if |vector a+ vector b| = |vector a- vector b|find the angle between a and b
let A+B=C, and c represent the magnatude of vectorC, or |vector a+ vector b| =c
then law of cosines
c^2=a^2+b^2-2ABcosAngleAB
Now consider |vector a- vector b|. The angle between A and B now is the supplement to the first angle AB. We can call this 180-AB.
law of cosines
c^2=a^2+b^2-2ABcosAngle(180-AB)
take the two equations, and subtract them.
0=0+0-2abcosAB -2abcos(180-AB)
or cosAB=-cos(180-AB)
cosAB= Cos180CosAB+sin180sinAB
cosAB= -cosAB+0
or 2cosAB=0 which means AB=90 deg.
To find the angle between two vectors, we can use the dot product of the vectors.
Let's start by considering the given equation:
|vector a + vector b| = |vector a - vector b|
The magnitude of a vector can be given by taking the square root of the dot product of the vector with itself. Using this information:
√((vector a + vector b) · (vector a + vector b)) = √((vector a - vector b) · (vector a - vector b))
Expanding the equation, we get:
√((vector a · vector a) + 2(vector a · vector b) + (vector b · vector b)) = √((vector a · vector a) - 2(vector a · vector b) + (vector b · vector b))
Simplifying, we have:
√(2(vector a · vector b)) = √(2(vector b · vector a))
Cancelling out the common factors of 2, we get:
√(vector a · vector b) = √(vector b · vector a)
Since the magnitudes on both sides are equal, we can conclude that:
vector a · vector b = vector b · vector a
This equation implies that the dot product of vector a and vector b is commutative, which means the angle between vector a and vector b is either 0 degrees (parallel) or 180 degrees (antiparallel).
Therefore, if |vector a + vector b| = |vector a - vector b|, the angle between vector a and vector b can be 0 degrees or 180 degrees.