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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.8

The vector equation of a line Problem set A Worked solutions

Skills

  1. The points A(2, 5) and B(8, −1) lie on a line.

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

      • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}8\\-1\end{pmatrix}-\begin{pmatrix}2\\5\end{pmatrix}=\begin{pmatrix}6\\-6\end{pmatrix}\)
      • \(\mathbf{r}=\begin{pmatrix}2\\5\end{pmatrix}+\lambda\begin{pmatrix}6\\-6\end{pmatrix}\)
      • Any point on the line with this direction vector works as the starting point.

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

      • \(\mathbf{r}=\begin{pmatrix}2\\5\end{pmatrix}+\frac{1}{2}\begin{pmatrix}6\\-6\end{pmatrix}=\begin{pmatrix}5\\2\end{pmatrix}\)
      • The point is (5, 2).

  2. A line passes through the point (3, −2, 4) and has direction vector \(\begin{pmatrix}1\\5\\-2\end{pmatrix}\).

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

      • \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}=\begin{pmatrix}3\\-2\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\5\\-2\end{pmatrix}\)

    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.

      • \(\lambda=1\Rightarrow \begin{pmatrix}3\\-2\\4\end{pmatrix}+\begin{pmatrix}1\\5\\-2\end{pmatrix}=\begin{pmatrix}4\\3\\2\end{pmatrix}\)
      • \(\mathbf{r}=\begin{pmatrix}4\\3\\2\end{pmatrix}+\mu\begin{pmatrix}1\\5\\-2\end{pmatrix}\). The direction vector stays the same.

  3. Find a vector equation of the line through A(1, −2, 5) and B(3, 4, 1). Hence determine whether C(4, 7, −1) lies on the line.

    • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2\\6\\-4\end{pmatrix},\quad \mathbf{r}=\begin{pmatrix}1\\-2\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\6\\-4\end{pmatrix}\)
    • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 1+2\lambda=4\); \((2)\ \ -2+6\lambda=7\); \((3)\ \ 5-4\lambda=-1\)
    • From equation (1): \(\lambda=\frac{3}{2}\)
    • Check (2): \(-2+6(\frac{3}{2})=7\) and this equals \(7\).
    • Check (3): \(5-4(\frac{3}{2})=-1\) and this equals \(-1\).
    • All three equations are satisfied, so the point is on the line.

  4. A line has vector equation \(\mathbf{r}=\begin{pmatrix}0\\4\\-3\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\5\end{pmatrix}\). Show that the point D(6, 1, 10) does not lie on the line.

    • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 2\lambda=6\); \((2)\ \ 4-\lambda=1\); \((3)\ \ -3+5\lambda=10\)
    • From equation (1): \(\lambda=3\)
    • Check (2): \(4-(3)=1\) and this equals \(1\).
    • Check (3): \(-3+5(3)=12\), but the equation needs \(10\). Not equal.
    • The values do not satisfy every equation, so the point is not on the line.

In context

  1. A drone flies in a straight line through P(0, 0, 30) and then Q(80, 60, 10), and keeps going on the same line. Coordinates are in metres, with the third value the height.

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

      • \(\overrightarrow{PQ}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}80\\60\\10\end{pmatrix}-\begin{pmatrix}0\\0\\30\end{pmatrix}=\begin{pmatrix}80\\60\\-20\end{pmatrix}\)
      • \(\mathbf{r}=\begin{pmatrix}0\\0\\30\end{pmatrix}+\lambda\begin{pmatrix}80\\60\\-20\end{pmatrix}\)

    2. (b) Determine whether the drone reaches a landing pad at (120, 90, 0).

      • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 80\lambda=120\); \((2)\ \ 60\lambda=90\); \((3)\ \ 30-20\lambda=0\)
      • From equation (1): \(\lambda=\frac{3}{2}\)
      • Check (2): \(60(\frac{3}{2})=90\) and this equals \(90\).
      • Check (3): \(30-20(\frac{3}{2})=0\) and this equals \(0\).
      • All three equations are satisfied, so the point is on the line.

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

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

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

    2. (b) Determine whether the beam passes through the target at (7, 9, −7).

      • Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 2+\lambda=7\); \((2)\ \ -1+2\lambda=9\); \((3)\ \ 3-2\lambda=-7\)
      • From equation (1): \(\lambda=5\)
      • Check (2): \(-1+2(5)=9\) and this equals \(9\).
      • Check (3): \(3-2(5)=-7\) and this equals \(-7\).
      • All three equations are satisfied, so the point is on the line.

  3. A zip wire runs from the top of a tower T(0, 0, 40) to a point G(120, 50, 0) on the ground. Coordinates are in metres, with the third value the height.

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

      • \(\overrightarrow{TG}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}120\\50\\0\end{pmatrix}-\begin{pmatrix}0\\0\\40\end{pmatrix}=\begin{pmatrix}120\\50\\-40\end{pmatrix}\)
      • \(\mathbf{r}=\begin{pmatrix}0\\0\\40\end{pmatrix}+\lambda\begin{pmatrix}120\\50\\-40\end{pmatrix}\)

    2. (b) Find the coordinates of the point on the wire where the height is 10 m.

      • Height: \(40-40\lambda=10\), so \(\lambda=\frac{3}{4}\)
      • \(\mathbf{r}=\begin{pmatrix}0\\0\\40\end{pmatrix}+\frac{3}{4}\begin{pmatrix}120\\50\\-40\end{pmatrix}=\begin{pmatrix}90\\\frac{75}{2}\\10\end{pmatrix}\)
      • The point is (90, 37.5, 10).

Stretch

  1. A line has vector equation \(\mathbf{r}=\begin{pmatrix}1\\0\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}\). The point P(k, −2, 10) lies on the line. Find the value of \(k\).

    • \((2)\ \ -\lambda=-2\), so \(\lambda=2\)
    • Check the other known component: \(4+3(2)=10=10\)
    • Put \(\lambda=2\) into the equation for the x-component: \(k=1+2(2)=5\)

  2. The points A(1, −3, 2) and B(5, 3, −6) lie on a line \(l\).

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

      • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}5\\3\\-6\end{pmatrix}-\begin{pmatrix}1\\-3\\2\end{pmatrix}=\begin{pmatrix}4\\6\\-8\end{pmatrix}\)
      • \(\mathbf{r}=\begin{pmatrix}1\\-3\\2\end{pmatrix}+\lambda\begin{pmatrix}4\\6\\-8\end{pmatrix}\)

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

      • Midpoint: \(\frac{1}{2}(\begin{pmatrix}1\\-3\\2\end{pmatrix}+\begin{pmatrix}5\\3\\-6\end{pmatrix})=\begin{pmatrix}3\\0\\-2\end{pmatrix}\)
      • \(\begin{pmatrix}1\\-3\\2\end{pmatrix}+\frac{1}{2}\begin{pmatrix}4\\6\\-8\end{pmatrix}=\begin{pmatrix}3\\0\\-2\end{pmatrix}\), so the midpoint is on \(l\) at \(\lambda=\frac{1}{2}\)

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

      • \(\mathbf{r}=\begin{pmatrix}1\\-3\\2\end{pmatrix}+3\begin{pmatrix}4\\6\\-8\end{pmatrix}=\begin{pmatrix}13\\15\\-22\end{pmatrix}\)

jmaths.xyz · AI HL · AHL 4.8 The vector equation of a line