Prove by vector method, that the angle subtended on semicircle is a right angle. - Mathematics and Statistics

Sum

Prove by vector method, that the angle subtended on semicircle is a right angle.

Solution

Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B.

Then ∠APB is an angle subtended on a semicircle.

Let bar"AC" = bar"CB" = bar"a" and bar"CP" = bar"r"

Then |bar"a"| = |bar"r"|       ....(1)

bar"AP" = bar"AC" + bar"CP"

= bar"a" + bar"r"

= bar"r" + bar"a"

bar"BP" = bar"BC" + bar"CP"

= - bar"CB" + bar"CP"

= - bar"a" + bar"r"

∴ bar"AP".bar"BP" = (bar"r" + bar"a").(bar"r" - bar"a")

= bar"r".bar"r" - bar"r".bar"a" + bar"a".bar"r" - bar"a".bar"a"

= |bar"r"|^2 - |bar"a"|^2

= 0    ....(∵ bar"r".bar"a" = bar"a".bar"r")

∴ bar"AP" ⊥ bar"BP"

∴ ∠APB is a right angle.

Hence, the angle subtended on a semicircle is the right angle.

Concept: Scalar Triple Product of Vectors
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