Vectors

Have a go first. Then open the working one line at a time.

1. Sketch

   \(\mathbf{v} = \begin{pmatrix}6\\-4\end{pmatrix}\)

   East 6 and south 4: the drone moves to the south-east. A sketch shows which quadrant you are in.

2. Angle from south

   \(\tan\alpha = \dfrac{6}{4},\quad \alpha = 56.3^\circ\)

   Opposite over adjacent in the right-angled triangle.

3. Bearing from north

   \(180^\circ - 56.3^\circ = 123.7^\circ\)

   South is 180°, and the drone is 56.3° short of south. Bearing 124° to the nearest degree.

Have a go first. Then open the working one line at a time.

1. The formula

   \(\text{speed} = |\mathbf{v}| = \sqrt{v_1^2+v_2^2+v_3^2}\)

   Speed is the magnitude of the velocity vector. It is never negative.

2. Substitute

   \(|\mathbf{v}| = \sqrt{2^2 + (-5)^2 + 6^2} = \sqrt{65}\)

   Square first, so the minus sign disappears.

3. Evaluate

   \(\sqrt{65} = 8.06\)

   Speed 8.06 m s⁻¹ (3 s.f.).

Have a go first. Then open the working one line at a time.

1. Length of the direction

   \(|\mathbf{d}| = \sqrt{3^2+4^2+12^2} = 13\)

   Find \(|\mathbf{d}|\) first.

2. Scale to the speed

   \(\mathbf{v} = \dfrac{26}{13}\,\begin{pmatrix}3\\4\\12\end{pmatrix} = \begin{pmatrix}6\\8\\24\end{pmatrix}\)

   Divide by \(|\mathbf{d}|\) to get a unit vector, then multiply by the speed 26. Here 26 ÷ 13 = 2.

3. Position at time t

   \(\mathbf{r} = \begin{pmatrix}5\\0\\2\end{pmatrix} + t\begin{pmatrix}6\\8\\24\end{pmatrix}\)

   \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\).

4. After 2 seconds

   \(x = 5 + 2(6) = 17\)

   Use the first component only.

Have a go first. Then open the working one line at a time.

1. Speed

   \(|\mathbf{v}| = \sqrt{1.5^2+2^2+(-0.5)^2} = \sqrt{6.5} = 2.55\)

   Speed 2.55 m s⁻¹.

2. Position at time t

   \(\mathbf{r} = \begin{pmatrix}4\\-2\\7\end{pmatrix} + t\begin{pmatrix}1.5\\2\\-0.5\end{pmatrix}\)

   Start position plus t times velocity.

3. Solve for t, one component

   \(4 + 1.5t = 10 \Rightarrow t = 4\)

   Use the i-component to find t.

4. Check the other components

   \(-2 + 2(4) = 6,\quad 7 - 0.5(4) = 5\)

   Both match (10, 6, 5) at the same t = 4. So the particle does pass through the point.

Have a go first. Then open the working one line at a time.

1. Both positions

   \(\mathbf{r}_1 = \begin{pmatrix}12\\-0.5\\1\end{pmatrix},\quad \mathbf{r}_2 = \begin{pmatrix}3\\5\\6.5\end{pmatrix}\)

   Put t = 2.5 into each vector.

2. The displacement between them

   \(\mathbf{r}_1 - \mathbf{r}_2 = \begin{pmatrix}9\\-5.5\\-5.5\end{pmatrix}\)

   Subtract one position from the other. The order does not change the distance.

3. Its magnitude

   \(\sqrt{141.5} = 11.9\)

   Distance about 11.9 m.

Have a go first. Then open the working one line at a time.

1. Positions

   \(\mathbf{r}_A = (-8 + 5t)\mathbf{i} + (3 + 2t)\mathbf{j},\quad \mathbf{r}_B = (7 + 2t)\mathbf{i} + (-4 - t)\mathbf{j}\)

   Start position plus t times velocity, for each ship.

2. Distance after 2 hours

   \(\mathbf{r}_A - \mathbf{r}_B = \begin{pmatrix}-9\\13\end{pmatrix},\quad d = \sqrt{250} = 15.8\text{ km}\)

   Put t = 2 in both, subtract, take the magnitude.

3. Due north means the same eastward position

   \(-8 + 5t = 7 + 2t \Rightarrow t = 5\)

   A is due north of B when their i-components are equal.

4. Check which is north

   \(y_A = 3 + 2(5) = 13,\quad y_B = -4 - 5 = -9\)

   13 > −9, so A is north of B. After 5 hours A is due north of B.

Have a go first. Then open the working one line at a time.

1. Different parameters for the paths

   \(\begin{pmatrix}2\\-3\\-3\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}0\\5\\6\end{pmatrix} + \mu\begin{pmatrix}2\\0\\1\end{pmatrix}\)

   A path is a line. Two paths cross if some λ and μ give the same point. They need not be equal.

2. Solve using i and j

   \(j: -3 + 2\lambda = 5 \Rightarrow \lambda = 4;\quad i: 2 + 4 = 2\mu \Rightarrow \mu = 3\)

   Two unknowns, so use two components.

3. Check k

   \(-3 + 3(4) = 9,\quad 6 + 3 = 9\)

   Equal, so the paths cross at (6, 5, 9).

4. Collision needs the same t

   \(i: 2 + t = 2t \Rightarrow t = 2;\quad j: -3 + 2t = 5 \Rightarrow t = 4\)

   The two values of t differ. Rocket 1 is at (6, 5, 9) when t = 4 and rocket 2 when t = 3. They do not collide.

Have a go first. Then open the working one line at a time.

1. Relative position

   \(\mathbf{r}_Q - \mathbf{r}_P = \begin{pmatrix}5\\6\end{pmatrix} + t\begin{pmatrix}2\\-5\end{pmatrix} = \begin{pmatrix}5 + 2t\\6 - 5t\end{pmatrix}\)

   Subtract the position vectors, collecting the constants and the t terms.

2. Distance as a function of t

   \(d = \sqrt{(5 + 2t)^2 + (6 - 5t)^2} = \sqrt{29t^2 - 40t + 61}\)

   Expand, then collect like terms.

3. From the calculator

   Minimum of d(t) for t ≥ 0: t = 0.690, d = 6.87

   Graph d(t) and use the minimum finder. The least distance is the y-value.

4. Due east

   \(-1 + 3t = 5 - 2t \Rightarrow t = 1.2\)

   Same j-component. Then check Q is east, not west: at t = 1.2 Q has x = 5.6 and P has x = −1.8.