Vectors：練習問題

The planes \(\Pi_1\) and \(\Pi_2\) have equations \(2x - y + 2z = 5\) and \(x + 2y - 2z = 7\) respectively. Find the acute angle between the two planes, giving your answer to the nearest degree.

Find the coordinates of the point where the line \(\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) intersects the plane \(x + 2y - z = 7\).

Find the equation of the plane that contains the point \((1, -1, 2)\) and the line \(\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\).

The plane \(\Pi\) has equation \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 6\). Find the coordinates of the point on the plane which is closest to the origin.

Find the shortest distance between the line \(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) and the line \(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).

Find the shortest distance from the point \(P(1, 2, 3)\) to the plane with equation \(2x - y + 2z = 5\).

Find the equation of the line of intersection of the planes with equations \(x + y + z = 1\) and \(2x - y + 3z = 4\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\).

Find the Cartesian equation of the plane containing the line \(\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{3}\) and parallel to the line \(\frac{x}{1} = \frac{y-2}{1} = \frac{z+1}{-2}\).