QUESIION7
Transforming a Vector from World Space to Local Space
Choose one $\cdot 5$ points
A vector coordinate in world space is $\vec{r}=(1,0,2)$, calculate the vector coordinate in body frame coordinate (rigid body) $\vec{r} 0$ if the orientation of the rigid body is 90 about the global z-axis. Note: use $\overrightarrow{r_{0}}=\boldsymbol{R}^{t} \cdot \vec{r}$
$\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)$
$\left(\begin{array}{l}2 \\ 0 \\ 1\end{array}\right)$
$\left(\begin{array}{c}0 \\ -1 \\ 2\end{array}\right)$

Step 1 ：The problem is asking to transform a vector from world space to local space. This involves rotating the vector by the orientation of the rigid body, which is given as 90 degrees about the global z-axis. The formula given to perform this transformation is \(\overrightarrow{r_{0}}=\boldsymbol{R}^{t} \cdot \vec{r}\), where \(\boldsymbol{R}^{t}\) is the transpose of the rotation matrix and \(\vec{r}\) is the vector in world space.

Step 2 ：The rotation matrix for a rotation of 90 degrees about the z-axis is: \[\boldsymbol{R} = \begin{bmatrix} cos(90) & -sin(90) & 0 \\ sin(90) & cos(90) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Step 3 ：The transpose of this matrix is: \[\boldsymbol{R}^{t} = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Step 4 ：So, we need to multiply this matrix with the vector \(\vec{r}=(1,0,2)\) to get the vector in the body frame coordinate.

Step 5 ：The result of the matrix multiplication gives the vector in the body frame coordinate as (0, -1, 2). This matches one of the options given in the question.

Step 6 ：Final Answer: The vector coordinate in body frame coordinate is \(\boxed{\left(\begin{array}{c}0 \\ -1 \\ 2\end{array}\right)}\).