AI HL · AHL 4
0 of 160 skills done
The acute angle between two lines
AHL 4.15
Screenshot and video to come
Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}2\\0\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\-2\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}4\\1\\3\end{pmatrix} + \mu\begin{pmatrix}2\\-3\\6\end{pmatrix} \]
Have a go first. Then open the working one line at a time.
\(\mathbf{d}_1 = \begin{pmatrix}1\\2\\-2\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}2\\-3\\6\end{pmatrix}\)
Read the numbers that multiply the parameters. The position vectors are not used.
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (2)(-3) + (-2)(6) = -16\)
Multiply matching components and add.
\(|\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3,\ |\mathbf{d}_2| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\)
Each magnitude is the square root of the sum of squares.
\(\cos\theta = -\frac{16}{21} = -0.7619\). From the calculator: \(\theta = 139.6\ldots^\circ\)
Use cos⁻¹ in degree mode.
\(180^\circ - 139.6^\circ = 40.4^\circ\)
The scalar product is negative, so this angle is obtuse. A question that asks for the acute angle wants 180° minus it.
Two straight paths cross in a park. Relative to the gate, in metres, path 1 is the line \(l_1\) and path 2 is the line \(l_2\). Find the acute angle at which the paths cross.\[ l_1:\ \mathbf{r} = \begin{pmatrix}0\\2\end{pmatrix} + \lambda\begin{pmatrix}4\\3\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}5\\1\end{pmatrix} + \mu\begin{pmatrix}1\\-2\end{pmatrix} \]
Have a go first. Then open the working one line at a time.
\(\mathbf{d}_1 = \begin{pmatrix}4\\3\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}1\\-2\end{pmatrix}\)
Read the numbers that multiply the parameters. The position vectors are not used.
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (4)(1) + (3)(-2) = -2\)
Multiply matching components and add.
\(|\mathbf{d}_1| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5,\ |\mathbf{d}_2| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\)
Each magnitude is the square root of the sum of squares.
\(\cos\theta = -\frac{2}{5 \sqrt{5}} = -0.1789\). From the calculator: \(\theta = 100.3\ldots^\circ\)
Use cos⁻¹ in degree mode.
\(180^\circ - 100.3^\circ = 79.7^\circ\)
The scalar product is negative, so this angle is obtuse. A question that asks for the acute angle wants 180° minus it.
Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}5\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}3\\-1\\2\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}-1\\4\\2\end{pmatrix} + \mu\begin{pmatrix}1\\2\\4\end{pmatrix} \]
Have a go first. Then open the working one line at a time.
\(\mathbf{d}_1 = \begin{pmatrix}3\\-1\\2\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}1\\2\\4\end{pmatrix}\)
Read the numbers that multiply the parameters. The position vectors are not used.
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(1) + (-1)(2) + (2)(4) = 9\)
Multiply matching components and add.
\(|\mathbf{d}_1| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14},\ |\mathbf{d}_2| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21}\)
Each magnitude is the square root of the sum of squares.
\(\cos\theta = \frac{9}{7 \sqrt{6}} = 0.5249\). From the calculator: \(\theta = 58.3\ldots^\circ\)
Use cos⁻¹ in degree mode.
\(\theta = 58.3^\circ\)
cos θ is positive, so θ is already acute.