# Math Help - Vectors

1. ## Vectors

1. AB is a diameter of a circle centered at the origin O, and P is any point on the circumference of the circle.
Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP.

2. Use a vector method to prove that the diagonals of the square OABC cross at right angles.

I’m confused, actually how to do these?

2. Originally Posted by geton
1. AB is a diameter of a circle centered at the origin O, and P is any point on the circumference of the circle.
Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP.

2. Use a vector method to prove that the diagonals of the square OABC cross at right angles.

I’m confused, actually how to do these?
Let $\alpha$ denote the angle $\angle(BOP)$

1. The position vector of A is (-r,0)
The position vector of B is (r, 0)
The position vector of P is $(r \cdot \cos(\alpha), r \cdot \sin(\alpha))$

2. Then

$\overrightarrow{AP}=(r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha))$ and

$\overrightarrow{BP}=(r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha))$

3. Now calculate

$\overrightarrow{AP} \cdot \overrightarrow{BP}= (r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) \cdot (r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha)) =$ $r^2\cos^2(\alpha)-r^2+r^2\sin^2(\alpha) = r^2-r^2 = 0$

And therefore the legs of a right triangle are perpendicular. (Thales theorem)

To the second question:

The cordinates of the squares vertices are O(0, 0), A(s, 0), B(s, s), C(0, s) where s means the side length.

Calculate the vectors of the diagonals and prove that the scalar product of these two vectors is zero.

3. Originally Posted by earboth
Let $\alpha$ denote the angle $\angle(BOP)$

1. The position vector of A is (-r,0)
The position vector of B is (r, 0)
The position vector of P is $(r \cdot \cos(\alpha), r \cdot \sin(\alpha))$

2. Then

$\overrightarrow{AP}=(r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha))$ and

$\overrightarrow{BP}=(r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha))$

3. Now calculate

$\overrightarrow{AP} \cdot \overrightarrow{BP}= (r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) \cdot (r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha)) =$ $r^2\cos^2(\alpha)-r^2+r^2\sin^2(\alpha) = r^2-r^2 = 0$

And therefore the legs of a right triangle are perpendicular. (Thales theorem)

To the second question:

The cordinates of the squares vertices are O(0, 0), A(s, 0), B(s, s), C(0, s) where s means the side length.

Calculate the vectors of the diagonals and prove that the scalar product of these two vectors is zero.

Thank you so much