2)\).
Solution

\(d\)	\(=\sqrt{ (1-2)^{2} + (0+1)^{2} + (-2-3)^{2} }\)
	\(= \sqrt{1+1+25} \)
	\(= \sqrt{27} \)
	\(=3 \sqrt{3}\)

Example 11.2.2 Finding the Equation for a Point on a Sphere's[3] Surface

Figure 11.2.5

Find the equation for any point on a Sphere's surface given the points \((5,-2,3)\) and \((0,4,-3)\) are on the sphere's surface and are exactly opposite each other. Solution A sphere centered at \((x_{0},y_{0},z_{0})\) with radius \(r\) has a surface such that all points \((x,y,z)\) have the distance \(r\) from \((x_{0},y_{0},z_{0})\). The formula for finding any point on the sphere's surface is called the standard equation for a sphere. The form for any arbitrary point is \( (x-x_{0})^{2} + (y-y_{0})^{2} + (z-z_{0})^{2} = r^{2}\:\:\:\:\) Equation for a Sphere as shown in Figure 11.2.5. The center at \((x_{0},y_{0},z_{0})\) can be found using both given points and the midpoint formula. $$\left ( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2},\frac{z_{1}+z_{2}}{2} \right )\:\:\:\: \color{Red} {\text{Midpoint Formula}} $$ Insert the points to find the center $$\left ( \frac{5+0}{2},\frac{-2+4}{2},\frac{3-3}{2} \right ) = \left(\frac{5}{2},1,0 \right) $$ Apply the Distance Formula to find the radius $$r=\sqrt{ (0-\frac{5}{2})^{2}+(4-1)^{2}+(-3-0)^{2} }= \sqrt{\frac{97}{4}} $$ The standard equation for this sphere is $$(x-\frac{5}{2})^{2}+(y-1)^{2}+z^{2} = \frac{97}{4}$$

Vectors in Three Dimensions

Figure 11.2.6 Figure 11.2.7

In three dimensions vectors are denoted by an ordered triple where \(\textbf{v}= \langle v_{1}, v_{2}, v_{3} \rangle\). The zero vector is denoted by \(\textbf{0}= \langle 0, 0, 0 \rangle\) using the unit vectors \( \textbf{i}= \langle 1, 0, 0 \rangle, \: \textbf{j}= \langle 0, 1, 0 \rangle, \text{ and } \textbf{k}= \langle 0, 0, 1 \rangle \) the standard unit vector notation for \(\textbf{v}\) is \( \textbf{v} = v_{1} \textbf{i}+v_{2} \textbf{j}+v_{3} \textbf{k} \) as shown in Figure 11.2.6. If \( \textbf{v} \) is represented by the directed line segment from \(P(p_{1},p_{2},p_{3})\) to \(Q(q_{1},q_{2},q_{3})\), as shown in Figure 11.2.7. The component form for \( \textbf{v} \) is written by subtracting the initial point from the terminal point: \( \textbf{v} = \langle v_{1}, v_{2}, v_{3} \rangle = \langle q_{1}-p_{1}, q_{2}-p_{2},q_{3}-p_{3} \rangle\) Vector Operations in Three Dimensions Let \( \textbf{u} = \langle u_{1}, u_{2}, u_{3} \rangle \) and \( \textbf{v} = \langle v_{1}, v_{2}, v_{3} \rangle \) be vectors in three dimensions. Let \(c\) be a scalar. 1. Vector Equality: \( \textbf{u} = \textbf{v} \) if and only if \( u_{1} = v_{1}, \: u_{2} = v_{2}, \text{ and } u_{3} = v_{3} \). 2. Component Form: If \( \textbf{v} \) is represented by the directed line segment from \(P(p_{1},p_{2},p_{3})\) to \(Q(q_{1},q_{2},q_{3})\), then \( \textbf{v} = \langle v_{1}, v_{2}, v_{3} \rangle = \langle q_{1}-p_{1}, q_{2}-p_{2},q_{3}-p_{3} \rangle\). 3. Length: \( \left \| \textbf{v} \right \| = \sqrt{(v_{1})^{2} +(v_{2})^{2}+ (v_{2})^{2}} \) 4. Unit Vector in \(\textbf{v}\)'s Direction: $$\frac{\textbf{v}}{ \| \textbf{v} \|} = \left (\frac{1}{ \| \textbf{v} \|} \right) \langle v_{1},v_{2},v_{3} \rangle , \: \textbf{v} \ne 0 $$ 5. Vector Addition: \( \textbf{v} + \textbf{u} = \langle v_{1}+u_{1}, v_{2}+u_{2}, v_{3}+u_{3} \rangle \) 6. Scalar Multiplication: \(c\textbf{v} = \langle cv_{1}, cv_{2}, cv_{3} \rangle \)

Example 11.2.3 Vector Component Form in Three Dimensions

Find the component form and magnitude for vector \( \textbf{v} \) having initial point \((-2,3,1)\) and terminal point \((0,-4,4)\). Then find a unit vector along \( \textbf{v} \)'s direction.
Solution The component form for \( \textbf{v} \) is

\( \textbf{v} = \langle q_{1}-p_{1}, q_{2}-p_{2},q_{3}-p_{3} \rangle = \langle 0-(-2),-4-3,4-1 \rangle = \langle 2, -7,3 \rangle \).

The magnitude for \( \textbf{v} \) is

\( \| \textbf{v} \| = \sqrt{ 2^{2} + (-7)^{2} +3^{2} } = \sqrt{62}\)

A unit vector along \( \textbf{v} \)'s direction is

\( \textbf{u} \)	$$ = \frac{\textbf{v}}{ \| \textbf{v} \|} $$
	$$ = \frac{1}{ \sqrt{67} } \langle 2, -7, 3 \rangle $$
	$$ = \langle \frac{2}{\sqrt{67}}, \frac{-7}{\sqrt{67}}, \frac{3}{\sqrt{67}} \rangle .$$

Parallel vectors Figure 11.2.8

A nonzero vector \( \textbf{v} \) multiplied by any positive scalar produces a new vector in the same direction. Multiplication by a negative scalar produces a new vector in the opposite direction. In general, two nonzero vectors \( \textbf{u} \) and \( \textbf{v} \) are parallel when there is some scalar \(c\) such that \( \textbf{u} = c \textbf{v} \). For example, the vectors \( \textbf{u} \), \( \textbf{v} \), and \( \textbf{w} \) are parallel because \( \textbf{u}=2 \textbf{v} \text{ and } \textbf{w}=-\textbf{v} \) as shown in Figure 11.2.8

Definition 11.2.1 Parallel Vectors

Two nonzero vectors \( \textbf{u} \) and \( \textbf{v} \) are parallel when there is some scalar \(c\) such that \( \textbf{u} = c \textbf{v} \).

Example 11.2.4 Parallel Vectors

Vector \( \textbf{w} \) has initial point \((2,-1,3)\) and terminal point \((-4,7,5)\). Are vectors \( \textbf{u} \) or \( \textbf{v} \) parallel to \( \textbf{w} \)?
a. \( \textbf{u} = \langle 3,-4,-1 \rangle \)
b. \( \textbf{v} = \langle 12,-16,4 \rangle \)
Solution Begin by writing \( \textbf{w} \) in component form.

\( \textbf{w}= \langle -4-2,7-(-1),5-3 \rangle = \langle -6,8,2 \rangle \)

a. Because \( \textbf{u} = \langle 3,-4,-1 \rangle = -1/2 \langle -6,8,2 \rangle = -1/2 \textbf{w} \), \( \textbf{u} \) is parallel to \( \textbf{w}\).
b. This requires a scalar \( \textbf{c} \) such that

\( \langle 12,-16,4 \rangle = c \langle -6,8,2 \rangle \).

To find \(c\), solve for a common multiplier

\(12= -6c\rightarrow c=-2\)

\(-16= 8c\rightarrow c=-2\)

\(4= 2c\rightarrow c=2\)

Since a single scalar does not solve the equation the vectors \( \textbf{u} \) and \( \textbf{v} \) are not parallel.

Example 11.2.5 Using Vectors to Determine Collinear Points

The points \(P\), \(Q\), and \(R\) lie on the same line. Figure 11.2.9

Determine whether the points \( P(1,-2,3),\: Q(2,1,0), \text{ and } R(4,7,-6)\) are collinear. Solution The component forms for \(\overrightarrow{PQ}\) and \( \overrightarrow{PR} \) are \( \overrightarrow{PQ} = \langle 2-1,1-(-2),0-3 \rangle = \langle 1,3,-3 \rangle \) and \( \overrightarrow{PR} = \langle 4-1,7-(-2),-6-3 \rangle = \langle 3,9,-9 \rangle \). These two vectors have a common initial point. Therefore, \(P\), \(Q\), and \(R\) lie on the same line if and only if \(\overrightarrow{PQ}\) and \( \overrightarrow{PR} \) are parallel. As shown in Figure 11.2.9, they are parallel because \(\overrightarrow{PR} = 3 \overrightarrow{PQ} \).

Example 11.2.6 Standard Unit Vector Notation

a. Write the vector \( \textbf{v} = 4 \textbf{i} - 5\textbf{k} \) in component form.
b. Find the terminal point for the vector \( \textbf{v} = 7 \textbf{i} - \textbf{j}+ 3\textbf{k} \), given the origin is \(P(-2,3,5)\).
c. Find the magnitude for the vector \( \textbf{v} = -6 \textbf{i} +2 \textbf{j}- 3\textbf{k} \). Then find a unit vector in \( \textbf{v} \)'s direction.
Solution
a. Because \( \textbf{j} \) is missing, its component is \(0\) and

\( \textbf{v} = 4 \textbf{i} - 5\textbf{k} = \langle 4,0,-5 \rangle \).

b. You need to find \(Q(q_{1},q_{2},q_{3})\) such that

\( \textbf{v} = \overrightarrow{PQ} = 7 \textbf{i} - \textbf{j}+ 3\textbf{k} \).

This implies that \(q_{1}-(-2) = 7\), \(q_{2}-3 = -1\), and \(q_{3}-5 = 3\). This resolves to \(q_{1} = 5\), \(q_{2} = 2\), and \(q_{3} = 8\), Thus \(Q\) is \((5,2,8)\).
c. Note that \(v_{1} = -6\), \(v_{2} = 2\), and \(v_{3} = -3\). The magnitude for \( \textbf{v} \) is

\( \left \| \textbf{v} \right \| = \sqrt{(-6)^{2} +2^{2}+ (-3)^{2}} = \sqrt{ 49} = 7 \).

The unit vector in \( \textbf{v} \)'s direction is

\(1/7(-6 \textbf{i} + 2\textbf{j} -3\textbf{k}) = -6/7\textbf{i} + 2/7\textbf{j} - 3/7 \textbf{k}\).

Example 11.2.7 Measuring Force

Figure 11.2.10

A television camera weighing 120 pounds is supported by a tripod, as shown in Figure 11.2.10. Represent the force exerted on each leg of the tripod as a vector. Solution Let the vectors \(F_{1}\), \(F_{2}\), and \(F_{3}\), represent the forces exerted on the three legs, as shown in Figure 11.2.10. The directions for \(F_{1}\), \(F_{2}\), and \(F_{3}\) are determined as follows. $$\overrightarrow{PQ}_{1} = \langle 0-0, -1-0,0-4 \rangle = \langle 0,-1,-4 \rangle $$ $$\overrightarrow{PQ}_{2} = \left \langle \frac{\sqrt{3}}{2} -0, \frac{1}{2} -0,0-4 \right \rangle = \left \langle \frac{\sqrt{3}}{2},\frac{1}{2},-4 \right \rangle $$ $$\overrightarrow{PQ}_{3} = \left \langle -\frac{\sqrt{3}}{2} -0, \frac{1}{2} -0,0-4 \right \rangle = \left \langle -\frac{\sqrt{3}}{2},\frac{1}{2},-4 \right \rangle $$ Each leg has the same length and the total force is distributed equally among the three legs, therefore \( \| F_{1} \| = \| F_{2} \| = \| F_{3} \| \). This means there exists some constant \(c\) such that \( F_{1} = c \langle 0,-1,-4 \rangle \) $$ F_{2} = c \left \langle \frac{\sqrt{3}}{2},\frac{1}{2},-4 \right \rangle $$ $$ F_{3} = c \left \langle -\frac{\sqrt{3}}{2},\frac{1}{2},-4 \right \rangle $$ Let the total force exerted by the camera be \(F = \langle 0,0,-120 \rangle \). Negative because the weight it pressing down. This means the force on each vector is -40. That means \(c(-4) \) \(=-40\) \(c \) \(=10\) Plugging this back into the \(F_{n}\) vectors produces. \( F_{1} = \langle 0,-10,-40 \rangle \) $$ F_{2} = \left \langle 5\sqrt{3},5,-40 \right \rangle $$ $$ F_{3} = \left \langle -5\sqrt{3},5,-40 \right \rangle $$

Example 11.2.8 Finding the Equation for a Sphere by Completing the Square

Complete the square to write the equation for the sphere in standard form. Given \(4x^{2}+4y^{2}+4z^{2}−24x−4y+8z−23=0\), find the center and radius.
Solution

\(0\)	\(=4x^{2}+4y^{2}+4z^{2}−24x−4y+8z−23\)
	\(=(4x^{2}−24x) +(4y^{2}−4y)+(4z^{2}+8z)−23\)	Group Like Items
	\(=4(x^{2}−6x) +4(y^{2}−y)+4(z^{2}+2z)−23\)	Extract the common multiplier, here 4.
	\(=4(x^{2}-3x-3x+9−9) +4(y^{2}-1/2y-1/2y+1/4−1/4)+4(z^{2}+z+z +1-1)−23\)	Complete the Square per variable using the \([b*1/2]^{2}\) method.
	\(=4[(x^{2}-3x-3x+9)−9] +4[(y^{2}-1/2y-1/2y+1/4)−1/4]+4[(z^{2}+z+z+1)-1]−23\)	Group by Squares.
	\(=4[(x-3)^{2}−9] +4[(y-1/2)^{2}−1/4]+4[(z+1)^{2}-1]−23\)	Reduce
	\(=4(x-3)^{2}−36+4(y-1/2)^{2}−1+4(z+1)^{2}-4−23\)	Distribute 4.
	\(=4(x-3)^{2}+4(y-1/2)^{2}+4(z+1)^{2}−64\)	Sum the stray integers.
	\(=(x-3)^{2}+(y-1/2)^{2}+(z+1)^{2}−16\)	Divide by 4.
\(16\)	\(=(x-3)^{2}+(y-1/2)^{2}+(z+1)^{2}\)	Set equal to the stray integer.
\(4\)	\(=(x-3)+(y-1/2)+(z+1)\)	Take the root.

The center is \((-3,-1/2,1)\) and the radius is 4.

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Parent Article: Calculus III 11 Vectors and Spacial Geometry