• Two vectors a and b


• Resultant of a and b is perpendicular to a


• Magnitude of resultant = 1/2 magnitude of b

• Angle between a and b

Let's assume that the angle between vectors a and b is theta.



Given that the magnitude of resultant vector is half of the magnitude of vector b.

Let's assume that the magnitude of vector b is 'x'.

Therefore, the magnitude of resultant vector is x/2.



Let's express vector a in terms of its magnitude and direction.

Let the magnitude of vector a be 'y'.

Let the angle between vector a and the positive x-axis be alpha.

Then vector a can be expressed as:

a = y cos alpha i + y sin alpha j

Let's express vector b in terms of its magnitude and direction.

Vector b can be expressed as:

b = x cos theta i + x sin theta j



The dot product of vectors a and b is given by:

a . b = (y cos alpha)(x cos theta) + (y sin alpha)(x sin theta)

a . b = xy(cos alpha cos theta + sin alpha sin theta)

a . b = xy(cos(alpha - theta))



Given that the resultant of vectors a and b is perpendicular to vector a.

Therefore, the dot product of vectors a and b should be equal to zero.

a . b = 0

xy(cos(alpha - theta)) = 0

cos(alpha - theta) = 0

alpha - theta = 90 degrees

theta = alpha - 90 degrees



We know that vector b can be expressed as:

b = x cos theta i + x sin theta j

Substituting the value of theta in the above equation, we get:

b = x cos (alpha - 90 degrees) i + x sin (alpha - 90 degrees) j

b = -x sin alpha i + x cos alpha j

Therefore, the angle between vectors a and b is 90 degrees - alpha.



The angle between vectors a and b is 90 degrees - alpha.

60 degree.