The vector equation of a line Problem set B One question at a time

The points C(−3, 4) and D(5, −2) lie on a line.

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

   • \(\overrightarrow{CD}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}5\\-2\end{pmatrix}-\begin{pmatrix}-3\\4\end{pmatrix}=\begin{pmatrix}8\\-6\end{pmatrix}\)
   • \(\mathbf{r}=\begin{pmatrix}-3\\4\end{pmatrix}+\lambda\begin{pmatrix}8\\-6\end{pmatrix}\)
   • Any point on the line with this direction vector works as the starting point.

2. (b) Find the point on the line when \(\lambda=\frac{1}{2}\).

   • \(\mathbf{r}=\begin{pmatrix}-3\\4\end{pmatrix}+\frac{1}{2}\begin{pmatrix}8\\-6\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}\)
   • The point is (1, 1).

A line passes through the point (−1, 6, 0) and has direction vector \(\begin{pmatrix}4\\-3\\2\end{pmatrix}\).

1. (a) Write down a vector equation of the line.

   • \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}=\begin{pmatrix}-1\\6\\0\end{pmatrix}+\lambda\begin{pmatrix}4\\-3\\2\end{pmatrix}\)

2. (b) Write down a second vector equation of the same line, using the point on the line where \(\lambda=1\) as the position vector.

   • \(\lambda=1\Rightarrow \begin{pmatrix}-1\\6\\0\end{pmatrix}+\begin{pmatrix}4\\-3\\2\end{pmatrix}=\begin{pmatrix}3\\3\\2\end{pmatrix}\)
   • \(\mathbf{r}=\begin{pmatrix}3\\3\\2\end{pmatrix}+\mu\begin{pmatrix}4\\-3\\2\end{pmatrix}\). The direction vector stays the same.

Find a vector equation of the line through P(2, 3, −4) and Q(4, −1, −2). Hence determine whether R(8, −9, 2) lies on the line.

• \(\overrightarrow{PQ}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2\\-4\\2\end{pmatrix},\quad \mathbf{r}=\begin{pmatrix}2\\3\\-4\end{pmatrix}+\lambda\begin{pmatrix}2\\-4\\2\end{pmatrix}\)
• Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 2+2\lambda=8\); \((2)\ \ 3-4\lambda=-9\); \((3)\ \ -4+2\lambda=2\)
• From equation (1): \(\lambda=3\)
• Check (2): \(3-4(3)=-9\) and this equals \(-9\).
• Check (3): \(-4+2(3)=2\) and this equals \(2\).
• All three equations are satisfied, so the point is on the line.

A line has vector equation \(\mathbf{r}=\begin{pmatrix}5\\-2\\1\end{pmatrix}+\lambda\begin{pmatrix}-3\\2\\4\end{pmatrix}\). Show that the point E(−4, 4, 12) does not lie on the line.

• Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 5-3\lambda=-4\); \((2)\ \ -2+2\lambda=4\); \((3)\ \ 1+4\lambda=12\)
• From equation (1): \(\lambda=3\)
• Check (2): \(-2+2(3)=4\) and this equals \(4\).
• Check (3): \(1+4(3)=13\), but the equation needs \(12\). Not equal.
• The values do not satisfy every equation, so the point is not on the line.

A zip wire is stretched in a straight line from the top of a tower T(−20, 10, 45) to a post U(40, −20, 15), and the line carries on beyond the post. Coordinates are in metres, with the third value the height.

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

   • \(\overrightarrow{TU}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}40\\-20\\15\end{pmatrix}-\begin{pmatrix}-20\\10\\45\end{pmatrix}=\begin{pmatrix}60\\-30\\-30\end{pmatrix}\)
   • \(\mathbf{r}=\begin{pmatrix}-20\\10\\45\end{pmatrix}+\lambda\begin{pmatrix}60\\-30\\-30\end{pmatrix}\)

2. (b) Determine whether this line passes through a hook at (100, −50, 10).

   • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ -20+60\lambda=100\); \((2)\ \ 10-30\lambda=-50\); \((3)\ \ 45-30\lambda=10\)
   • From equation (1): \(\lambda=2\)
   • Check (2): \(10-30(2)=-50\) and this equals \(-50\).
   • Check (3): \(45-30(2)=-15\), but the equation needs \(10\). Not equal.
   • The values do not satisfy every equation, so the point is not on the line.

A spotlight is fixed at S(−3, 4, 1) and points in the direction \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\). Coordinates are in metres.

1. (a) Write down a vector equation of the beam.

   • \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}=\begin{pmatrix}-3\\4\\1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}\)

2. (b) Determine whether the beam passes through the stage mark at (5, 0, 12).

   • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ -3+2\lambda=5\); \((2)\ \ 4-\lambda=0\); \((3)\ \ 1+3\lambda=12\)
   • From equation (1): \(\lambda=4\)
   • Check (2): \(4-(4)=0\) and this equals \(0\).
   • Check (3): \(1+3(4)=13\), but the equation needs \(12\). Not equal.
   • The values do not satisfy every equation, so the point is not on the line.

A cable runs from the top of a mast M(0, 0, 50) to an anchor N(80, −60, 0) on the ground. Coordinates are in metres, with the third value the height.

1. (a) Find a vector equation of the cable.

   • \(\overrightarrow{MN}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}80\\-60\\0\end{pmatrix}-\begin{pmatrix}0\\0\\50\end{pmatrix}=\begin{pmatrix}80\\-60\\-50\end{pmatrix}\)
   • \(\mathbf{r}=\begin{pmatrix}0\\0\\50\end{pmatrix}+\lambda\begin{pmatrix}80\\-60\\-50\end{pmatrix}\)

2. (b) Find the coordinates of the point on the cable where the height is 20 m.

   • Height: \(50-50\lambda=20\), so \(\lambda=\frac{3}{5}\)
   • \(\mathbf{r}=\begin{pmatrix}0\\0\\50\end{pmatrix}+\frac{3}{5}\begin{pmatrix}80\\-60\\-50\end{pmatrix}=\begin{pmatrix}48\\-36\\20\end{pmatrix}\)
   • The point is (48, −36, 20).

A line has vector equation \(\mathbf{r}=\begin{pmatrix}-2\\5\\1\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\-1\end{pmatrix}\). The point Q(7, k, −2) lies on the line. Find the value of \(k\).

• \((1)\ \ -2+3\lambda=7\), so \(\lambda=3\)
• Check the other known component: \(1-(3)=-2=-2\)
• Put \(\lambda=3\) into the equation for the y-component: \(k=5+2(3)=11\)

The points C(−2, 4, 6) and D(2, −2, −2) lie on a line \(l\).

1. (a) Find a vector equation of \(l\).

   • \(\overrightarrow{CD}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2\\-2\\-2\end{pmatrix}-\begin{pmatrix}-2\\4\\6\end{pmatrix}=\begin{pmatrix}4\\-6\\-8\end{pmatrix}\)
   • \(\mathbf{r}=\begin{pmatrix}-2\\4\\6\end{pmatrix}+\lambda\begin{pmatrix}4\\-6\\-8\end{pmatrix}\)

2. (b) The midpoint of CD is N. Show that it lies on \(l\) and state the value of \(\lambda\) there.

   • Midpoint: \(\frac{1}{2}(\begin{pmatrix}-2\\4\\6\end{pmatrix}+\begin{pmatrix}2\\-2\\-2\end{pmatrix})=\begin{pmatrix}0\\1\\2\end{pmatrix}\)
   • \(\begin{pmatrix}-2\\4\\6\end{pmatrix}+\frac{1}{2}\begin{pmatrix}4\\-6\\-8\end{pmatrix}=\begin{pmatrix}0\\1\\2\end{pmatrix}\), so the midpoint is on \(l\) at \(\lambda=\frac{1}{2}\)

3. (c) Find the coordinates of the point on \(l\) where \(\lambda=3\).

   • \(\mathbf{r}=\begin{pmatrix}-2\\4\\6\end{pmatrix}+3\begin{pmatrix}4\\-6\\-8\end{pmatrix}=\begin{pmatrix}10\\-14\\-18\end{pmatrix}\)