AI HL · AHL 4
0 of 160 skills done
The angle between two vectors; perpendicular and parallel vectors
AHL 4.14
Screenshot and video to come
Drag the tip of a and the tip of b. Both vectors leave the origin. Watch the sign of a · b and the angle between them.
Click a tip, then use the arrow keys to move it one unit. The tips stay on whole numbers, so you can check each line by hand.
Drag a tip to change a vector.
Find the angle between \(\mathbf{a} = \begin{pmatrix}3\\-2\\6\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\4\\-2\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
Combine the two forms of the scalar product.
\(\mathbf{a}\cdot\mathbf{b} = (3)(1) + (-2)(4) + (6)(-2) = -17\)
Multiply matching components.
\(|\mathbf{a}| = \sqrt{49},\quad |\mathbf{b}| = \sqrt{21}\)
Square, add, root.
\(\cos\theta = \dfrac{-17}{\sqrt{49}\sqrt{21}} = -0.5300\) \(\theta = \cos^{-1}(-0.5300) = 122^\circ\)
A negative cosine gives an obtuse angle.
Find the angle between \(\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}\) and \(\mathbf{b} = 6\mathbf{i} - \mathbf{j}\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a} = \begin{pmatrix}2\\5\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}6\\-1\end{pmatrix}\)
In 2D there is no third term.
\(\mathbf{a}\cdot\mathbf{b} = (2)(6) + (5)(-1) = 7\)
\(|\mathbf{a}| = \sqrt{29},\quad |\mathbf{b}| = \sqrt{37}\)
\(\cos\theta = \dfrac{7}{\sqrt{29}\sqrt{37}} = 0.2137\) \(\theta = \cos^{-1}(0.2137) = 77.7^\circ\)
Acute: the scalar product is positive.
The points A, B and C have coordinates (2, −1, 3), (5, 1, −1) and (0, 4, 2). Find angle \(BAC\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{AB} = \begin{pmatrix}5\\1\\-1\end{pmatrix} - \begin{pmatrix}2\\-1\\3\end{pmatrix} = \begin{pmatrix}3\\2\\-4\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}0\\4\\2\end{pmatrix} - \begin{pmatrix}2\\-1\\3\end{pmatrix} = \begin{pmatrix}-2\\5\\-1\end{pmatrix}\)
Draw the triangle. Both vectors leave A, so subtract A from each.
\(\overrightarrow{AB}\cdot\overrightarrow{AC} = (3)(-2) + (2)(5) + (-4)(-1) = 8\)
\(|\overrightarrow{AB}| = \sqrt{29},\quad |\overrightarrow{AC}| = \sqrt{30}\)
\(\cos\theta = \dfrac{8}{\sqrt{29}\sqrt{30}} = 0.2712\) \(\theta = \cos^{-1}(0.2712) = 74.3^\circ\)
Angle BAC = 74.3°.
Find the value of \(t\) for which \(\mathbf{a} = \begin{pmatrix}3t\\ t+2\\ 1-2t\end{pmatrix}\) is perpendicular to \(\mathbf{b} = \begin{pmatrix}2\\-3\\4\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a}\cdot\mathbf{b} = 0\)
cos 90° = 0.
\((3t)(2) + (t+2)(-3) + (1-2t)(4) = 0\)
Multiply matching components.
\(-2 - 5t = 0\)
6t − 3t − 6 + 4 − 8t = −5t − 2.
\(t = -0.4\)
A linear equation in t.
The vectors \(\mathbf{a} = \begin{pmatrix}t\\ 3\\ t-1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}t-5\\ 2\\ -1\end{pmatrix}\) are perpendicular. Find the possible values of \(t\).
Have a go first. Then open the working one line at a time.
\(t(t-5) + (3)(2) + (t-1)(-1) = 0\)
Perpendicular means \(\mathbf{a}\cdot\mathbf{b} = 0\).
\(t^2 - 5t + 6 - t + 1 = t^2 - 6t + 7 = 0\)
This time t² appears: a quadratic.
Polynomial solver: t = 1.59 or t = 4.41
Solve t² − 6t + 7 = 0 on the calculator. Both values work.
\(t = 3 - \sqrt{2} = 1.59,\quad t = 3 + \sqrt{2} = 4.41\)
There are two vectors of this form that are perpendicular.
Show that \(\mathbf{a} = \begin{pmatrix}2\\-4\\6\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-3\\6\\-9\end{pmatrix}\) are parallel, by finding \(\cos\theta\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a}\cdot\mathbf{b} = (2)(-3) + (-4)(6) + (6)(-9) = -84\)
\(|\mathbf{a}||\mathbf{b}| = \sqrt{56}\sqrt{126} = 84.0\)
Square roots of 56 and 126 multiply to 84.
\(\cos\theta = \dfrac{-84}{84.0} = -1\)
θ = 180°. The vectors are parallel, pointing in opposite directions. Parallel means \(|\mathbf{a}\cdot\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\).