AI HL · AHL 4
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The vector equation of a line
AHL 4.8
Drag the point A, then the tip of the direction vector d (orange). Move λ and watch R slide along the line.
The points snap to whole numbers. Select one (Tab) and use the arrow keys to move it.
Find a vector equation of the line through A(4, −1, 2) and B(−2, 3, 8). Enter the second component of the direction vector \(\overrightarrow{AB}\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}-2\\3\\8\end{pmatrix}-\begin{pmatrix}4\\-1\\2\end{pmatrix}=\begin{pmatrix}-6\\4\\6\end{pmatrix}\)
End minus start gives a vector along the line.
\(\mathbf{r}=\begin{pmatrix}4\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}-6\\4\\6\end{pmatrix}\)
Use A as the starting point. \(\lambda=0\) gives A and \(\lambda=1\) gives B.
\(\lambda=1:\ \begin{pmatrix}4\\-1\\2\end{pmatrix}+\begin{pmatrix}-6\\4\\6\end{pmatrix}=\begin{pmatrix}-2\\3\\8\end{pmatrix}\)
The line reaches B at \(\lambda=1\), so the equation is right.
The line through A(3, 0, −2) and B(1, 4, 4) has equation \(\mathbf{r}=\begin{pmatrix}3\\0\\-2\end{pmatrix}+\lambda\begin{pmatrix}-2\\4\\6\end{pmatrix}\). Determine whether C(2, 2, 1) lies on the line. Find the value of \(\lambda\) at C.
Have a go first. Then open the working one line at a time.
Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 3-2\lambda=2\); \((2)\ \ 4\lambda=2\); \((3)\ \ -2+6\lambda=1\)
Each component of the point must match.
From equation (1): \(\lambda=\frac{1}{2}\)
Use any one equation to find \(\lambda\).
Check (2): \(4(\frac{1}{2})=2\) and this equals \(2\). Check (3): \(-2+6(\frac{1}{2})=1\) and this equals \(1\).
The value must work in every component.
All three equations are satisfied, so the point is on the line.
Write the conclusion in words.
Use the same line \(\mathbf{r}=\begin{pmatrix}3\\0\\-2\end{pmatrix}+\lambda\begin{pmatrix}-2\\4\\6\end{pmatrix}\). Determine whether D(0, 6, 1) lies on the line. Find the value of \(\lambda\) from the first equation.
Have a go first. Then open the working one line at a time.
Put \(\mathbf{r}=\mathbf{c}\) and compare components: \((1)\ \ 3-2\lambda=0\); \((2)\ \ 4\lambda=6\); \((3)\ \ -2+6\lambda=1\)
Same line, new point.
From equation (1): \(\lambda=\frac{3}{2}\)
Take \(\lambda\) from the first equation.
Check (2): \(4(\frac{3}{2})=6\) and this equals \(6\). Check (3): \(-2+6(\frac{3}{2})=7\), but the equation needs \(1\). Not equal.
The second works but the third does not.
The values do not satisfy every equation, so the point is not on the line.
One failed equation is enough. D is not on the line.
A line passes through the point (2, −1, 5) and has direction vector \(\begin{pmatrix}3\\0\\-2\end{pmatrix}\). Write down a vector equation of the line. Find the x-coordinate of the point where \(\lambda=2\).
Have a go first. Then open the working one line at a time.
\(\mathbf{r}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\-2\end{pmatrix}\)
Put the point as \(\mathbf{a}\) and the direction as \(\mathbf{d}\).
\(\mathbf{r}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+2\begin{pmatrix}3\\0\\-2\end{pmatrix}=\begin{pmatrix}8\\-1\\1\end{pmatrix}\)
Multiply the direction by 2, then add.
The point is (8, −1, 1)
The y-coordinate stays at −1 because the direction has 0 there.
Find a vector equation of the line through (2, 7) and (6, 1). Hence find the gradient of the line.
Have a go first. Then open the working one line at a time.
\(\mathbf{b}-\mathbf{a}=\begin{pmatrix}6\\1\end{pmatrix}-\begin{pmatrix}2\\7\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}\)
The same method works with two components.
\(\mathbf{r}=\begin{pmatrix}2\\7\end{pmatrix}+\lambda\begin{pmatrix}4\\-6\end{pmatrix}\)
Start at the first point.
\(\text{gradient}=\dfrac{-6}{4}=-1.5\)
A direction vector \((p, q)\) means gradient \(q \div p\). Any multiple of the direction gives the same gradient.