To determine the degree of rotation and direction given the points \( P(-4, 10) \) and \( P'(-10, -4) \), we can follow these steps:
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Calculate the angle of rotation using the dot product: The angle of rotation can be found using the dot product and magnitudes of the original vector and the rotated vector.
Let \( \vec{P} = (-4, 10) \) and \( \vec{P'} = (-10, -4) \).
The dot product \( \vec{P} \cdot \vec{P'} \) is calculated as: \[ \vec{P} \cdot \vec{P'} = (-4)(-10) + (10)(-4) = 40 - 40 = 0 \]
The magnitude of \( \vec{P} \) and \( \vec{P'} \) are: \[ |\vec{P}| = \sqrt{(-4)^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29} \] \[ |\vec{P'}| = \sqrt{(-10)^2 + (-4)^2} = \sqrt{100 + 16} = \sqrt{116} = 2\sqrt{29} \]
Using the dot product formula, we relate them: \[ \vec{P} \cdot \vec{P'} = |\vec{P}| |\vec{P'}| \cos(\theta) \] Since the dot product is \( 0 \), we have: \[ 0 = (2\sqrt{29})(2\sqrt{29}) \cos(\theta) \] This implies \( \cos(\theta) = 0 \), which means \( \theta = 90^\circ \) or \( \theta = 270^\circ \).
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Determine the direction of rotation: To find the rotation direction, we compute the angles of the vectors from the positive x-axis.
The angle of \( \vec{P} = (-4, 10) \): \[ \tan(\alpha) = \frac{10}{-4} \implies \alpha = \tan^{-1}\left(-\frac{5}{2}\right) \text{ (2nd quadrant)} \]
The angle of \( \vec{P'} = (-10, -4) \): \[ \tan(\beta) = \frac{-4}{-10} = \frac{2}{5} \implies \beta = \tan^{-1}\left(\frac{2}{5}\right) \text{ (3rd quadrant)} \]
Calculating the actual angles:
- For \( P(-4, 10) \), since it is in the 2nd quadrant, the reference angle is: \[ \theta_P = 180^\circ + \tan^{-1} \left( \frac{10}{4} \right) \approx 180^\circ + 68.2^\circ \approx 248.2^\circ \]
- For \( P'(-10, -4) \), since it is in the 3rd quadrant: \[ \theta_{P'} = 180^\circ + \tan^{-1} \left( \frac{-4}{-10} \right) \approx 180^\circ + 21.8^\circ \approx 201.8^\circ \]
To find the rotation angle, compute the difference: \[ \text{Rotation angle} = \theta_{P'} - \theta_P = 201.8^\circ - 248.2^\circ = -46.4^\circ \]
Since we want the positive angle, we perform: \[ 360^\circ - 46.4^\circ = 313.6^\circ \]
Thus, the degree of rotation is \( 90^\circ \) counterclockwise (or \( 270^\circ \) clockwise, but typically we refer to counterclockwise).
Final Answer:
The degree of rotation is \( 90^\circ \) in a counterclockwise direction.