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13.2 Vectors in 3D (part 2)

Vectors can be used to build shapes.

For example, sphere:

A sketch of a sphere with center C(h, k, l), a point (x, y, z) on the surface, and radius r.

Vector from center \( C(h, k, l) \) to point on sphere \( (x, y, z) \) must have a magnitude of \( r \) (radius).

Vector from center to point: \( \langle x-h, y-k, z-l \rangle \)

\[ |\langle x-h, y-k, z-l \rangle| = r \]

\[ \sqrt{(x-h)^2 + (y-k)^2 + (z-l)^2} = r \]

\[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]

Standard form of eq. of sphere

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Example

Find eq. of sphere whose diameter has endpoints \( P(-2, 3, 6) \), \( Q(4, -7, 5) \).

A sphere with diameter PQ passing through center C.

Need center, radius.

Center: midpoint between \( P, Q \)

average of \( x, y, z \) coordinates

\[ C \left( \frac{-2+4}{2}, \frac{3-7}{2}, \frac{6+5}{2} \right) \]

\[ = C \left( 1, -2, \frac{11}{2} \right) \]

Radius: half distance of \( \vec{PQ} \)

\[ \vec{PQ} = \langle 6, -10, -1 \rangle \]

\[ |\vec{PQ}| = \sqrt{36 + 100 + 1} = \sqrt{137} \]

radius \( = \frac{\sqrt{137}}{2} \)

eq: \[ (x-1)^2 + (y+2)^2 + \left( z - \frac{11}{2} \right)^2 = \left( \frac{\sqrt{137}}{2} \right)^2 \]

\[ = \frac{137}{4} \]

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Example: Finding the Center and Radius of a Sphere

Find center and radius of

\[x^2 + y^2 + z^2 - 14x + 16y - 10z + 102 = 0\]

Not in standard form, let's turn it into the standard form.

Complete the Square:

\[x^2 - 14x + 49 + y^2 + 16y + 64 + z^2 - 10z + 25 = -102 + 49 + 64 + 25\]

Note: To complete the square for each variable, take half of the linear coefficient and square it.

  • For \(x\): half of \(-14\) is \(-7\), squared is \(49\).
  • For \(y\): half of \(16\) is \(8\), squared is \(64\).
  • For \(z\): half of \(-10\) is \(-5\), squared is \(25\).
\[x^2 - 14x + 49 + y^2 + 16y + 64 + z^2 - 10z + 25 = 36\]
\[(x - 7)^2 + (y + 8)^2 + (z - 5)^2 = 6^2\]
  • Center: \((7, -8, 5)\)
  • Radius: \(6\)

Complete the square: coefficient of squared terms must be 1

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\[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]

All vectors from center to surface have length \(r\).

What about \((x - h)^2 + (y - k)^2 + (z - l)^2 \le r^2\)?

All vectors from center have length \(\le r\).

Sphere + all space inside \(\rightarrow\) solid ball

What about \((x - h)^2 + (y - k)^2 + (z - l)^2 < r^2\)?

Ball w/o the spherical "skin"

\((x - h)^2 + (y - k)^2 + (z - l)^2 > r^2\)

Vectors have lengths \(> r\).

Space outside NOT including sphere

A dashed sphere representing a boundary, with radiating lines indicating the infinite space outside of it.
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\[ 9 \le (x-1)^2 + (y+2)^2 + (z-5)^2 \le 25 \]

radius 3

center: \( (1, -2, 5) \)

radius 5

vectors from center have magnitudes between 3 and 5

A sketch of a spherical shell showing an inner sphere of radius 3 and an outer sphere of radius 5.
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Collinearity in 3D Space

Given 3 points in 3D space \( (\mathbb{R}^3) \), how do we know if they lie along a line? (collinear)

Three points P, Q, and R connected by a single straight line segment, labeled collinear.
Three points P, Q, and R not forming a straight line, labeled not collinear.

note: distances between them are not relevant for collinear

Vector diagram showing vector PQ and vector PR starting from point P and passing through Q to R.

if collinear, \( \vec{PQ} \parallel \vec{PR} \)

or equivalently, \( \vec{PQ} = k \vec{PR} \)

k = some constant

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Example: Collinear Points

Points \( P(1, 3, 2) \), \( Q(5, 1, 3) \), and \( R(x, 2, z) \) are collinear. Find \( x \) and \( z \).

\[ \vec{PQ} = \langle 4, -2, 1 \rangle \]\[ \vec{PR} = \langle x-1, -1, z-2 \rangle \]

Since the points are collinear, the vectors are parallel:

\[ \vec{PQ} \parallel \vec{PR} \implies \vec{PQ} = k \vec{PR} \]\[ \langle 4, -2, 1 \rangle = k \langle x-1, -1, z-2 \rangle \]

The vector equation turns into scalar equations by equating components:

\( x: \quad 4 = k(x-1) \)

\( y: \quad -2 = k(-1) \implies k = 2 \)

\( z: \quad 1 = k(z-2) \)

Now use this in the other equations to find \( x \) and \( z \).

\( x: \quad 4 = 2(x-1) \implies \)

\( x = 3 \)

\( z: \quad 1 = 2(z-2) \implies \)

\( z = \frac{5}{2} \)