- #1

brotherbobby

- 700

- 163

- Homework Statement
- Vector ##\vec A = 3.5 \hat x+2 \hat y-2.5 \hat z##. Find out the vector ##\vec B## that is obtained by rotating ##\vec A## about the ##x##-axis (in the positive sense according to the right-hand cork screw rule) by an angle of ##60^{\circ}##.

- Relevant Equations
- Dot product of two vectors : ##\vec A \cdot \vec B = AB \cos \theta##

The sketch above shows the situation of the problem. Clearly, as the rotation is taking place in the ##y-z## plane, the x-components of the two vectors remain unchanged : ##A_x = B_x##.

Let the projection of the vector ##\vec B## on to the

*y-z*plane be vector ##(\vec B)_{yz} = B_y \hat y + B_z \hat z##. It is these two vectors ##B_y \;\text{and}\; B_z## that I need to find.

The dot product ##(\vec A)_{yz} \cdot (\vec B)_{yz} = A_{yz} B_{yz} \cos \theta = \sqrt{2^2+2.5^2} \sqrt{B_y^2+B_z^2} \cos 60^{\circ} = 1.6 \sqrt{B_y^2+B_z^2} ## .

But again, ##(\vec A)_{yz} \cdot (\vec B)_{yz} = A_y B_y + A_z B_z = 2 B_y - 2.5 B_z \Rightarrow 2 B_y - 2.5 B_z = 1.6 \sqrt{B_y^2+B_z^2}## which simplifies to (after squaring and some algebra) : ##\mathbf{1.44 B_y^2 - 10 B_y B_z + 3.69 B_z^2 = 0}##

Also, as the length of the vectors remain invariant (property of pure rotation), we can write ##A_y^2+A_z^2 = B_y^2 + B_z^2 \Rightarrow B_y^2 + B_z^2 = 2^2 + 2.5^2 \Rightarrow \mathbf{B_y^2 + B_z^2 = 10.25}##.

*I could not solve (algebraically) the two equations above in bold for ##B_y\; \text{and} \;B_z##. Any help would be welcome.*