If O be the origin and the coordinates of A and B are (x1,y1) and (x2,y2)respectively then prove that OA*OB*cos(angleAOB)=x1*x2+y1*y2?
Answers
Step-by-step explanation:
Let OA and OB makes angles θ₁ and θ₂ with x axis respectively and ∠AOB=α
Now OA=√{(x₁-0)²+(y₁-0)²}=√(x₁²+y₁²)
OB=√{(x₂-0)²+(y₂-0)²}=√(x₂²+y₂²)
Slope of OA=(y₁-0)/(x₁-0) =y₁/x₁
tanθ₁=y₁/x₁
Similarly, tanθ₂=y₂/x₂
Now, tanθ₁ -tanθ₂
tan(θ₁-θ₂)= --------------------------
1+tanθ₁ tanθ₂
(y₁/x₁) - (y₂/x₂)
= --------------------------
1 + (y₁/x₁)(y₂/x₂)
(y₁x₂-x₁y₂)/x₁x₂
= -------------------------
(x₁x₂+y₁y₂)/x₁x₂
(y₁x₂-x₁y₂)
tanα =---------------------
(x₁x₂+y₁y₂)
Νow squaring both sides
(y₁x₂-x₁y₂)²
tan²α = ---------------------
(x₁x₂+y₁y₂)²
(y₁x₂-x₁y₂)²
Sec²α - 1 =-----------------------
(x₁x₂+y₁y₂)²
(y₁x₂ -x₁y₂)²
Sec²α= ------------------- + 1
(x₁x₂+y₁y₂)²
(y₁x₂-x₁y₂)² +(x₁x₂+y₁y₂)²
=--------------------------------------
(x₁x₂ + y₁ y₂)²
y₁²x₂²+x₁²y₂²-2x₁x₂y₁y₂+x₁²x₂²+y₁²y₂²+2x₁x₂y₁
y₂
=--------------------------------------------------------------
(x₁x₂+y₁y₂)²
x₁²x₂²+x₁y₂²+y₁²x₂²+y₁²y₂²
=--------------------------------------------
(x₁x₂+y₁y₂)²
x₁²(x₂²+y₂²)+y₁²(x₂²+y₂²)
=-------------------------------------------
(x₁x₂ + y₁²y₂²)
(x₁²+y₁²)(x₂²+y₂²)
Sec²α = ------------------------------
(x₁x₂+y₁y₂)²
√(x₁²+y₁²) √(x₂²+y₂²)
Secα = --------------------------------
(x₁x₂+y₁y₂)
OA × OB
Secα= -------------------
(x₁x₂+y₁y₂)
(x₁x₂+y₁y₂)
Cosα= --------------------
OA ×OB
OA × OB Cos (∠AOB)= (x₁x₂+y₁y₂)
Given, O is the origin, So O is ( 0, 0)
Co-ordinates : A is (x₁, y₁), B is (x₂, y₂)
In vector form,
OA = x₁ i + y₁ j
OB = x₂ i + y₂ j
Let's say OA, OB are represented by Vectors a, b.
⇒ So,
⇒a. b = ( x₁ i + y₁ j) (x₂ i + y₂ j)
⇒ a. b = x₁x₂ + y₁y₂ - - - - (1)
By definition of Dot product,
⇒ a. b = |a| |b| cos(a, b)
⇒a. b = OA × OB ×cos ( ∠AOB) - - - - (2)
Now, From (1) & (2)
OA × OB × cos ( ∠AOB) = x₁x₂ + y₁y₂
Hence proved