Let \(\vec v\) be the column vector in \(\mathbb R^4\) which points from the point \(P = (2,-2,1,3)\) to \(Q = (0,-4,3,1)\).
Calculate the magnitude \(\lVert \vec v \rVert\)
The vector that points from \(Q\) to \(P\) is the vector \(Q - P\) (Check: \(P + (Q - P) = Q\)). So
\[\vec v = \begin{pmatrix}-2\\-2\\2\\-2\end{pmatrix}.\]The magnitude \(\lVert \vec v \rVert = \sqrt{4 + 4 + 4 + 4} = 4\).
Calculate the angle between \(\vec v\) and the vector,
\[\vec w = \begin{pmatrix}1\\-1\\0\\1\end{pmatrix}\]We just use the formula,
\[\theta = \cos^{-1}\left(\frac{\vec v \cdot \vec w}{\lVert \vec v \rVert \lVert \vec w \rVert}\right) = \cos^{-1}\left(\frac {-2+2+0-2}{4\sqrt 3}\right) \cos^{-1}\left(\frac {-1}{2\sqrt 3}\right).\]
This question pertains to an example of something called the cross product of two vectors in \(\mathbb R^3\). The cross product \(\vec u \times \vec v\) is defined as,
\[\begin{pmatrix}u^1\\u^2\\u^3\end{pmatrix} \times \begin{pmatrix}v^1\\v^2\\v^3\end{pmatrix} = \begin{pmatrix}a\\b\\c\end{pmatrix}\]where \(a,b,c\) are the coefficients of the variables \(i,j,k\) in the determinant,
\[\det \begin{pmatrix} i & j & k \\ u^1 & u^2 & u^3 \\ v^1 & v^2 & v^3 \end{pmatrix}.\]This problem has you work out a specific example.
If \(i, j, k\) are arbitrary variables, calculate:
\[\det \begin{pmatrix} i & j & k \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{pmatrix}\]The determinant is \((2-0)i + (0-2)j + (6-0)k\).
Let \(\vec v\) be the vector
\[\vec v = \begin{pmatrix} a\\b\\c \end{pmatrix}\]where \(a\) is the coefficient of \(i\) in your answer to (a), \(b\) is the coefficient of \(j\) in your answer to (a), and \(b\) is the coefficient of \(k\) in your answer to (a).
Calculate the dot product,
\[\vec v \cdot \begin{pmatrix}2\\1\\0\end{pmatrix}\]We get that \(\vec v\) is the vector
\[\vec v = \begin{pmatrix}2\\-2\\6\end{pmatrix}\]So the dot product is \(2-2+0=0\).
Calculate the dot product,
\[\vec v \cdot \begin{pmatrix}0\\3\\1\end{pmatrix}\]The dot product is \(0-6+6=0\).
In fact, the cross product \(\vec w \times \vec u\) is always orthogonal to the two vectors \(\vec w\) and \(\vec u\), so these dot products will always be 0!
Consider the vectors,
\[\vec v = \begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix} \quad\textrm{and}\quad \vec u = \begin{pmatrix}4 \\ 1 \\ 3\end{pmatrix}\]Find the set of all vectors \(\vec w\) which are orthogonal to both of \(\vec v\) and \(\vec u\).
The condition that a vector \(\vec x\) is orthogonal to these vectors yields the linear system of equations, \(\vec v^T \vec x = 0\) and \(\vec u^T \vec x = 0\), which becomes the matrix equation,
\[\begin{pmatrix} 2 & -1 & 3 \\ 4 & 1 & 3 \end{pmatrix} \vec x = 0.\]This is equivalent to the augmented matrix,
\[\begin{pmatrix} 2 & -1 & 3 & 0 \\ 4 & 1 & 3 & 0 \end{pmatrix},\]which row reduces to,
\[\begin{pmatrix} 2 & -1 & 3 \\ 0 & 3 & -3 \end{pmatrix} \vec x = 0.\]This means the solutions are, with \(z\) free,
\(y = z\) and \(2x - y + 3z = 0\) so \(2x = -2z\). As a span this is,
\[\mathop{\text{span}}\left\{ \begin{pmatrix}-1 \\ 1 \\ 1\end{pmatrix} \right\}.\]
Consider the matrix,
\[S = \begin{pmatrix} 4 & 3 \\ 0 & -2 \end{pmatrix}\]The determinant is -8.
The matrix,
\[S - \lambda I = \begin{pmatrix} 4-\lambda & 3 \\ 0 & -2 - \lambda\end{pmatrix},\]which has determinant \((4-\lambda)(-2-\lambda)\). The solutions to this being 0 are then \(\lambda = 4, \lambda = -2\) (it’s already factored!)
For any angle \(\varphi\), multiplication by the matrix,
\[M_\varphi = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix}\]defines a linear transformation (a.k.a. a linear function) that rotates vectors \(\vec v\) in \(\mathbb R^3\) around the \(x\)-axis by \(\varphi\) radians counterclockwise.
If you have any vector \(\vec v\) and you know that \(\lVert \vec v \rVert = 3\), calculate \(\lVert M_\varphi \vec v \rVert\). Hint: Think about how rotating something changes its length. The same principles apply here!
Length doesn’t change under rotation. So \(\lVert M_\varphi \vec v \rVert = \lVert \vec v \rVert = 3\)
Calculate the angle between \(\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).
We get the angle \(\cos^{-1}\left( \frac{2}{(2)(\sqrt{2})} \right).\)
Calculate the angle between \(M_{\pi/3} \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}\) and \(M_{\pi/3} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). Hint: You’re welcome to calculate this using matrix multiplication, but that might get a little exhausting. I encourage you to think about how this rotation will affect the angle between the vectors.
Rotating two vectors at the same time won’t change their angle. So it’s the same as before:
We get the angle \(\cos^{-1}\left( \frac{2}{(2)(\sqrt{2})} \right).\)