Let O, A, B, C be four points such that OA=a, OB=b, OC=1/5(3b-a), then
(A) point C lies inside triangle OAB
(B) point C lies outside triangle OAB but inside angle OAB
(C) point C lies outside triangle OAB but inside angle ABO
(D) point C lies outside triangle OAB but inside angle BOA
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Class 12
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>>Section Formula Related to Vectors
>>In a triangle OAB, E is the mid - point
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In a triangle OAB, E is the mid-point of OB and D is a point on AB such that AD: DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using vector methods.
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Correct option is B)
With O as origin let a and b be the position vectors of A and B respectively.
Then the position vector of E, the mid-point of OB, is
2
b
Again, since AD:DB=2:1, the position vector of D is
1+2
1⋅a+2b
=
3
a+2b
Equation of OD and AE are r=t
3
a+2b
...(1)
and r=a+s(
2
b
−a) or r=(1−s)a+s
2
b
...(2)
If they intersect at p, then we will have identical values of r.
Hence comparing the coefficients of a and b, we get
3
t
=1−s,
3
2t
=
2
s
∴t=
5
3
or s=
5
4
.
Putting for t in (1) or for s in (2), we get the position vector of point of intersection P as
5
a+2b
...(3)
Now let P divide OD in the ratio λ:1.
Hence by ratio formula the P.V. of P is
λ+1
3
λ(a+2b)
+1.0
=
3(λ+1)
λ
(a+2b) ....(4)
Comparing (3) and (4), we get
3(λ+1)
λ
=
5
1
⇒5λ=3λ+3⇒2λ=3⇒λ=
2
3
∴OP:PD=3:2
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