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Note: The 3-D coordinate axes divides the coordinate system into 3 distinct planes ... The following picture graphs the isosceles triangle in 3D space.
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Practice HW from Stewart Textbook (not to hand in) p. 641 # 1, 2, 3, 7, 10, 11, 13, 14, 15b, 16
In this chapter, we want to consider the 3 dimensional coordinate axes.
Points are located using ordered triples ( x , y , z ).
Example 1: Plot the points (1, 1, 1), (2, 5, 0), (-2, 3, 4), (1, 1, -4), and (2, -5, 3).
Solution:
Note: The 3-D coordinate axes divides the coordinate system into 3 distinct planes โ the x-y plane z = 0, the y-z plane x = 0, and the x-z plane y = 0. (see next page)
Suppose we are now given the two points P = ( x 1 , y 1 , z 1 )and Q = ( x 2 , y 2 , z 2 )in 3D space.
Then
2 2 1
2 2 1
2 2 1 1 1 1 2 2 2
( , , )and( , , )
Distance between thepoints d x x y y z z x y z x y z
( , , )and( , , ) 2
Midpoint between thepoints 1 2 1 2 1 2 1 1 1 2 2 2
x x y y z z M x y z x y z
P =( x 1 , y 1 , z 1 )
Q =( x 2 , y 2 , z 2 )
Example 2: Find the lengths of the sides of the triangle PQR if P = (1, -3, -2), Q = (5, -1, 2) and R = (-1, 1, 2). Deterrmine if the resulting triangle is an isosceles or a right triangle.
Solution: We will use the distance formula to find the length of the triangles sides PQ, QR, and RP. We obtain the following results:
Length of side PQ = ( 5 โ 1 )^2 +(โ 1 โโ 3 )^2 +( 2 โโ 2 )^2 = 42 + 22 + 42 = 36 = 6.
Length of side QR = ( โ 1 โ 5 )^2 +( 1 โโ 1 )^2 +( 2 โ 2 )^2 = (โ 6 )^2 + 22 + 02 = 40 = 4 โ 10 = 2 10
Length of side RP = ( 1 โ โ 1 )^2 +(โ 3 โ 1 )^2 +(โ 2 โ 2 )^2 = 22 +(โ 4 )^2 +(โ 4 )^2 = 36 = 6.
Since sides PQ and RP are equal, the triangle is isosceles. It is not a right triangle since the sum of squares of the shorter sides PQ and RP does not equal the square of the longest side QR. That is,
The following picture graphs the isosceles triangle in 3D space.
Standard Equation of a Sphere
The standard equation of a sphere with radius r and center ( x 0 (^) , y 0 , z 0 )is given by
2 2 0
2 0
2 ( x โ x 0 ) +( y โ y ) +( z โ z ) = r
In particular, if the center of the sphere is at the origin, that is, if ( x (^) 0 , y 0 , z 0 )=( 0 , 0 , 0 ),
then the equation becomes
x^2 + y^2 + z^2 = r^2
Fact: When the standard equation of a sphere is expanded and simplify, we obtain the general equation of a sphere
Standard Equation of a Sphere
Ax^2 + Ay^2 + Az^2 + Bx + Cy + Dz + E = 0
Note: To convert the general form of the sphere equation to standard form, we must complete the square.
Example 4: Find the center and radius of the sphere
4 x^2 + 4 y^2 + 4 z^2 โ 4 x โ 32 y + 8 z + 33 = 0
Solution:
