Question

In the figure, the vectors $\mathbf{u$ and $\mathbf{v}$ are related as: $A\mathbf{u} = \mathbf{v}$ by a transformation matrix $A$. The correct choice of $A$ is:} \begin{center} \includegraphics[width=0.5\textwidth]{09.jpeg} \end{center}

• A. $\begin{bmatrix} \tfrac{4}{5} & \tfrac{3}{5}
  [6pt] -\tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}$
• B. $\begin{bmatrix} \tfrac{4}{5} & -\tfrac{3}{5}
  [6pt] \tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}$
• C. $\begin{bmatrix} \tfrac{4}{5} & \tfrac{3}{5}
  [6pt] \tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}$
• D. $\begin{bmatrix} \tfrac{4}{5} & -\tfrac{3}{5}
  [6pt] -\tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}$

Hint:

When a transformation maps one vector to another with the same magnitude, the transformation is a rotation. Always compute the angle from the dot product or slope, and check whether it's clockwise or anticlockwise.

Answer 1

The Correct Option is B

Step 1: Identify the given vectors. \\ From the figure, \[ \mathbf{u} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}, \mathbf{v} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}. \] We want a transformation matrix $A$ such that $A \mathbf{u} = \mathbf{v}$.

Step 2: Interpret geometrically. \\ Vector $\mathbf{u} = (4,3)$ has magnitude: \[ |\mathbf{u}| = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5. \] Vector $\mathbf{v} = (5,0)$ also has magnitude $5$. Thus, $A$ must be a rotation matrix that maps $(4,3)$ onto $(5,0)$.

Step 3: Find the rotation angle. \\ Direction of $\mathbf{u}$: \[ \theta_{u} = \tan^{-1}\!\left(\frac{3}{4}\right) \approx 36.87^\circ. \] Direction of $\mathbf{v}$: \[ \theta_{v} = 0^\circ. \] Therefore, the required rotation is clockwise by $36.87^\circ$.

Step 4: General form of 2D rotation matrix. \\ For a clockwise rotation by angle $\theta$, the rotation matrix is: \[ R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ [6pt] -\sin \theta & \cos \theta \end{bmatrix}. \] Here $\cos \theta = \tfrac{4}{5}$ and $\sin \theta = \tfrac{3}{5}$. So, \[ R = \begin{bmatrix} \tfrac{4}{5} & \tfrac{3}{5} \\ [6pt] -\tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}. \] But this corresponds to anticlockwise rotation. For clockwise, the correct form is: \[ R = \begin{bmatrix} \tfrac{4}{5} & -\tfrac{3}{5} \\ [6pt] \tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}. \]

Step 5: Verification. \\ \[ A \mathbf{u} = \begin{bmatrix} \tfrac{4}{5} & -\tfrac{3}{5} \\ [6pt] \tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix} \begin{bmatrix} 4 \\ 3 \end{bmatrix} = \begin{bmatrix} \tfrac{16}{5} - \tfrac{9}{5} \\ [6pt] \tfrac{12}{5} + \tfrac{12}{5} \end{bmatrix} = \begin{bmatrix} \tfrac{7}{5} \\ \tfrac{24}{5} \end{bmatrix}. \] Oops—this does not yield $(5,0)$. Let us carefully check again. Actually, the clockwise rotation matrix is: \[ R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ [6pt] -\sin \theta & \cos \theta \end{bmatrix}. \] Substituting $\cos \theta = 4/5$, $\sin \theta = 3/5$: \[ R = \begin{bmatrix} 4/5 & 3/5 \\ [6pt] -3/5 & 4/5 \end{bmatrix}. \] Now, \[ R \begin{bmatrix} 4 \\ 3 \end{bmatrix} = \begin{bmatrix} \tfrac{16}{5} + \tfrac{9}{5} \\ [6pt] -\tfrac{12}{5} + \tfrac{12}{5} \end{bmatrix} = \begin{bmatrix} 25/5 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}. \] This matches perfectly!

Step 6: Match with options. \\ The correct option is: \[ \begin{bmatrix} \tfrac{4}{5} & \tfrac{3}{5} \\ [6pt] -\tfrac{3}{5} & \tfrac{4}{5} \end{bmatrix}. \] This is option (A). % Final Answer \[ \boxed{\text{Option (A)}} \]