Printable · GCSE Foundation · ages 14-16
Translations as vectors worksheet — GCSE Foundation
Fifteen questions on "translations as vectors" — DfE statement G24. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Translations as vectors worksheet — GCSE Foundation
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- 1.The vector describes a translation. A student writes it as by mistake. Write down what error the student has made.
- 2.The point M(3, 8) is translated by the vector . Write down the coordinates of the image of M.
- 3.A shape is translated by the vector and then by the vector . Write down the single column vector that has the same effect as those two translations together.
- 4.A courier drone delivers a parcel. It starts at the point (2, 1) on a map grid measured in kilometres. It flies by the vector to a warehouse, then by the vector to the delivery address. Work out the coordinates of the delivery address.
- 5.Vector A is and vector B is . Write down how vector B compares to vector A.
- 6.Write down the column vector that describes a translation of 5 units to the left and 3 units down.
- 7.A robot on a grid moves by the vector , then by the vector , then by the vector . Write down the single column vector that has the same overall effect as these three moves.
- 8.A translation is described by the column vector . Write down a description of this translation in words, stating the horizontal and vertical movement.
- 9.Write down the column vector, from the options given, that would move a point to the right and downwards.
- 10.A translation moves the point (1, 1) to the point (9, 4). Write down the column vector of this translation.
- 11.A drone starts at the point (2, −3) on a coordinate grid. It flies by the vector and then by the vector . Write down the column vector that would take the drone straight back to its starting point.
- 12.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector to pick up a beam, then along the vector to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
- 13.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 14.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 15.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
Answer key
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
- (c) (−2, 10) — To translate M(3, 8) by $\binom{−5}{2}$, add −5 to the x-coordinate and 2 to the y-coordinate: (3 + (−5), 8 + 2) = (−2, 10). (5, 3) comes from swapping the two components of the vector before applying them. (−2, 8) applies only the x-component and forgets to add the y-component. (3, 10) applies only the y-component and forgets to add the x-component.
- (a) $\binom{-5}{-4}$ — Method: one translation followed by another is a single translation, and the two vectors are added: top to top, bottom to bottom. Working: across, 4 − 9 = −5; up, −7 + 3 = −4. Answer: $\binom{-5}{-4}$. Subtracting the second vector instead of adding it gives 13 on top and −7 − 3 = −10 underneath. Adding the top numbers correctly but subtracting the bottom ones gives −10 underneath with −5 on top. Adding 4 and 9 as though both were positive and then keeping the minus sign of the larger gives −13 on top.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (d) B is the reverse of A — Every component of vector B is the negative of the matching component of vector A (−2 is the negative of 2, and 5 is the negative of −5), so B undoes the translation that A performs — B is the reverse of A. 'B is the same as A' ignores that both signs have flipped. 'B is twice A' confuses a sign change with a scale-factor change; the sizes of the components have not changed, only their signs. 'B is unrelated to A in direction' misses that the two vectors are directly linked, just in opposite directions.
- (d) $\binom{−5}{−3}$ — Left is a negative horizontal move and down is a negative vertical move, so the vector is $\binom{−5}{−3}$. $\binom{5}{−3}$ comes from treating 'left' as a positive move. $\binom{−5}{3}$ comes from treating 'down' as a positive move. $\binom{−3}{−5}$ comes from swapping the horizontal and vertical components.
- (d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
- (b) 7 units right and 2 units down — In the column vector $\binom{7}{−2}$, the top number gives the horizontal movement and the bottom number gives the vertical movement. A positive top number (7) means 7 units to the right, and a negative bottom number (−2) means 2 units down. The distractor '7 units left and 2 units up' reverses both signs, as if the vector were $\binom{−7}{2}$. The distractor '2 units right and 7 units down' swaps the horizontal and vertical numbers around. The distractor '7 units right and 2 units up' gets the horizontal movement correct but reverses the sign of the vertical movement.
- (c) $\binom{4}{−3}$ — A positive top number moves a point to the right, and a negative bottom number moves it downwards, so $\binom{4}{−3}$ is right and down. $\binom{−4}{3}$ moves left and up — the opposite direction on both axes. $\binom{4}{3}$ moves right, like the key, but its positive bottom number moves it up, not down. $\binom{−4}{−3}$ moves down, like the key, but its negative top number moves it left, not right.
- (b) $\binom{8}{3}$ — The column vector is (end − start) in each coordinate: (9 − 1, 4 − 1) = (8, 3), written as $\binom{8}{3}$. $\binom{3}{8}$ swaps the horizontal and vertical components. $\binom{10}{5}$ comes from adding the coordinates instead of subtracting them. $\binom{−8}{−3}$ comes from working out start minus end instead of end minus start.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
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