MATH 251 Lecture 2

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Vectors and 3D Space

Given a center point $P$ and the radius $r$ , the defined sphere is:

S_{P} (r) = {p o i n t s x | d (X, P) = r}

where

d (X, P)

represents the distance between points

X

and

P

.

And its (standard) equation is

S_{P} (r) = {(x - P_{x})}^{2} + {(y - P_{y})}^{2} + {(z - P_{z})}^{2} = r^{2}

Find the center and radius of the sphere with the equation $x^{2} + 6 x + y^{2} - 4 y + z^{2} + 2 z = 0$

complete the square for all variables: $(x^{2} + 6 x + 9) + (y^{2} - 4 y + 4) + (z + 2 z + 1) = 0 + 9 + 4 + 1$
Factor: $(x + 3)^{2} + (y - 2)^{2} + (z + 1)^{2} = 14$
The center is at $(- 3, 2, - 1)$ , and the radius is $\sqrt{14}$

Prof's notation: ${\vec{v}}^{2} = \vec{v} \cdot \vec{v} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} = {| \vec{v} |}^{2}$

Dot product gives distances via

d (A, B)^{2} = (\vec{B - A}) \cdot (\vec{B - A})

θ = \arccos (\frac{\vec{v} \cdot \vec{w}}{| \vec{v} | | \vec{w} |})