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Size (magnitude)
Direction

","mcqOptions":[{"text":"

Size (magnitude)
Direction

","correct":true,"explanation":"A vector is fully described by both its magnitude, meaning its size, and its direction. A scalar has magnitude only and no direction."},{"text":"

Size (magnitude)
Position

","correct":false,"explanation":"Position does not distinguish vectors from scalars. The defining extra property is direction, while both vectors and scalars can have a magnitude or size."},{"text":"

Direction
Units

","correct":false,"explanation":"Units are not the distinguishing property because both vector and scalar quantities have units. A vector specifically requires magnitude and direction."},{"text":"

Size (magnitude)
Starting point

","correct":false,"explanation":"A vector can be moved without changing its value, so its starting point is not essential. The required second property is direction."}]},{"id":"80167c15-4d8a-4aa4-b269-4541e9508a22","cardType":"cloze","points":null,"front":"In 2D vector notation, $\\text{[...]}$ represents one unit in the positive horizontal direction and $\\mathbf{j}$ represents one unit in the positive vertical direction.","back":"In 2D vector notation, $\\boxed{\\mathbf{i}}$ represents one unit in the positive horizontal direction and $\\mathbf{j}$ represents one unit in the positive vertical direction.","mcqOptions":[{"text":"$$\\mathbf{i}$","correct":true,"explanation":"The unit vector $\\mathbf{i}$ points one unit in the positive horizontal direction. The unit vector $\\mathbf{j}$ similarly represents one unit vertically upwards."},{"text":"$$\\mathbf{j}$","correct":false,"explanation":"The vertical unit vector is $\\mathbf{j}$, not the horizontal one. In standard 2D notation, $\\mathbf{i}$ represents one positive horizontal unit."},{"text":"$$-\\mathbf{i}$","correct":false,"explanation":"The negative horizontal unit vector points one unit in the negative horizontal direction. Positive horizontal direction is represented by $\\mathbf{i}$."},{"text":"$$-\\mathbf{j}$","correct":false,"explanation":"The negative vertical unit vector points downwards, not in the positive horizontal direction. The positive horizontal unit vector is $\\mathbf{i}$."}]},{"id":"4aff49ef-58db-4b21-b47f-5995f4c39381","cardType":"cloze","points":null,"front":"In 2D vector notation, $\\mathbf{i}$ represents one unit in the positive horizontal direction and $\\text{[...]}$ represents one unit in the positive vertical direction.","back":"In 2D vector notation, $\\mathbf{i}$ represents one unit in the positive horizontal direction and $\\boxed{\\mathbf{j}}$ represents one unit in the positive vertical direction.","mcqOptions":[{"text":"$$\\mathbf{j}$","correct":true,"explanation":"The unit vector $\\mathbf{j}$ represents one unit in the positive vertical direction. The unit vector $\\mathbf{i}$ represents one unit horizontally to the right."},{"text":"$$\\mathbf{i}$","correct":false,"explanation":"The horizontal unit vector is $\\mathbf{i}$, so it cannot represent positive vertical movement. Positive vertical direction uses $\\mathbf{j}$."},{"text":"$$-\\mathbf{i}$","correct":false,"explanation":"This represents one unit in the negative horizontal direction, rather than positive vertical direction. The required vertical unit vector is $\\mathbf{j}$."},{"text":"$$-\\mathbf{j}$","correct":false,"explanation":"This points one unit vertically downwards, reversing the required direction. Positive vertical movement is represented by $\\mathbf{j}$."}]},{"id":"2a081bfb-3b26-48ff-a1cc-40d8073ac636","cardType":"basic","points":1,"front":"What is the rule for adding two vectors written in component form?","back":"Add like components separately: $\\mathbf{i}$ with $\\mathbf{i}$, and $\\mathbf{j}$ with $\\mathbf{j}$.","mcqOptions":[{"text":"Add like components separately: $\\mathbf{i}$ with $\\mathbf{i}$, and $\\mathbf{j}$ with $\\mathbf{j}$.","correct":true,"explanation":"Component addition combines the coefficients of matching unit vectors separately. The $\\mathbf{i}$ components are added together, and the $\\mathbf{j}$ components are added together."},{"text":"Add unlike components separately: $\\mathbf{i}$ with $\\mathbf{j}$, and $\\mathbf{j}$ with $\\mathbf{i}$.","correct":false,"explanation":"Unit-vector components must be combined with matching unit vectors. Pairing $\\mathbf{i}$ with $\\mathbf{j}$ mixes horizontal and vertical directions instead of adding corresponding components."},{"text":"Add like components together: the $\\mathbf{i}$ and $\\mathbf{j}$ components form one total.","correct":false,"explanation":"The horizontal and vertical components must remain separate because they represent perpendicular directions. Only matching $\\mathbf{i}$ or matching $\\mathbf{j}$ terms can be combined."},{"text":"Multiply like components separately: $\\mathbf{i}$ with $\\mathbf{i}$, and $\\mathbf{j}$ with $\\mathbf{j}$.","correct":false,"explanation":"Adding vectors requires adding corresponding components, not multiplying them. Multiplication would produce different quantities and does not give the vector sum."}]},{"id":"ae1e6921-64c9-4fa8-8afe-30baf3282b47","cardType":"basic","points":1,"front":"How is a position vector of a point $A$ defined?","back":"The vector $\\vec{OA}$, which is the vector from the origin $O$ to point $A$.","mcqOptions":[{"text":"The vector $\\vec{OA}$, which is the vector from the origin $O$ to point $A$.","correct":true,"explanation":"A position vector locates a point relative to the origin. It is defined as the directed vector from $O$ to the point, written $\\vec{OA}$."},{"text":"The vector $\\vec{AO}$, which is the vector from the point $A$ to origin $O$.","correct":false,"explanation":"Reversing the order gives the opposite vector, pointing from $A$ to $O$. A position vector is directed from the origin to the point."},{"text":"The vector $\\vec{AB}$, which is the vector from the origin $O$ to point $A$.","correct":false,"explanation":"The notation $\\vec{AB}$ describes a vector from $A$ to $B$, not from the origin to $A$. The position vector is written $\\vec{OA}$."},{"text":"The vector $\\vec{OA}$, which is the vector from the point $A$ to origin $O$.","correct":false,"explanation":"Although the notation starts at $O$, the stated direction is reversed. A position vector travels from the origin $O$ to point $A$."}]},{"id":"79346dc0-d415-4d5b-85f7-3e22b1556030","cardType":"basic","points":1,"front":"State the formula for finding the vector $\\vec{AB}$ using the position vectors of $A$ and $B$.","back":"$$$\\vec{AB} = \\vec{OB} - \\vec{OA}$$","mcqOptions":[{"text":"$$$\\vec{AB} = \\vec{OB} - \\vec{OA}$$","correct":true,"explanation":"To travel from $A$ to $B$, subtract the position vector of $A$ from the position vector of $B$. This gives the displacement $\\vec{AB}$."},{"text":"$$$\\vec{AB} = \\vec{OA} - \\vec{OB}$$","correct":false,"explanation":"This subtraction reverses the direction and gives $\\vec{BA}$ instead. The vector from $A$ to $B$ requires the position of $A$ to be subtracted from that of $B$."},{"text":"$$$\\vec{AB} = \\vec{OA} + \\vec{OB}$$","correct":false,"explanation":"Adding the two position vectors does not calculate the displacement between the points. The required displacement is found by subtracting $\\vec{OA}$ from $\\vec{OB}$."},{"text":"$$$\\vec{AB} = \\vec{OA} - \\vec{AB}$$","correct":false,"explanation":"This expression contains the unknown vector on both sides and does not use the correct pair of position vectors. Subtracting $\\vec{OA}$ from $\\vec{OB}$ gives $\\vec{AB}$."}]},{"id":"e24b6122-313d-4378-a2ed-658233f08d4d","cardType":"basic","points":1,"front":"What does reversing the order of a vector (e.g. $\\vec{BA}$ instead of $\\vec{AB}$) do to its components?","back":"It reverses the signs of both the $\\mathbf{i}$ and $\\mathbf{j}$ components.","mcqOptions":[{"text":"It reverses the signs of both the $\\mathbf{i}$ and $\\mathbf{j}$ components.","correct":true,"explanation":"Reversing a vector reverses its direction, which is equivalent to multiplying every component by $-1$. Therefore both the $\\mathbf{i}$ and $\\mathbf{j}$ signs change."},{"text":"It keeps the signs of both the $\\mathbf{i}$ and $\\mathbf{j}$ components.","correct":false,"explanation":"Keeping both signs unchanged would leave the vector pointing in the original direction. Reversing the order changes the sign of every component."},{"text":"It reverses the sign of only the $\\mathbf{i}$ component.","correct":false,"explanation":"Changing only the horizontal sign alters the direction but does not produce the complete opposite vector. Both components must have their signs reversed."},{"text":"It reverses the sign of only the $\\mathbf{j}$ component.","correct":false,"explanation":"Changing only the vertical sign is insufficient to reverse the whole vector. Reversing the order changes the signs of both components."}]},{"id":"bc93573e-409e-49eb-8754-4e3338f07df1","cardType":"basic","points":1,"front":"How is the magnitude of vector $\\mathbf{v} = a\\mathbf{i} + b\\mathbf{j}$ calculated?","back":"Use Pythagoras: $$|\\mathbf{v}| = \\sqrt{a^2 + b^2}$$","mcqOptions":[{"text":"Use Pythagoras: $$|\\mathbf{v}| = \\sqrt{a^2 + b^2}$$","correct":true,"explanation":"The components form perpendicular horizontal and vertical sides of a right-angled triangle. Pythagoras therefore gives the magnitude as $\\sqrt{a^2+b^2}$."},{"text":"Use Pythagoras: $$|\\mathbf{v}| = \\sqrt{a + b}$$","correct":false,"explanation":"Pythagoras requires the squares of the perpendicular components before adding them. Omitting the squares gives an incorrect magnitude except in special cases."},{"text":"Use Pythagoras: $$|\\mathbf{v}| = a^2 + b^2$$","correct":false,"explanation":"The sum of the squares is the square of the magnitude, not the magnitude itself. A square root must be taken after adding $a^2$ and $b^2$."},{"text":"Use Pythagoras: $$|\\mathbf{v}| = \\sqrt{a^2 - b^2}$$","correct":false,"explanation":"The perpendicular sides contribute positively in Pythagoras, so their squares are added. Subtracting one square is a common rearrangement error and gives the wrong length."}]},{"id":"90cbfd61-4f6e-471b-a826-beb6ff943365","cardType":"basic","points":2,"front":"If $|5\\mathbf{i} + k\\mathbf{j}| = 13$, why are there two possible values for $k$?","back":"Squaring $k$ in the magnitude formula means both positive and negative values ($\\pm 12$) result in the same length.","mcqOptions":[{"text":"Squaring $k$ in the magnitude formula means both positive and negative values ($\\pm 12$) result in the same length.","correct":true,"explanation":"The magnitude is calculated using $\\sqrt{5^2+k^2}=13$, so $k^2=144$ and $k=12$ or $k=-12$. Both signs produce the same squared value and therefore the same length."},{"text":"Setting $k$ to zero makes the horizontal component sufficient to produce the required length.","correct":false,"explanation":"The value $k=0$ would give a magnitude of $5$, not $13$. The two solutions arise because solving $k^2=144$ gives both $k=12$ and $k=-12$."},{"text":"Taking the square root of the magnitude gives only one positive value, $k=12$, because lengths cannot be negative.","correct":false,"explanation":"The magnitude formula uses $k^2$, not $|k|$ alone, so the two possible values are $12$ and $-12$, rather than just one positive value."},{"text":"Both components can be positive or negative, so changing the signs of $5$ and $k$ creates two possible lengths.","correct":false,"explanation":"The signs of the components do not both need to vary. The fixed component is $5$, while solving $k^2=144$ allows the vertical component to be either $12$ or $-12$."}]},{"id":"349e5fda-d4f0-45cb-848b-6c38c05e9d9d","cardType":"basic","points":1,"front":"Define a resultant vector.","back":"A single vector that has the same overall effect as multiple vectors acting together.","mcqOptions":[{"text":"A single vector that has the same overall effect as multiple vectors acting together.","correct":true,"explanation":"A resultant vector combines the effects of several vectors into one vector, while preserving the same overall effect as the original vectors acting together."},{"text":"The largest vector in a group of multiple vectors acting together.","correct":false,"explanation":"A resultant is not the largest vector in a collection. It is the vector sum of all the vectors and may be smaller than some individual vectors if they oppose one another."},{"text":"A single vector pointing in the same direction as the first vector acting on an object.","correct":false,"explanation":"A resultant vector is not restricted to the direction of one original vector. It combines both magnitudes and directions to represent their total effect."},{"text":"A zero vector produced whenever multiple vectors act together on an object.","correct":false,"explanation":"The resultant is not necessarily zero. It is zero only when the vectors exactly balance; otherwise, it represents their non-zero combined effect."}]},{"id":"441abc5e-586d-4816-886f-dbc277e320ac","cardType":"basic","points":1,"front":"How do you calculate the resultant force when multiple force vectors are acting on a particle?","back":"Calculate the sum of all individual force vectors.","mcqOptions":[{"text":"Calculate the sum of all individual force vectors.","correct":true,"explanation":"Adding the vectors component by component gives the total force, including the effects of opposing directions. This vector sum is the resultant force on the particle."},{"text":"Calculate the product of all individual force vectors.","correct":false,"explanation":"Multiplying the forces ignores their directions and is not how vector forces combine. The resultant force is found by adding the individual force vectors."},{"text":"Calculate the sum of the magnitudes of all individual force vectors.","correct":false,"explanation":"Adding only magnitudes treats every force as if it acted in the same direction. Vector addition must account for direction as well as size."},{"text":"Calculate the difference between all individual force vectors.","correct":false,"explanation":"Subtracting every force would reverse the contribution of forces that act in their stated directions. The correct method is to add all force vectors algebraically."}]},{"id":"43cd23dc-0e28-49c4-abec-44aea302267b","cardType":"cloze","points":null,"front":"To find the vector between two endpoints in a triangle, $\\vec{BC} = \\vec{AC} - \\text{[...]}.","back":"To find the vector between two endpoints in a triangle, $\\vec{BC} = \\vec{AC} - \\boxed{\\vec{AB}}.","mcqOptions":[{"text":"$$\\vec{AB}$","correct":true,"explanation":"Using the triangle relation $\\vec{AC}=\\vec{AB}+\\vec{BC}$ and rearranging gives $\\vec{BC}=\\vec{AC}-\\vec{AB}$."},{"text":"$$\\vec{BC}$","correct":false,"explanation":"Subtracting $\\vec{BC}$ would not isolate $\\vec{BC}$ in the triangle relation. The required rearrangement subtracts $\\vec{AB}$ from $\\vec{AC}$."},{"text":"$$\\vec{CA}$","correct":false,"explanation":"The vector $\\vec{CA}$ points opposite to $\\vec{AC}$ and is not the first part of the path from $A$ to $C$. The triangle relation uses $\\vec{AB}$."},{"text":"$$\\vec{BA}$","correct":false,"explanation":"The vector $\\vec{BA}$ has the opposite direction to $\\vec{AB}$. Subtracting it would produce a different expression, whereas the correct subtraction is $\\vec{AB}$."}]},{"id":"0286c548-6a5d-4e21-84ca-27336ad5f8bb","cardType":"basic","points":2,"front":"In triangle $ABC$, explain the vector path taken to find $\\vec{BC}$ if starting from $B$.","back":"Travel from $B$ back to $A$ ($-\\vec{AB}$), then from $A$ to $C$ ($+\\vec{AC}$).","mcqOptions":[{"text":"Travel from $B$ back to $A$ ($-\\vec{AB}$), then from $A$ to $C$ ($+\\vec{AC}$).","correct":true,"explanation":"To travel from $B$ to $C$ through $A$, first reverse $\\vec{AB}$ to get $-\\vec{AB}$, then follow $\\vec{AC}$ from $A$ to $C$."},{"text":"Travel from $B$ to $A$ ($+\\vec{AB}$), then from $A$ to $C$ ($+\\vec{AC}$).","correct":false,"explanation":"The first leg must go from $B$ to $A$, which is $-\\vec{AB}$, not $+\\vec{AB}$. Using the positive vector starts at $A$ and points away from the required path."},{"text":"Travel from $B$ back to $A$ ($-\\vec{AB}$), then from $C$ to $A$ ($-\\vec{AC}$).","correct":false,"explanation":"The second leg from $A$ to $C$ is $+\\vec{AC}$, not $-\\vec{AC}$. Reversing it would take the path from $C$ back to $A$."},{"text":"Travel from $B$ directly to $C$ ($+\\vec{BC}$), then from $C$ back to $A$ ($-\\vec{AC}$).","correct":false,"explanation":"The path should pass through $A$, but this version uses $\\vec{BC}$ itself as the first step and then reverses the direction of $\\vec{AC}$."}]},{"id":"58ff625a-e910-4da0-9f37-91edd0427f66","cardType":"cloze","points":null,"front":"When working with velocity vectors, the speed is defined as the [ ] of the velocity vector.","back":"When working with velocity vectors, the speed is defined as the magnitude of the velocity vector.","mcqOptions":[{"text":"magnitude","correct":true,"explanation":"Speed is a scalar quantity equal to the magnitude, or length, of the velocity vector. Direction is removed when the magnitude is taken."},{"text":"direction","correct":false,"explanation":"The direction of velocity describes the direction of motion, not its speed. Speed is the magnitude of the velocity vector and has no direction."},{"text":"component","correct":false,"explanation":"The component of velocity is only one part of the vector, whereas speed depends on all components through the vector’s magnitude."},{"text":"gradient","correct":false,"explanation":"The gradient is not the measure of a velocity vector’s size. Speed is obtained by calculating the magnitude of the velocity vector."}]},{"id":"b3d39c24-8a62-458c-ad20-6bdd1706980d","cardType":"basic","points":2,"front":"What are the two defining characteristics of a bearing?","back":"

Measured clockwise from North
Written as a three-figure angle

","mcqOptions":[{"text":"

Measured clockwise from North
Written as a three-figure angle

","correct":true,"explanation":"Bearings are measured clockwise from North and written using three figures, such as $045^\\circ$ or $270^\\circ$. Both conventions are required for a standard bearing."},{"text":"

Measured anticlockwise from North
Written as a three-figure angle

","correct":false,"explanation":"Bearings are measured clockwise from North, not anticlockwise from North. They are still written as three figures, so the reference direction is the error."},{"text":"

Measured clockwise from East
Written as a three-figure angle

","correct":false,"explanation":"A bearing is measured from North, not East. Although the clockwise convention and three-figure format are retained here, the starting direction is incorrect."},{"text":"

Measured clockwise from North
Written as a one-figure angle

","correct":false,"explanation":"Bearings use three figures, including leading zeroes where necessary. They are measured clockwise from North, not written as ordinary one- or two-figure angles."}]},{"id":"c3b1e79c-770d-4ea2-a1db-465cc56b0c46","cardType":"basic","points":1,"front":"If $\\mathbf{i}$ is East and $\\mathbf{j}$ is North, what direction is a vector with positive $\\mathbf{i}$ and negative $\\mathbf{j}$ components?","back":"South-East","mcqOptions":[{"text":"South-East","correct":true,"explanation":"A positive $\\mathbf{i}$ component points East and a negative $\\mathbf{j}$ component points South. Together these components place the vector in the South-East direction."},{"text":"North-East","correct":false,"explanation":"North-East requires both the $\\mathbf{i}$ and $\\mathbf{j}$ components to be positive. A negative $\\mathbf{j}$ component instead points South."},{"text":"South-West","correct":false,"explanation":"South-West requires a negative $\\mathbf{i}$ component as well as a negative $\\mathbf{j}$ component. Here, positive $\\mathbf{i}$ points East."},{"text":"North-West","correct":false,"explanation":"North-West requires a negative $\\mathbf{i}$ component and a positive $\\mathbf{j}$ component. The stated signs point East and South instead."}]},{"id":"65538fc7-305c-41b2-9074-2202dd3b27b2","cardType":"basic","points":2,"front":"A boat has velocity $6\\mathbf{i} - 8\\mathbf{j}$. How would you calculate the angle $\\theta$ below the East line to find the bearing?","back":"$$$\\theta = \\tan^{-1}\\left(\\frac{8}{6}\\right)$$ (using the absolute values of the components).","mcqOptions":[{"text":"$$$\\theta = \\tan^{-1}\\left(\\frac{8}{6}\\right)$$ (using the absolute values of the components).","correct":true,"explanation":"The angle below the East line is found from the absolute vertical and horizontal components, so opposite over adjacent gives $\\tan^{-1}(8/6)$."},{"text":"$$$\\theta = \\tan^{-1}\\left(\\frac{6}{8}\\right)$$ (using the absolute values of the components).","correct":false,"explanation":"The components have been reversed, giving the angle from the North-South direction instead. For an angle below East, use the southward component divided by the eastward component: $\\tan^{-1}(8/6)$."},{"text":"$$$\\theta = \\tan^{-1}\\left(\\frac{8}{14}\\right)$$ (using the absolute values of the components).","correct":false,"explanation":"The denominator incorrectly combines the components rather than using the horizontal component as the adjacent side. The required ratio is vertical over horizontal, giving $\\tan^{-1}(8/6)$."},{"text":"$$$\\theta = \\tan^{-1}\\left(\\frac{6}{14}\\right)$$ (using the absolute values of the components).","correct":false,"explanation":"This uses the horizontal component over the sum of the components, which is not the tangent ratio for the angle below East. Use opposite over adjacent: $\\tan^{-1}(8/6)$."}]},{"id":"0e208f14-8bdd-4f1c-b5ca-ede8e6a0a958","cardType":"basic","points":1,"front":"If an object starts at position $\\mathbf{r}_0$ and moves through vector $\\mathbf{m}$, what is its new position vector?","back":"$$$\\mathbf{r}_{new} = \\mathbf{r}_0 + \\mathbf{m}$$","mcqOptions":[{"text":"$$$\\mathbf{r}_{new} = \\mathbf{r}_0 + \\mathbf{m}$$","correct":true,"explanation":"A displacement vector changes the position by adding its components to the original position, so the new position is the vector sum of $\\mathbf{r}_0$ and $\\mathbf{m}$."},{"text":"$$$\\mathbf{r}_{new} = \\mathbf{r}_0 - \\mathbf{m}$$","correct":false,"explanation":"Subtracting the movement reverses its direction, giving the position reached by moving through $-\\mathbf{m}$. A movement through $\\mathbf{m}$ requires addition: $\\mathbf{r}_{new} = \\mathbf{r}_0 + \\mathbf{m}$."},{"text":"$$$\\mathbf{r}_{new} = \\mathbf{m} - \\mathbf{r}_0$$","correct":false,"explanation":"This reverses the order of the subtraction and does not represent adding the displacement to the starting position. The correct position is found by adding $\\mathbf{m}$ to $\\mathbf{r}_0$."},{"text":"$$$\\mathbf{r}_{new} = \\mathbf{m}$$","correct":false,"explanation":"The starting position has been omitted, leaving only the movement vector. The new position must include both the original position and the displacement, added together."}]},{"id":"1e7fc1c9-22f3-4c3d-9a42-b6431b2eea95","cardType":"cloze","points":null,"front":"A bearing between $180^{\\circ}$ and $270^{\\circ}$ corresponds to a velocity vector with a [ ] $\\mathbf{i}$ component and a negative $\\mathbf{j}$ component.","back":"A bearing between $180^{\\circ}$ and $270^{\\circ}$ corresponds to a velocity vector with a negative $\\mathbf{i}$ component and a negative $\\mathbf{j}$ component.","mcqOptions":[{"text":"negative","correct":true,"explanation":"Bearings from $180^{\\circ}$ to $270^{\\circ}$ point towards the west, so the horizontal or $\\mathbf{i}$ component is negative."},{"text":"positive","correct":false,"explanation":"A positive $\\mathbf{i}$ component points east, placing the vector in the eastern half-plane rather than between bearings $180^{\\circ}$ and $270^{\\circ}$. The correct component is negative."},{"text":"zero","correct":false,"explanation":"A zero $\\mathbf{i}$ component would point directly north or south, not into the south-west quadrant covered by these bearings. The correct $\\mathbf{i}$ component is negative."},{"text":"variable","correct":false,"explanation":"A changing $\\mathbf{i}$ component is not a directional sign and cannot describe the westward component of a fixed velocity vector. The correct term is negative."}]},{"id":"f475f0c1-7911-4b22-9ead-0ecbc6aac504","cardType":"cloze","points":null,"front":"A bearing between $180^{\\circ}$ and $270^{\\circ}$ corresponds to a velocity vector with a negative $\\mathbf{i}$ component and a [ ] $\\mathbf{j}$ component.","back":"A bearing between $180^{\\circ}$ and $270^{\\circ}$ corresponds to a velocity vector with a negative $\\mathbf{i}$ component and a negative $\\mathbf{j}$ component.","mcqOptions":[{"text":"negative","correct":true,"explanation":"Bearings from $180^{\\circ}$ to $270^{\\circ}$ point towards the south, so the vertical or $\\mathbf{j}$ component is negative."},{"text":"positive","correct":false,"explanation":"A positive $\\mathbf{j}$ component points north, placing the vector outside the south-west quadrant. Between bearings $180^{\\circ}$ and $270^{\\circ}$, the $\\mathbf{j}$ component is negative."},{"text":"zero","correct":false,"explanation":"A zero $\\mathbf{j}$ component would indicate purely horizontal motion, corresponding to an east or west direction rather than this range of south-west bearings."},{"text":"variable","correct":false,"explanation":"A changing $\\mathbf{j}$ component does not specify a fixed direction or its sign. The southward direction in this bearing range requires a negative $\\mathbf{j}$ component."}]},{"id":"17180196-1ddd-4ee6-a838-9f1800b5f32b","cardType":"basic","points":2,"front":"How do you find unknown components $x$ and $y$ of vector $\\mathbf{b}$ if the resultant $\\mathbf{a} + \\mathbf{b}$ is known?","back":"Set up and solve separate equations for the $\\mathbf{i}$ and $\\mathbf{j}$ components.","mcqOptions":[{"text":"Set up and solve separate equations for the $\\mathbf{i}$ and $\\mathbf{j}$ components.","correct":true,"explanation":"Vector addition acts component by component, so equating the known resultant's $\\mathbf{i}$ and $\\mathbf{j}$ components separately provides two equations for $x$ and $y$."},{"text":"Set up and solve one combined equation for the $\\mathbf{i}$ and $\\mathbf{j}$ components.","correct":false,"explanation":"One combined equation loses the independent horizontal and vertical information needed to determine both unknowns. The resultant must be matched using separate equations for the $\\mathbf{i}$ and $\\mathbf{j}$ components."},{"text":"Set up and solve one equation for the $\\mathbf{i}$ component.","correct":false,"explanation":"Comparing only the $\\mathbf{i}$ components cannot determine the unknown $\\mathbf{j}$ component. Both directions must be handled independently using separate component equations."},{"text":"Set up and solve one equation for the $\\mathbf{j}$ component.","correct":false,"explanation":"Comparing only the $\\mathbf{j}$ components cannot determine the unknown $\\mathbf{i}$ component. Both directions must be handled independently using separate component equations."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"Vectors","title":"Vectors Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"8ca17b28-92ab-4be8-9426-0d87b763783a","walled":"$L38"}],"$L39","$L3a"]