IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

Ashish

If O is the origin, and if the coordinates of any two pts. P₁and P₂ are (x₁ ,y₁ ) and(x₂ ,y₂ ) respectively ,prove that

OP₁ .OP₂ .cos P₁OP₂ = x₁x₂ + y₁y₂

Please do not use the properties of straight lines to solve this problem.

I would appreciate it very much anybody could give the full solution.

[Taken from ?The Elements Of Coordinate Geometry? by S.L.Loney.]

Thanks.

krish

In ur question ,

P₁(x1,y1) and P₂(x2,y2)

OP₁= sqrt(x1²+y1²)

OP₂ =sqrt(x2²+y2²)

slope of OP₁ (m₁)= y1/x1

slope of OP₂ (m₂)= y2/x2

tan (P₁OP₂) = (m₁-m₂)/(1+m₁m₂)

cos(P₁OP₂) = (1+m₁m₂)/(sqrt((m₁-m₂)²+(1+m₁m₂)²))

OP₁.OP₂. cos(P₁OP₂) =sqrt(x1²x2²+x1²y2²+y1²x2²+y1²y2²) . (x1x2+y1y2)/

sqrt(x1²x2²+x1²y2²+y1²x2²+y1²y2²)

Therefore , OP₁.OP₂.cos (P₁OP₂) = x1x2+y1y2

Ashish

Thanks a lot Krish,but,as pointed out in the question itself, one is not supposed to use the properties of straight lines.Anyway,thanks for the effort and the time that you spent.

saurabh_hr23

Consider O as Origin, P1(x1,y1) P2(x2,y2) as two points. OP1, OP2 are two vectors. When their magnitude multiplied by the cosine of the angle between them you call it dot product.

Now (x1i + y1j) dot product (x2i + y2j)

Now you know it better!

rashmi_jain

hiiiiiiiiii

saurabh is right,but there can be one more way also.

just use distance formula.

in

ABC cosB=(a²+c²-b²)/2ac

use this formula to calculate the cos term

solution:

OP₁=sqrt(x₁²+y₁²)

OP₂=sqrt(x₂²+y₂²)

cosP₁OP₂=(x₁x₂+y₁y₂)/sqrt(x₁²+y₁²)sqrt(x₂²+y₂²)

I have taken a=OP₁ ,b=P₁P₂ c=OP₂

P₁P₂=sqrt((x₁-x₂)²+(y₁-y₂)²)²

so now u'll get the answer

Ashish

Thanks to both Saurabh hr and Ms Rashmi for giving the answers.

With regards,
Ashish

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