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Question
Prove by vector method that an angle in a semi-circle is a right angle
Solution
Let AB be the diameter of the circle with centre ‘O’
Let P be any point on the semi-circle.
To prove ∠APB = 90°
We have OA = OB = OP .......(radii)
Now `bar"PA" - bar"PO" + bar"OA"`
`bar"PB" = bar"PO" + bar"OB"`
= `bar"PO" - bar"OA"` .....`("Since" bar"OB" = - bar"OA")`
∴ `bar"PA" * bar"PB" = (bar"PO" + bar"OA") * (bar"PO" - bar"OA")`
= `(bar"PO")^2 - (bar"OA")^2`
= (PO)2 – (OA)2
`bar"PA" * bar"PB"` = 0
∴ `bar"PA"` ⊥' to `bar"PB"`
This gives ∠APB = 90°.
Hence the result.
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