prove by vector method that the angle in Craft in a semicircle in a right angle
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I thought that you could only use dot product to find the angle between two vectors if you have the vectors directed away from the angle (and it says so in the Heinemann text book too). But in the following question:
AB is a diameter of a circle centred 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 (i.e. the angle in the semicircle is a right angle).
The solution's working uses AP and BP, i.e. the vectors are going INTO the angle in question, at P - here's a copy of just one version of this proof on the internet (and the answers CD from the text book does the same thing)
AP = p - a
BP = p - b
AP.BP
= (p-a).(p-b)
= |p|^2 - a.p -p.b + a.b
= r^2 - p.(a+b) + |a||b|cos 180
= r^2 - p.0 + r^2*-1
= 0
Hence, AP.BP = 0 => AP perpendicular to BP.
AB is a diameter of a circle centred 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 (i.e. the angle in the semicircle is a right angle).
The solution's working uses AP and BP, i.e. the vectors are going INTO the angle in question, at P - here's a copy of just one version of this proof on the internet (and the answers CD from the text book does the same thing)
AP = p - a
BP = p - b
AP.BP
= (p-a).(p-b)
= |p|^2 - a.p -p.b + a.b
= r^2 - p.(a+b) + |a||b|cos 180
= r^2 - p.0 + r^2*-1
= 0
Hence, AP.BP = 0 => AP perpendicular to BP.
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