Question

graph the image of v(-6, 3) after a rotation 90° clockwise around the origin.

Explanation:

Step1: Recall rotation rule

For a \(90^\circ\) clockwise rotation around the origin, the rule is \((x,y)\to(y, -x)\).

Step2: Apply the rule to \(V(-6,3)\)

Here, \(x = - 6\) and \(y = 3\). Substituting into the rule, we get the new coordinates as \((3,-(-6))=(3,6)\)? Wait, no, wait. Wait, the correct rule for \(90^\circ\) clockwise rotation about the origin is \((x,y)\to(y, -x)\). So for \(V(-6,3)\), \(x=-6\), \(y = 3\). So the new \(x\)-coordinate is \(y = 3\), and the new \(y\)-coordinate is \(-x=-(-6)=6\)? Wait, no, that's a mistake. Wait, no, the correct rule for \(90^\circ\) clockwise rotation: Let's derive it. A point \((x,y)\) in the coordinate plane. When we rotate \(90^\circ\) clockwise, the \(x\)-coordinate of the image is the \(y\)-coordinate of the original, and the \(y\)-coordinate of the image is the negative of the \(x\)-coordinate of the original. So \((x,y)\to(y, -x)\). So for \(V(-6,3)\), \(x=-6\), \(y = 3\). So the image \(V'\) has coordinates \((3,-(-6))=(3,6)\)? Wait, no, \(-x\) when \(x=-6\) is \(-(-6)=6\)? Wait, no, let's take an example. Let's take a point \((1,0)\), rotating \(90^\circ\) clockwise around origin should go to \((0,-1)\). Using the rule \((x,y)\to(y, -x)\), \((1,0)\to(0,-1)\), which is correct. Another example: \((0,1)\) rotating \(90^\circ\) clockwise should go to \((1,0)\). Using the rule \((0,1)\to(1, - 0)=(1,0)\), correct. Another example: \((2,3)\) rotating \(90^\circ\) clockwise should go to \((3,-2)\). Let's check: the vector from origin to \((2,3)\) is \(\vec{v}=(2,3)\). Rotating \(90^\circ\) clockwise, the rotation matrix is \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\) (wait, no, the standard rotation matrix for \(90^\circ\) clockwise is \(

$$\begin{pmatrix}\cos(-90^\circ)&-\sin(-90^\circ)\\\sin(-90^\circ)&\cos(-90^\circ)\end{pmatrix}$$

=

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\)? Wait, no, the rotation matrix for \(\theta\) clockwise is \(

$$\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}$$

\). For \(\theta = 90^\circ\), \(\cos90^\circ = 0\), \(\sin90^\circ=1\). So the matrix is \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\). So applying to vector \((x,y)\), we get \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}y\\-x\end{pmatrix}$$

\). So yes, the rule is \((x,y)\to(y, -x)\). So for \(V(-6,3)\), \(x=-6\), \(y = 3\). So the image is \((3,-(-6))=(3,6)\)? Wait, no, \(-x\) is \(-(-6)=6\)? Wait, no, \(x=-6\), so \(-x = -(-6)=6\). So the new coordinates are \((3,6)\)? Wait, but let's plot the original point \(V(-6,3)\). It's in the second quadrant. Rotating \(90^\circ\) clockwise should move it to the first quadrant? Wait, no, rotating \(90^\circ\) clockwise from second quadrant: the second quadrant is \(x<0,y>0\). Rotating \(90^\circ\) clockwise, the \(x\) becomes \(y\) (positive) and \(y\) becomes \(-x\) (positive, since \(x\) is negative). So yes, it should be in the first quadrant. So the image of \(V(-6,3)\) after \(90^\circ\) clockwise rotation around origin is \((3,6)\)? Wait, no, wait, I think I made a mistake. Wait, let's take the point \((-1,2)\). Rotating \(90^\circ\) clockwise: using the rule \((x,y)\to(y, -x)\), we get \((2,1)\). Let's check with the rotation matrix. The vector is \((-1,2)\). Applying the matrix \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\), we get \((0\times(-1)+1\times2, - 1\times(-1)+0\times2)=(2,1)\), which is correct. So for \((-6,3)\), applying the matrix, we get \((0\times(-6)+1\times3, - 1\times(-6)+0\times3)=(3,6)\). So that's correct.

Answer:

The image of \(V(-6,3)\) after a \(90^\circ\) clockwise rotation around the origin is the point with coordinates \((3,6)\). To graph it, plot the point \((3,6)\) on the coordinate plane.