- #1
kramer733
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Question is:
1. True or False (Explain answer): A sphere with center P (Xo,Yo,Zo) and radius r consists of all points (x, y, z) that satisfy the inequality
(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2
My answer:
True because if we pick a vector where the "coordinate" of that vector is an element of the natural numbers (for ease) such as (7,23,16) and let that become the radius of the circle. Let the centre of the point become point of origin (0,0,0) and we can plug in the numbers for the inequality. Looking at the centre of the sphere, it's defined in this situation as (0,0,0). Since the inequality is
(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2
the (Xo ,Yo , Zo) are equal to zero in this case here.
We will then obtain something that equals r^2 because radius = (7,23,16)
To gain magnitude of vector, we use
7^2 + 23^2 + 16^2
After that, we can square root 7^2 + 23^2 + 16^2
Essentially that will give you the magnitude of the radius which is (834)^(0.5)
We can then just move the sphere around the R^3 plane. It should be the same all around right? I HONESTLY DON'T KNOW AT ALL AND AM MOSTLY PULLING IT OUT OF MY ***. To be honest i' don't even know if it works that way with Inequalities =/
Please correct me if I'm wrong.
1. True or False (Explain answer): A sphere with center P (Xo,Yo,Zo) and radius r consists of all points (x, y, z) that satisfy the inequality
(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2
My answer:
True because if we pick a vector where the "coordinate" of that vector is an element of the natural numbers (for ease) such as (7,23,16) and let that become the radius of the circle. Let the centre of the point become point of origin (0,0,0) and we can plug in the numbers for the inequality. Looking at the centre of the sphere, it's defined in this situation as (0,0,0). Since the inequality is
(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2
the (Xo ,Yo , Zo) are equal to zero in this case here.
We will then obtain something that equals r^2 because radius = (7,23,16)
To gain magnitude of vector, we use
7^2 + 23^2 + 16^2
After that, we can square root 7^2 + 23^2 + 16^2
Essentially that will give you the magnitude of the radius which is (834)^(0.5)
We can then just move the sphere around the R^3 plane. It should be the same all around right? I HONESTLY DON'T KNOW AT ALL AND AM MOSTLY PULLING IT OUT OF MY ***. To be honest i' don't even know if it works that way with Inequalities =/
Please correct me if I'm wrong.