Find solutions to your problems with the help of IDNLearn.com's knowledgeable users. Our platform is designed to provide accurate and comprehensive answers to any questions you may have.
Sagot :
To rotate the vector [tex]\(\langle -3, 5 \rangle\)[/tex] by [tex]\(270^\circ\)[/tex] clockwise about the origin, we need to apply a rotation matrix appropriate for a clockwise rotation.
1. Angle Conversion:
The angle of rotation given is [tex]\(270^\circ\)[/tex].
2. Rotation Matrix:
The general form of the rotation matrix for a clockwise rotation by angle [tex]\(\theta\)[/tex] is:
[tex]\[ \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} \][/tex]
For [tex]\(\theta = 270^\circ\)[/tex], we need to use the trigonometric values for [tex]\(270^\circ\)[/tex].
3. Trigonometric Values:
We know that:
[tex]\[ \cos(270^\circ) = 0 \quad \text{and} \quad \sin(270^\circ) = -1 \][/tex]
4. Substitute Trigonometric Values:
Substituting these values into the rotation matrix, we get:
[tex]\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \][/tex]
5. Matrix Multiplication:
Now, we multiply this rotation matrix by our vector [tex]\(\langle -3, 5 \rangle\)[/tex]:
[tex]\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} -3 \\ 5 \end{bmatrix} \][/tex]
Calculating the resulting vector:
[tex]\[ \begin{bmatrix} 0 \cdot -3 + (-1) \cdot 5 \\ 1 \cdot -3 + 0 \cdot 5 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \end{bmatrix} \][/tex]
So, the vector [tex]\(\langle -3, 5 \rangle\)[/tex] rotated [tex]\(270^\circ\)[/tex] clockwise about the origin is [tex]\(\langle -5, -3 \rangle\)[/tex].
Therefore, the rotated vector [tex]\(\langle -3, 5 \rangle\)[/tex] is [tex]\(\langle -4.999999999999999, -3.000000000000001 \rangle\)[/tex].
1. Angle Conversion:
The angle of rotation given is [tex]\(270^\circ\)[/tex].
2. Rotation Matrix:
The general form of the rotation matrix for a clockwise rotation by angle [tex]\(\theta\)[/tex] is:
[tex]\[ \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} \][/tex]
For [tex]\(\theta = 270^\circ\)[/tex], we need to use the trigonometric values for [tex]\(270^\circ\)[/tex].
3. Trigonometric Values:
We know that:
[tex]\[ \cos(270^\circ) = 0 \quad \text{and} \quad \sin(270^\circ) = -1 \][/tex]
4. Substitute Trigonometric Values:
Substituting these values into the rotation matrix, we get:
[tex]\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \][/tex]
5. Matrix Multiplication:
Now, we multiply this rotation matrix by our vector [tex]\(\langle -3, 5 \rangle\)[/tex]:
[tex]\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} -3 \\ 5 \end{bmatrix} \][/tex]
Calculating the resulting vector:
[tex]\[ \begin{bmatrix} 0 \cdot -3 + (-1) \cdot 5 \\ 1 \cdot -3 + 0 \cdot 5 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \end{bmatrix} \][/tex]
So, the vector [tex]\(\langle -3, 5 \rangle\)[/tex] rotated [tex]\(270^\circ\)[/tex] clockwise about the origin is [tex]\(\langle -5, -3 \rangle\)[/tex].
Therefore, the rotated vector [tex]\(\langle -3, 5 \rangle\)[/tex] is [tex]\(\langle -4.999999999999999, -3.000000000000001 \rangle\)[/tex].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.