Roam around in the Space

Geometry Level 3

The surface area of a surface in three dimensional space represented by the equation x 2 + y 2 + z 2 − 2 α x − 4 α y − 6 α z − 8 α 2 = 0 x^2+y^2+z^2-2 \alpha x-4\alpha y-6 \alpha z-8\alpha^2=0 for some positive real α , \alpha, is 88. If α = π a \alpha = \pi^a then find 16 a 16a .


The answer is -8.

View wiki
Sanjeet Raria by Sanjeet Raria , 27, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Since the coefficient of x 2 , y 2 , z 2 x^2 , y^2 , z^2 are equal, the given three dimensinal surface is a sphere.

Radius = r = α 2 + 4 α 2 + 9 α 2 + 8 α 2 = 22 α 2 r = \sqrt{\alpha^2 + 4\alpha^2 + 9\alpha^2 + 8\alpha^2} = \sqrt{22\alpha^2}

Surface Area = 4 π r 2 = 4 π 22 α 2 = 88 4 \pi r^2 = 4 \pi 22 \alpha^2 = 88

α 2 = 1 π ⇒ α = π − 1 2 \alpha^2 = \dfrac{1}{\pi} \Rightarrow \alpha = \pi^{\frac{-1}{2}}

16 a = − 8 \boxed{16a = \color{#D61F06}{-8} }

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...