Mixed practice: lines in 3D, parametric form and intersections Problem set B One question at a time

The points \(A(-1, 4, 6)\), \(B(-3, 7, 7)\) and \(C(-7, 13, 11)\) are in three dimensions.

1. (a) Find a vector equation of the line \(AB\).

   • Direction: \(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)
   • Use \(A\) as the point: \(\mathbf{r} = \begin{pmatrix} -1 \\ 4 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)

2. (b) Determine whether \(C\) lies on the line \(AB\).

   • Put the coordinates of \(C\) into the line and find \(\lambda\) from each: \(x: -1 - 2\lambda = -7 \Rightarrow \lambda = 3\); \(y: 4 + 3\lambda = 13 \Rightarrow \lambda = 3\); \(z: 6 + \lambda = 11 \Rightarrow \lambda = 5\)
   • The coordinates do not give the same \(\lambda\) (or one fails outright), so \(C\) does not lie on the line.

Write down a vector equation of each line.

1. (a) The line through the point \((0, 5, -2)\) with direction \(\begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix}\).

   • Use \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), with \(\mathbf{a}\) the point and \(\mathbf{b}\) the direction.
   • \(\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix}\)

2. (b) The line through the point \((6, 1, -3)\) with direction \(\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}\).

   • The point has position vector \(6\mathbf{i} + \mathbf{j} - 3\mathbf{k}\).
   • \(\mathbf{r} = (6\mathbf{i} + \mathbf{j} - 3\mathbf{k}) + \lambda(\mathbf{i} + 5\mathbf{j} - 2\mathbf{k})\)

3. (c) The line through the point \((-4, 1)\) with direction \(\begin{pmatrix} 6 \\ -1 \end{pmatrix}\).

   • The same in two dimensions.
   • \(\mathbf{r} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 6 \\ -1 \end{pmatrix}\)

Write each line in the form asked.

1. (a) Write \(\mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ -6 \\ 3 \end{pmatrix}\) in parametric form.

   • Each row of the vector equation gives one coordinate.
   • \(x = -3 + 4\lambda, y = 1 - 6\lambda, z = 5 + 3\lambda\)

2. (b) Write \(\mathbf{r} = (4\mathbf{j} - 2\mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{k})\) in parametric form.

   • A missing \(\mathbf{i}\), \(\mathbf{j}\) or \(\mathbf{k}\) means a zero coordinate.
   • \(x = 3\lambda, y = 4, z = -2 - \lambda\)

3. (c) A line has parametric equations \(x = -1 + 3\lambda\), \(y = 7\), \(z = \frac{\lambda - 6}{3}\). Write it as a vector equation.

   • Constant parts form the point; the multiples of \(\lambda\) form the direction.
   • Split \(z\): \(z = -2 + \frac{1}{3}\lambda\)
   • \(\mathbf{r} = \begin{pmatrix} -1 \\ 7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 0 \\ \frac{1}{3} \end{pmatrix}\)

A line \(l\) has equation \(\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ -4 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}\). Find parametric equations of the line parallel to \(l\) that passes through the point \((-2, 6, 0)\).

• Parallel lines have the same direction vector: \(\begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}\).
• Use the point \((-2, 6, 0)\) and this direction.
• \(x = -2 - \lambda, y = 6 + 4\lambda, z = 3\lambda\)

A line has parametric equations \(x = 3 + 2\lambda, y = -1 - \lambda, z = 4 + 3\lambda\).

1. (a) Determine whether the point \((-1, 2, -2)\) lies on the line.

   • Find \(\lambda\) from each coordinate: \(x: 3 + 2\lambda = -1 \Rightarrow \lambda = -2\); \(y: -1 - \lambda = 2 \Rightarrow \lambda = -3\); \(z: 4 + 3\lambda = -2 \Rightarrow \lambda = -2\)
   • The values of \(\lambda\) do not agree, so the point is not on the line.

2. (b) The point \((9, p, q)\) lies on the line. Find \(p\) and \(q\).

   • The known coordinate gives \(\lambda\): \(3 + 2\lambda = 9 \Rightarrow \lambda = 3\)
   • Put this into the other two equations.
   • \(p = -4, q = 13\)

Find the point of intersection of the lines \(\mathbf{r} = (-2\mathbf{i} + 9\mathbf{k}) + \lambda(2\mathbf{i} - \mathbf{j} + 3\mathbf{k})\) and \(\mathbf{r} = \begin{pmatrix} -7 \\ -5 \\ 12 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\).

• Equate the coordinates: \(-2 + 2\lambda = -7 + \mu\) (1), \(-\lambda = -5 + 2\mu\) (2), \(9 + 3\lambda = 12 - 2\mu\) (3)
• Solve (1) and (2) together. From the calculator: \(\lambda = -1, \mu = 3\)
• Check (3): \(9 + 3\lambda = 6\) and \(12 - 2\mu = 6\). They are equal, so the lines meet.
• Put \(\lambda = -1\) into the first line: \((-4, 1, 6)\)

Decide about the lines in each part.

1. (a) Show that the lines \(\mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} -5 \\ 1 \\ 7 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}\) do not intersect.

   • Equate the coordinates: \(-1 + \lambda = -5 + 3\mu\) (1), \(2 + 2\lambda = 1 - \mu\) (2), \(7 + 2\lambda = 7 + \mu\) (3)
   • Solve (1) and (2) together. From the calculator: \(\lambda = -1, \mu = 1\)
   • Check (3): left side \(5\), right side \(8\). They are not equal, so the lines do not intersect.

2. (b) The lines \(l_1\) and \(l_2\) have parametric equations \(l_1: x = 3 - \lambda, y = 1 + 2\lambda, z = -9 + 3\lambda\) and \(l_2: x = 3 + 2\mu, y = 6 + \mu, z = -5 - 2\mu\). Show that they intersect and find the point of intersection.

   • Equate the coordinates: \(3 - \lambda = 3 + 2\mu\) (1), \(1 + 2\lambda = 6 + \mu\) (2), \(-9 + 3\lambda = -5 - 2\mu\) (3)
   • Solve (1) and (2) together. From the calculator: \(\lambda = 2, \mu = -1\)
   • Check (3): \(-9 + 3\lambda = -3\) and \(-5 - 2\mu = -3\). They are equal, so the lines meet.
   • Put \(\lambda = 2\) into the first line: \((1, 5, -3)\)

The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\).

1. (a) Show that the equations \(x = -5 + 9\mu, y = 4 + 3\mu, z = 3 - 6\mu\) describe the same line as \(l\).

   • The direction \(\begin{pmatrix} 9 \\ 3 \\ -6 \end{pmatrix}\) is \(3\) times \(\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\), so the lines are parallel.
   • Is the point \((-5, 4, 3)\) on \(l\)? \(x: -2 + 3\lambda = -5 \Rightarrow \lambda = -1\); \(y: 5 + \lambda = 4 \Rightarrow \lambda = -1\); \(z: 1 - 2\lambda = 3 \Rightarrow \lambda = -1\)
   • It is on \(l\), so the lines are the same.

2. (b) Show that \(\mathbf{r} = (5\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}) + t(-3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) describes a different line from \(l\).

   • The direction \(-3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) is \(-1\) times \(\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\), so the lines are parallel.
   • Is the point \((5, 7, -3)\) on \(l\)? \(x: -2 + 3\lambda = 5 \Rightarrow \lambda = \frac{7}{3}\); \(y: 5 + \lambda = 7 \Rightarrow \lambda = 2\); \(z: 1 - 2\lambda = -3 \Rightarrow \lambda = 2\)
   • It is not on \(l\), so this is a different (parallel) line.

Two straight tunnels are being planned in a mine. One tunnel passes through the points \(A\) and \(B\), the other through \(C\) and \(D\). Coordinates are in metres. \(A(-1, 9, 0)\), \(B(0, 11, -1)\), \(C(-7, 7, -4)\), \(D(-5, 6, -1)\). Find the coordinates of the point where the two lines cross.

• Directions: \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\overrightarrow{CD} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\).
• Lines: \(\mathbf{r} = \begin{pmatrix} -1 \\ 9 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} -7 \\ 7 \\ -4 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\).
• Equate the coordinates: \(-1 + \lambda = -7 + 2\mu\) (1), \(9 + 2\lambda = 7 - \mu\) (2), \(-\lambda = -4 + 3\mu\) (3)
• Solve (1) and (2) together. From the calculator: \(\lambda = -2, \mu = 2\)
• Check (3): \(-\lambda = 2\) and \(-4 + 3\mu = 2\). They are equal, so the lines meet.
• Put \(\lambda = -2\) into the first line: \((-3, 5, 2)\)

A drone flies along a straight path through waypoint \(P\) and waypoint \(Q\). Coordinates are in metres. A landing marker \(R\) is on the line through \(P\) and \(Q\), with \(PR = 2PQ\). The waypoints are at \(P(-3, 6, 2)\) and \(Q(-1, 3, -2)\).

1. (a) Find a vector equation of the line \(PQ\).

   • Direction: \(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}\)
   • \(\mathbf{r} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}\)

2. (b) Find the possible coordinates of \(R\).

   • \(PR = 2PQ\) means \(R\) is 2 lots of \(\overrightarrow{PQ}\) from \(P\), in either direction: \(\lambda = 2\) or \(\lambda = -2\).
   • \(\lambda = 2: (1, 0, -6)\)
   • \(\lambda = -2: (-7, 12, 10)\)

A line has equation \(\mathbf{r} = \begin{pmatrix} 10 \\ -4 \\ 10 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix}\).

1. (a) Find the coordinates of the point where the line meets the \(x\)-axis.

   • On the \(x\)-axis, \(y\) and \(z\) are both 0.
   • \(y: -4 - 2\lambda = 0 \Rightarrow \lambda = -2\); \(z: 10 + 5\lambda = 0 \Rightarrow \lambda = -2\)
   • Both give \(\lambda = -2\). Put it into the line: \((4, 0, 0)\)

2. (b) Show that the line does not meet the \(y\)-axis.

   • On the \(y\)-axis, \(x\) and \(z\) are both 0.
   • \(x: 10 + 3\lambda = 0 \Rightarrow \lambda = -\frac{10}{3}\); \(z: 10 + 5\lambda = 0 \Rightarrow \lambda = -2\)
   • The values of \(\lambda\) are different, so no point of the line has both coordinates 0.

The lines \(\mathbf{r} = (5\mathbf{j} + 5\mathbf{k}) + \lambda(2\mathbf{i} + \mathbf{j} + 2\mathbf{k})\) and \(\mathbf{r} = (-5\mathbf{i} + 7\mathbf{j} + 9\mathbf{k}) + \mu(\mathbf{i} - \mathbf{j} + p\mathbf{k})\) intersect. Find the value of \(p\) and the coordinates of the point of intersection.

• Equate the coordinates: \(2\lambda = -5 + \mu\) (1), \(5 + \lambda = 7 - \mu\) (2), \(5 + 2\lambda = 9 + \mu\,p\) (3)
• Equations (1) and (2) do not contain \(p\). From the calculator: \(\lambda = -1, \mu = 3\)
• Put these into (3): \(3 = 9 + 3p\), so \(p = -2\)
• Put \(\lambda = -1\) into the first line: \((-2, 4, 3)\)

A fishing line \(l\) is modelled as a straight line through the point \(A\), parallel to the vector \(3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}\). Distances are in centimetres. The point \(A\) has coordinates \((-4, 3, 1)\).

1. (a) Calculate the magnitude of \(3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}\).

   • \(|3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\)

2. (b) Find the possible coordinates of the point \(P\) on \(l\) such that \(AP = 39\).

   • The line is \(\mathbf{r} = (-4\mathbf{i} + 3\mathbf{j} + \mathbf{k}) + \lambda(3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k})\).
   • A step of \(\lambda\) along \(l\) moves a distance \(|\lambda|\times 13\). Set this equal to 39: \(|\lambda| = 39 \div 13 = 3\)
   • \(\lambda = 3: (5, -9, 37)\), \(\lambda = -3: (-13, 15, -35)\)