Question

Prove that semicircular angle is 90°.(Solve this numerical using vector algebra.)
If you answer this correctly I will mark you as the brainliest and will also follow you. No spams please. Please answer this fast.

Answer 1

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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Answer 2

Answer:

Sorry...... But I am much disciplined than I look.... I study in nights... But thanks for that.... I know even a second is important in IIT...... So OK.....