In a triangle ΔOAB, . If P and Q are points of trisection of AB, prove that
Given:- , P and Q are trisection of AB
i.e. AP = PQ = QB or 1:1:1 division of line AB
To Prove:-
Proof:- Let be position vector of O, A and B respectively
Now, Find position vector of P, we use section formulae of internal division: Theorem given below
“Let A and B be two points with position vectors
respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by
By above theorem, here P point divides AB in 1:2, so we get
⇒
⇒
Similarly, Position vector of Q is calculated
By above theorem, here Q point divides AB in 2:1, so we get
⇒
⇒
Length OA and OB in vector form
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⇒
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⇒
Now length/distance OP in vector form
⇒
⇒
⇒
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length/distance OQ in vector form
⇒
⇒
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Taking LHS
OP2 + OQ2
=
=
as we know in case of dot product
Angle between OA and OB is 90°,
⇒
⇒
Therefore, OP2 + OQ2
=
=
=
=
As from figure OA2 + OB2 = AB2
=
= RHS
Hence, Proved.