10.5 Vectors in pure mathematics and in context — Practise

Work through these on your own. In a proof, write the two vectors you are comparing and name the shared point. In a context question, sketch the situation, choose axes, and interpret your answer with its units.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Show that $A(1, 2)$ , $B(4, 8)$ and $C(7, 14)$ are collinear.
The quadrilateral has vertices $A(0, 0)$ , $B(3, 1)$ , $C(5, 5)$ and $D(2, 4)$ . Show that it is a parallelogram.
The points are $A(1, 3)$ and $B(7, 9)$ . Find the point $P$ on $AB$ with $AP : PB = 1 : 2$ .
The points are $A(2, 1)$ and $B(7, 6)$ . Find the point $P$ on $AB$ with $AP : PB = 3 : 2$ .
Forces $\mathbf{F}_1 = 3\mathbf{i} - 2\mathbf{j}$ N and $\mathbf{F}_2 = -\mathbf{i} + 4\mathbf{j}$ N act at a point. Find the third force that keeps the point still.
Two forces $\mathbf{F}_1 = 5\mathbf{i} + \mathbf{j}$ N and $\mathbf{F}_2 = -2\mathbf{i} + 4\mathbf{j}$ N act at a point. Find the resultant and its magnitude.
A particle starts at $\mathbf{r}_0 = 2\mathbf{i} - \mathbf{j}$ m and moves with velocity $\mathbf{v} = 3\mathbf{i} + 2\mathbf{j}$ m/s. Find its position after $3$ s and its distance from the origin then.
A boat has velocity $5\mathbf{i}$ m/s in still water. A current of $2\mathbf{j}$ m/s flows. Find the resultant velocity, the speed and the direction from the positive $x$ -axis.
A plane flies at $240$ km/h due east. A wind blows from the north at $70$ km/h, pushing the plane due south. Find the resultant velocity, the speed and the direction.
Show that $A(1, 1)$ , $B(3, 5)$ and $C(5, 9)$ are collinear.
The quadrilateral has vertices $A(0, 0)$ , $B(4, 0)$ , $C(5, 3)$ and $D(1, 3)$ . Show that the diagonals $AC$ and $BD$ bisect each other.
A hiker starts at $B(1, 2)$ km, walks to a first waypoint at $(4, 6)$ km, then to a second at $(8, 3)$ km. Find the single displacement from the start, the distance from the start, and the direction of that displacement from the positive $x$ -axis.

Answers and concise solutions

1. The two displacements

Answer: $\vec{AB} = 3\mathbf{i} + 6\mathbf{j}$ and $\vec{BC} = 3\mathbf{i} + 6\mathbf{j}$ . So $\vec{AB} = \vec{BC}$ , and the two vectors share the point $B$ . The three points lie on one straight line.

2. The two opposite sides

Answer: $\vec{AB} = 3\mathbf{i} + \mathbf{j}$ and $\vec{DC} = (5 - 2)\mathbf{i} + (5 - 4)\mathbf{j} = 3\mathbf{i} + \mathbf{j}$ . The opposite sides are equal, so $ABCD$ is a parallelogram.

3. The position vector of $P$

Answer: With $m = 1$ and $n = 2$ , $\vec{OP} = \dfrac{2\mathbf{a} + \mathbf{b}}{3} = \dfrac{2(1, 3) + (7, 9)}{3} = \dfrac{(9, 15)}{3} = (3, 5)$ . So $P$ is $(3, 5)$ , and $AP = (2, 2)$ while $PB = (4, 4)$ , the ratio $1 : 2$ .

4. The position vector of $P$

Answer: With $m = 3$ and $n = 2$ , $\vec{OP} = \dfrac{2\mathbf{a} + 3\mathbf{b}}{5} = \dfrac{(4, 2) + (21, 18)}{5} = \dfrac{(25, 20)}{5} = (5, 4)$ . Then $AP = (3, 3)$ and $PB = (2, 2)$ , the ratio $3 : 2$ .

5. The third force

Answer: The three forces sum to zero, so the third force is $-(\mathbf{F}_1 + \mathbf{F}_2)$ . Now $\mathbf{F}_1 + \mathbf{F}_2 = 2\mathbf{i} + 2\mathbf{j}$ N, so the third force is $-2\mathbf{i} - 2\mathbf{j}$ N.

6. The resultant and its magnitude

Answer: The resultant is $\mathbf{F}_1 + \mathbf{F}_2 = 3\mathbf{i} + 5\mathbf{j}$ N. Its magnitude is $\sqrt{3^2 + 5^2} = \sqrt{34} \approx 5.83$ N.

7. The position and the distance

Answer: After $3$ s, $\mathbf{r} = \mathbf{r}_0 + 3\mathbf{v} = (2\mathbf{i} - \mathbf{j}) + (9\mathbf{i} + 6\mathbf{j}) = 11\mathbf{i} + 5\mathbf{j}$ m. The distance from the origin is $\sqrt{11^2 + 5^2} = \sqrt{146} \approx 12.08$ m.

8. The velocity, the speed and the direction

Answer: The resultant is $5\mathbf{i} + 2\mathbf{j}$ m/s. The speed is $\sqrt{5^2 + 2^2} = \sqrt{29} \approx 5.39$ m/s, and the direction is $\tan^{-1}\left(\dfrac{2}{5}\right) = 21.8^\circ$ to $1$ d.p. from the positive $x$ -axis.

9. The velocity, the speed and the direction

Answer: Take east as $\mathbf{i}$ : the resultant is $240\mathbf{i} - 70\mathbf{j}$ km/h. The speed is $\sqrt{240^2 + 70^2} = \sqrt{62500} = 250$ km/h. The direction is $\tan^{-1}\left(\dfrac{70}{240}\right) = 16.3^\circ$ to $1$ d.p. south of east, which is $-16.3^\circ$ from the positive $x$ -axis.

10. The two displacements

Answer: $\vec{AB} = 2\mathbf{i} + 4\mathbf{j}$ and $\vec{BC} = 2\mathbf{i} + 4\mathbf{j}$ . The two displacements are equal and share the point $B$ , so the three points are collinear.

11. The two midpoints

Answer: The midpoint of $AC$ is $\dfrac{1}{2}((0, 0) + (5, 3)) = \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$ . The midpoint of $BD$ is $\dfrac{1}{2}((4, 0) + (1, 3)) = \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$ . The two midpoints are the same, so the diagonals bisect each other.

12. The displacement, the distance and the direction

Answer: The displacement from the start is $(8 - 1)\mathbf{i} + (3 - 2)\mathbf{j} = 7\mathbf{i} + \mathbf{j}$ km, whatever route was taken. The distance from the start is $\sqrt{7^2 + 1^2} = \sqrt{50} = 5\sqrt{2} \approx 7.07$ km, and the direction is $\tan^{-1}\left(\dfrac{1}{7}\right) = 8.1^\circ$ to $1$ d.p. from the positive $x$ -axis.