Show that the resultant of two vectors of equal magnitude bisect the angle between the two vector
Answers
Answer:
THE RESULTANT OF TWO VECTORS OF EQUAL MAGNITUDE
THAT ARE INCLINED AT AN ANGLE θ TO EACH OTHER,bisects θ.
② Let the two vectors be OA,OB ,∠AOB=θ
Let |OA|=|OB|=m
∴Let //gram of forces be AOBR, OR=resultant=r
In △ of forces isosceles △AOR, AR=OB ,
let∠AOR=α,∠OAR=β
α+α+β=180°
β=180-2α↙
But∠OAR=β=180-θ→
↘β=180-θ=180–2α
θ=2α
Let A = Ai and B = Aj (where i and j are unit vectors) and C=A+B.
Then C = A(i + j).
A . C = Ai .A(i + j) = A^2 +A^2(i . j)
B . C = Aj . A(i + j) = A^2(j . i) + A^2 = A . C
Since the dot products are the same and the lengths of the corresponding vectors are the same, the angle between A and C must be the same as the angle between B and C. Therefore, C bisects the angle between A and B.
You would have to apply graphical vector addition, if you wanted to depict it. You would place the tip of one vector onto the tail the the other. You would then draw a vector from the free tail of one vector to the free tip of the other. You could carry out the same procedure using vector algebra. After you complete the problem, you would be able to determine the unit vectors and show that the resultant is NOT colinear with either vector when the vectors are not themselves colinear.
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Draw the vectors as with the parallelogram rule. Observe that if the vectors have equal magnitude, the figure is a rhombus, then show that the diagonal of a rhombus bisects the angle.
If you combine two vectors then the resultant is the diagonal of the parallelogram formed by them. In the case of two vectors of equal magnitude the parallelogram becomes a rhombus. The resultant then always bisects the angle between them, and its magnitude varies from twice one of them to zero as the angle changes from 0 to 180 degrees.
By symmetry. Since the two vectors are equal the resultant can’t be closer to one than the other.
How do you prove two unit vectors (a and b), a + b bisects the angle between vector a and vector b?
Let’s try to clarify the question with an example. Say 2 dimensions, polar coordinates, a=30 degrees and b=90 degrees.
In complex numbers,
a=cos30∘+isin30∘=12(3–√+i)
b=cos90∘+isin90∘=i
a+b=12(3–√+3i)
That’s in the first quadrant, angle
arctan33–√=arctan3–√=60∘
So the claim is some generalization of: if |a|=|b|=1 ,
arg(a+b)=12(arga+argb)
Let’s rewrite that
2arg(a+b)=arga+argb
arg(a+b)+arg(a+b)=arga+argb
arg(a+b)−arga=argb−arg(a+
Does the result of two vectors always lie along their angle bisector?
No. This will be the case only when the two vectors have equal magnitude.
R→=A→+B→
|R→|2=|A→|2+|B→|2+2|A→||B→|cosθ
Consider two vectors A→ and B→
We can move a vector any where in space, but only when we do not change its direction or magnitude.
According to head to tail method the resultant of the two vectors can be represented as :
If we draw a perpendicular from the point where the head of B→
How do we find a vector that bisects the angle between two vectors?
That depends on what you mean.
Do you mean bisects the angle when tail-to-tail or tail-to-head?
Let's start with a tail-to-tail bisector.
If our vectors were the same magnitude it would be easy, just add them. But then, your question says nothing of magnitude does it? So… if we had two vectors of the same direction as our vectors, but of the same magnitude as each other, we could simply use those!
Let our two vectors be a⃗ and b⃗ . Let's convert them to unit vectors of the same direction, and add them:
x(a⃗ |a⃗ |+b⃗ |b⃗ |)
That x is to show that th
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