Two vectors A and B have precisely equal magnitudes. For the magnitude of vector A + vector B to be larger th

Question

Two vectors A and B have precisely equal magnitudes. For the magnitude of vector A + vector B to be larger than the magnitude of vector A - vector B by the factor n, what must be the angle between them? (Use n for n as appropriate.)

Answer 1

From the definition of the dot product, we have that for two vectors, A = a1, a2, a3, a4…aN) and B = (b1, b2, b3, b4…bN), in N-dimensional space:

(A dot B)/(|A|*|B|) = cos(theta),

Where A dot B is the dot (inner) product of the vectors, |A| and |B| are the norms (lengths) of the vectors, and theta is the angle between them. For vectors in spaces with dimensions > 3, this relationship is commonly used to define the meaning of an angle between the two vectors. The dot product is calculated as:

A dot B = SUM[a_i * b_i], I = 1 to N

The problem asks for the value of theta for two vectors given that:

|A| = |B| (or, equivalently, |A|^2 = |B|^2

and

|A+B| = n*|A-B| (or, equivalently |A+B|^2 = n^2 * |A-B|^2)


The norm of an N-dimensional vector is given by: |X|^2 = SUM(x_i)^2, i = 1, N.
Using this to expand the equation relating |A+B| and |A-B| yields:

SUM[(a_i + b_i)^2] = n^2 * SUM[(a_i – b_i)^2], i = 1 to N
= SUM[a_i^2 + 2a_i*b_i + b_i^2] = n^2 * SUM[a_i^2 - 2a_i*b_i + b_i^2]
= SUM[a_i^2 + 2a_i*b_i + b_i^2 – n^2* a_i^2 + n^2*2a_i*b_i – n^2* b_i^2]
= SUM[(1-n^2)*a_i^2 + (1-n^2)*b_i^2) + 2*(1+n^2)*a_i*b_i)]
= (1-n^2)*|A|^2 + (1-n^2)*|B|^2 + 2*(1+n^2)* (A dot B)

Rearranging this last equation to solving for (A dot B)

(A dot B) =
[(n^2-1)/(2*(n^2+1))]*(|A|^2 + |B|^2)

But we know that |A|^2 = |B|^2, so:

(A dot B) =
[(n^2-1)/((n^2+1))]*|A|^2

Plug this into very first expression, which relates the angle between the vectors to their dot product:

cos(theta)=
[(n^2-1)/(n^2+1)]*|A|^2/(|A||B|)

Because |A| = |B|, |A||B| = |A|^2, so:
cos(theta)=[(n^2-1)/(n^2+1)]

theta = acos((n^2-1)/((n^2+1)))

Answer 2

Magnitude of vector A+B = (|A|^2+|B|^2+2*|A|*|B|*Cos T) ^1/2
If |A|=|B|, = (2*|A|^2 + 2*|A|^2*Cos T) ^1/2
Magnitude of vector A-B = (2*|A|^2 - 2*|A|^2*Cos T) ^1/2

Since |A+B| > n* |A-B|
(2*|A|^2 + 2*|A|^2*Cos T) ^1/2 > n*(2*|A|^2 - 2*|A|^2*Cos T) ^1/2

|A|*(2+2*Cos T) ^1/2 > n*|A|*(2-2*Cos T) ^1/2

(1+Cos T) > n ^2*(1-Cos T)

Cos T > (n^2-1)/ (n^2+1)

Example If n=2 , Cos T > 0.616
T< 53.13

Let T=52 deg.& |A|=|B|=1, |A+B| =1.80
|A-B| = 0.88

Therefore, |A+B|>2*|A-B|

Answer 3

Assume | A | = | B | = 1, and | A + B | = n| A – B |
Squaring:
A^2 + B^2 + 2AB = n^2( A^2 + B^2 – 2AB )
=> 2 + 2AB = n^2 ( 2 – 2AB )
=> 2AB ( 1 + n^2 ) = 2 ( n^2 - 1 )
=> AB = ( n^2 - 1 ) / ( n^2 + 1 )
But AB = |A| |B| CosT = CosT
Hence T = ArcCos [ (n^2 - 1) / (n^2 + 1) ]

Answer 4

|A|+|B|>n(|A|-|B|)
|A|=|B| are given.

Hence
|A|+|B|>n*0
or
|A|+|B|>0

therefore
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