Unit 22 Review Exercises

Vectors III

These questions mix the ideas from the whole unit. All of them are HL only. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 AHL The points \(A(1, -1, 2)\), \(B(3, 0, 1)\) and \(C(2, 2, 0)\) lie on the plane \(\Pi\).

(a) Find \(\overrightarrow{AB}\times\overrightarrow{AC}\).

(b) Hence find the Cartesian equation of \(\Pi\).

Solution

(a) \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}\).

\[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix} (1)(-2) - (-1)(3) \\ (-1)(1) - (2)(-2) \\ (2)(3) - (1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}. \]

(b) \(d = 1 - 3 + 10 = 8\), so \(\;x + 3y + 5z = 8\). (Check: \(B\) gives \(3 + 0 + 5 = 8\) ✓.) See 22.1 and 22.2.

2 Paper 1 AHL The plane \(\Pi\) has vector equation \(\boldsymbol{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}\).

(a) Find the Cartesian equation of \(\Pi\).

(b) Find the points where \(\Pi\) meets the coordinate axes.

Solution

(a) \(\boldsymbol{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\times\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} (1)(-1) - (0)(0) \\ (0)(2) - (1)(-1) \\ (1)(0) - (1)(2) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}\). Use \(\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\): \(\;d = 0 - 1 + 4 = 3\), so \(\;x - y + 2z = 3\).

(b) \((3, 0, 0)\), \(\;(0, -3, 0)\), \(\;\left(0, 0, \dfrac{3}{2}\right)\). See 22.2.

3 Paper 1 AHL Find the coordinates of the point where the line \(\boldsymbol{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) meets the plane \(2x + y + 3z = 9\).

Solution

\(2(3 + t) + (-1 + 2t) + 3(2 - t) = 9 \Rightarrow 11 + t = 9 \Rightarrow t = -2\). The point is \((1, -5, 4)\). See 22.3.

4 Paper 2 AHL Find the acute angle between the line and the plane in Question 3.

Solution

\(\boldsymbol{d}\cdot\boldsymbol{n} = 2 + 2 - 3 = 1\), \(|\boldsymbol{d}| = \sqrt{6}\), \(|\boldsymbol{n}| = \sqrt{14}\).

\(\sin\theta = \dfrac{1}{\sqrt{6}\sqrt{14}} = \dfrac{1}{\sqrt{84}}\), so \(\theta = 6.26^\circ\) (3 s.f.). See 22.3.

5 Paper 1 AHL The point \(P\) is \((4, 1, 5)\) and the plane \(\Pi\) is \(x + 2y + 2z = 7\).

(a) Find the foot of the perpendicular from \(P\) to \(\Pi\).

(b) Find the distance from \(P\) to \(\Pi\).

(c) Find the image of \(P\) in \(\Pi\).

Solution

(a) \((4 + \lambda) + 2(1 + 2\lambda) + 2(5 + 2\lambda) = 7 \Rightarrow 16 + 9\lambda = 7 \Rightarrow \lambda = -1\). Foot \(F(3, -1, 3)\).

(b) \(|\lambda|\,|\boldsymbol{n}| = 1 \times 3 = 3\).

(c) \(\overrightarrow{OP'} = \overrightarrow{OP} + 2\lambda\boldsymbol{n} = \begin{pmatrix} 4 \\ 1 \\ 5 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\). See 22.4.

6 Paper 1 AHL Find a vector equation of the line of intersection of the planes \(x + y - z = 1\) and \(2x - y + 3z = 8\).

Solution

\(\boldsymbol{n}_1\times\boldsymbol{n}_2 = \begin{pmatrix} (1)(3) - (-1)(-1) \\ (-1)(2) - (1)(3) \\ (1)(-1) - (1)(2) \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}\).

\(z = 0\): \(x + y = 1\), \(2x - y = 8\), so \(x = 3\), \(y = -2\).

\[ \boldsymbol{r} = \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} \]

See 22.5.

7 Paper 2 AHL Find the acute angle between the two planes in Question 6.

Solution

\(\boldsymbol{n}_1\cdot\boldsymbol{n}_2 = 2 - 1 - 3 = -2\), \(|\boldsymbol{n}_1| = \sqrt{3}\), \(|\boldsymbol{n}_2| = \sqrt{14}\).

\(\cos\theta = \dfrac{|-2|}{\sqrt{3}\sqrt{14}} = \dfrac{2}{\sqrt{42}}\), so \(\theta = 72.0^\circ\) (3 s.f.). See 22.5.

8 Paper 2 AHL Find the common point of the planes

\[ 1.5x - 2y + z = 3.2, \qquad x + y + 2z = 4, \qquad 3x - y - z = 1.6 . \]

Solution

linSolve({1.5x-2y+z=3.2, x+y+2z=4, 3x-y-z=1.6}, {x,y,z}) gives

\(x = 1.02\), \(y = -0.0727\), \(z = 1.53\) (3 s.f.). See 22.6.

9 Paper 1 AHL The planes are \(x + y + z = 1\), \(\;x - y + 2z = 3\) and \(\;3x + y + kz = 5\).

(a) Find the value of \(k\) for which the planes do not have exactly one common point.

(b) For this value of \(k\), show that the planes meet in a line, and find a vector equation of the line.

Solution

(a) \(\boldsymbol{n}_1\times\boldsymbol{n}_2 = \begin{pmatrix} (1)(2) - (1)(-1) \\ (1)(1) - (1)(2) \\ (1)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ -2 \end{pmatrix}\), and \(\begin{pmatrix} 3 \\ -1 \\ -2 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \\ k \end{pmatrix} = 8 - 2k = 0\), so \(k = 4\).

(b) \(2 \times\) (first) \(+\) (second) \(= 3x + y + 4z = 5\), which is the third equation. So the third plane contains the line of the first two, and the planes form a sheaf. With \(z = 0\): \(x + y = 1\), \(x - y = 3\), so \(x = 2\), \(y = -1\).

\[ \boldsymbol{r} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -1 \\ -2 \end{pmatrix} \]

See 22.6.

10 Paper 1 AHL The line \(l\) is \(\boldsymbol{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) and the plane \(\Pi\) is \(x + 2z = 10\).

(a) Show that \(l\) is parallel to \(\Pi\) and does not lie in \(\Pi\).

(b) Find the distance between \(l\) and \(\Pi\).

(c) Find the Cartesian equation of the plane that contains \(l\) and is perpendicular to \(\Pi\).

Solution

(a) \(\boldsymbol{d}\cdot\boldsymbol{n} = 2 + 0 - 2 = 0\), and \((1, 0, 2)\) gives \(1 + 4 = 5 \neq 10\). So \(l\) is parallel to \(\Pi\), not in it.

(b) \(\dfrac{|5 - 10|}{\sqrt{1 + 0 + 4}} = \dfrac{5}{\sqrt{5}} = \sqrt{5}\).

(c) The new plane contains the direction \(\boldsymbol{d}\) and is perpendicular to \(\Pi\), so it also contains the direction \(\boldsymbol{n}\):

\[ \boldsymbol{d}\times\boldsymbol{n} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\times\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} (2)(2) - (-1)(0) \\ (-1)(1) - (2)(2) \\ (2)(0) - (2)(1) \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \\ -2 \end{pmatrix}. \]

\(d = 4 - 0 - 4 = 0\), so the plane is \(\;4x - 5y - 2z = 0\). See 22.3, 22.4 and 22.5.