Vector `vec(A)` is 2`cm` long and is `60^(@)` above the x-axis in the first quadrant. Vector `vec(B)` is `2 cm` long and is `60^(@)` below the x-axis in the fourth quadrant. The sum `vec(A)+vec(B)` is a vector of magnitudes

Vector vec(A) is 2 cm long and is 60^(@) above the x-axis in the first quadrant. Vactor vec(B) is 2 cm long and is 60^(@) below the x-axis in the fourth quadrant. The sum vec(A)+vec(B) is a vector of magnitudes

A vector vec(A) is along the positive x - axis . If B is another vector such that vec(A) xx vec(B) is zero , then B could be

If for two vectors vec(A) and vec(B), vec(A)xxvec(B)=0 , the vectors

A vector vec(B) which has a magnitude 8.0 is added to a vector vec(A) which lies along the x-axis. The sum of these two vector is a third vector which lies along the y-axis and has a magnitude that is twice the magnitude of vec(A) . Find the magnitude of vec(A)

A vector vec(A) of length 10 units makes an angle of 60^(@) with a vector vec(B) of length 6 units. Find the magnitude of the vector difference vec(A)-vec(B) & the angles with vector vec(A) .

Two vector vec(A) and vec(B) have equal magnitudes. Then the vector vec(A) + vec(B) is perpendicular :

If the angel between unit vectors vec a and vec b60^(@) ,then find the value of |vec a-vec b|

Let vec(AB) be a vector in two dimensional plane with magnitude 4 units. And making an anle of 60^0 with x-axis, and lying in first quadrant. Find the components of vec(AB) in terms of unit vectors hati and hatj . so verify that calculation of components is correct.

Solve the vector equation in vec(x): vec(x) + vec(x) xx vec(a) = vec(b) .