Consider a sphere of radius R > 0. Take two antipodal points A and B on the sphere, and another point C on the sphere. Check that the angle ACB is a right angle.

Use only vectors. No solution involving classical geometry or coordinates will be accepted.

You may want to draw a picture, and introduce a good coordinate system.

To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?

Step-by-step explanation:

I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.

Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).

Now just form the dot product of those two vectors:

(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)

Now the truth of your claim will be embodied in that dot product being zero:

x^2 + y^2 - (1+z)(1-z) = 0

x^2 + y^2 - (1-z^2) = 0

x^2 + y^2 + z^2 = 1

But that last line is just the definition of points on the surface of a sphere of radius 1, so the claim is proven.

Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.

Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐