Express \(\vec v\) as a column vector.
\[\vec v = Q-P = \begin{pmatrix}-1\\-1\\-3\\-6\end{pmatrix}.\]
Find the magnitude of \(\vec v\).
\[\lVert \vec v\rVert = \sqrt{(-1)^2+(-1)^2+(-3)^2+(-6)^2} = \sqrt{47}.\]
Find the angle from \(\vec v\) to the vector,
\[\vec w = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}.\]Because \(\vec v \cdot \vec w = \lVert \vec v \rVert \lVert \vec w \rVert\cos\theta\), we get that
\[\begin{aligned} \frac{-3}{\sqrt{47}\sqrt{1}} &= \cos \theta \\ \cos^{-1}{\left(\frac{-3}{\sqrt{47}}\right)} = \theta. \end{aligned}\]
The matrix \(R_\theta\)
\[R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\]has a nice graphical explanation. If \(\vec v\) is a 2-vector in \(\mathbb R^2\), the vector \(R_\theta \vec v\) (that is, the product of the matrix multiplication) has the same length as \(\vec v\), but has been rotated by \(\theta\) degrees counterclockwise (\(\theta\) can be any angle).
Let \(\vec v\) be the 2-vector
\[\vec v = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\]Without actually computing the vector \(R_{\pi/2} \vec v\), compute the following:
The dot product,
\[\vec v \cdot \left(R_{\pi/2} \vec v\right).\]The vector \(R_{\pi/2}\vec v\) is the same as \(\vec v\) but rotated by \(\pi/2\) radians. So, it is orthogonal to \(\vec v\), meaning that the dot product is \(0\).
The magnitude,
\[\lVert R_{\pi/2} \vec v \rVert.\]The vector \(R_{\pi/2}\vec v\) is the same as \(\vec v\) but rotated by \(\pi/2\) radians. So, it is the same length as \(\vec v\), and its magnitude \(\lVert R_{\pi/2}\vec v \rVert = \lVert \vec v \rVert = \sqrt{1^2 + (-2)^2} = \sqrt{5}\).
Let \(\vec {r_0}\) be a fixed vector in \(\mathbb R^2\). For each part, describe in words the set of all vectors \(\vec r\) that satisfy the stated condition. Hint: Think about nice shapes. An answer which just re-writes the math in English will not receive full credit.
\(\lVert \vec {r} - \vec {r_0} \rVert = 1\).
If the vector \(\vec {r_0}\) is drawn starting at the origin and ending at a point \(C\), then the set of all \(\vec r\) is the set of all vectors which, when drawn starting at he origin, end in the unit circle around \(C\).
\(\lVert \vec {r} - \vec {r_0} \rVert \ge 1\).
If the vector \(\vec {r_0}\) is drawn starting at the origin and ending at a point \(C\), then the set of all \(\vec r\) is the set of all vectors which, when drawn starting at he origin, end in or outside of the unit circle around \(C\).
Explain why the line of 3-vectors,
\[L = \left\{ \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix} + t\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} \;\middle|\; t \in \mathbb R \right\}\]is not a vector space.
You only have to show that \(L\) fails one of the vector space axioms. The easiest in this case is that it fails to have the zero vector \(\vec 0\):
Vectors in \(L\) look like
\[\begin{pmatrix}0\\2\\1\end{pmatrix} + t\begin{pmatrix}1\\-1\\0\end{pmatrix} = \begin{pmatrix}t\\2-t\\1\end{pmatrix},\]and always have a 1 in the third component. But the zero vector is
\[\vec 0 = \begin{pmatrix}0\\0\\0\end{pmatrix},\]which has a zero (not a 1) in the third component, and so can’t be in \(L\).
Does there exist a linear transformation \(T: \mathbb R^2 \to \mathbb R^3\) such that,
\[T\left(\begin{pmatrix}1\\0\end{pmatrix}\right) = \begin{pmatrix}1\\2\\3\end{pmatrix},\] \[T\left(\begin{pmatrix}0\\1\end{pmatrix}\right) = \begin{pmatrix}-1\\2\\1\end{pmatrix},\]and,
\[T\left(\begin{pmatrix}5\\1\end{pmatrix}\right) = \begin{pmatrix}4\\12\\1\end{pmatrix}?\]Justify your answer.
No. If \(T\) were linear then
\[\begin{aligned} T\left(\begin{pmatrix}5\\1\end{pmatrix}\right) &= T\left(\begin{pmatrix}0\\1\end{pmatrix} + 5\begin{pmatrix}1\\0\end{pmatrix}\right) \\ &= T\left(\begin{pmatrix}0\\1\end{pmatrix}\right) + 5T\left(\begin{pmatrix}1\\0\end{pmatrix}\right) \\ &= \begin{pmatrix}-1\\2\\1\end{pmatrix} + 5\begin{pmatrix}1\\2\\3\end{pmatrix} \\ &= \begin{pmatrix}4\\12\\16\end{pmatrix} \end{aligned}\]But this is different from what the problem says it should be:
\[T\left(\begin{pmatrix}5\\1\end{pmatrix}\right) = \begin{pmatrix}4\\12\\1\end{pmatrix}.\]So \(T\) can’t be linear.