Rotating a given vector about an axis

[brotherbobby]

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.

 

[hutchphd]

Your logic seems fine. Let me eliminate the dead wood:
$$ \Rightarrow \mathbf{B_y^2 + B_z^2 = 10.25} $$
$$\Rightarrow2 B_y - 2.5 B_z = 1.6 \sqrt{B_y^2+B_z^2} $$
You really can't solve this?i

 

[kuruman]

An alternative way is to skip the algebra and use simple trig. The initial vector can be written as the sum of three vectors along the principal axes:$$\vec A=A_x~\hat x+A_y~\hat y+A_z~\hat z=\vec {A_x}+\vec {A_y}+\vec {A_z}.$$As you correctly pointed out $\vec{A_x}$ is unchanged upon rotation, but the other two rotate in the $zy$ plane by $60^o$ away from their respective principal axes. Make a drawing if you need to and then find the new $y$ and $z$ components of each. Add everything to get $\vec B$.

 

[brotherbobby]

Your logic seems fine. Let me eliminate the dead wood:
$$ \Rightarrow \mathbf{B_y^2 + B_z^2 = 10.25} $$
$$\Rightarrow2 B_y - 2.5 B_z = 1.6 \sqrt{B_y^2+B_z^2} $$
You really can't solve this?i

Thank you very much. I didn't realize that it would have been wiser to keep the quadratic expression under the square root as it is and use it for the other equation where it reappears. Am trying not to be embarrassed.

 

[hutchphd]

Better to ask a question. And graciously return the favor when possible.