Consider the four dimensional space R4 with coordinates x1 x

Consider the four dimensional space R^4 with coordinates (x_1, x_2, x_3, x_4). A hyperplane is the set of points whose coordinates satisfy an equation a x_1 + b x_2 + c x_3 + d x_4 = k, where a, b, c, d, and k are fixed real numbers. Find the coordinates od a vector which is perpendicular to a plane a x_1 + b x_2 + c x_3 + d x_4 = k? What is the intersection of two hyperplanes x_1 = 1 and x_2 = 2? Describe the intersection as a geometric object in words and by an equation. What is the intersection of a four dimensional sphere x^2 _1 + x^2 _2 + x^2 _3 + x^2 _4 = R^2 with a hyperplane x_1 = r for r R? Find the distance between the hypersphere x^2 _1 + x^2 _2 + x^2 _3 + x^2 _4 = R^2 and the hyperplane a x_1 + b x_2 + c x_3 + d x_4 = f where a, b, c, d, f are real numbers. Derive a formula for the volume of an n-dimensional sphere sigma^n k = 0 |z_k|^2 = R^2 and an n-dimensional ball sigma^n k = 1 |z_k|^2

Solution

(1) Given plane ax₁+bx₂+cx₃+dx₄ = k

So we can express it in terms of vectors ---- r.n = k

So perpendicular coordinates are (a,b,c,d).

(2) Now x₁ = 1, x₂ = 2

So a+2b+cx₃+dx₄ = k

cx₃+dx₄ = k-a-2b

cx₃+dx₄ = p (a new constant equat to k-a-2b)

so It represents the equation of a line in the plane of x₃ and x_4.

(3). x₁² + x₂² + x₃² + x₄² = R²

   x₁ = R

   for r < R,

x₂² + x₃² + x₄² = R² - r²

It represents a three dimensional sphere because R² - r² > 0 .

for r = R,

  x₂² + x₃² + x₄² = R² - R²

  x₂² + x₃² + x₄² = 0

It means x₂ = 0, x₃ = 0, x₄= 0

It represents a point (R,0,0,0).

for r > R

  x₂² + x₃² + x₄² = R² - r²

It doesn\'t represent any geometry bacause right hand side is negative whereas left hand side is positive.