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Size (magnitude)
Direction

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

Size (magnitude)
Direction

","correct":true,"explanation":"A vector is fully defined by its magnitude, which describes its size, and its direction, which describes the way it acts or points."},{"text":"

Mass
Speed

","correct":false,"explanation":"Mass and speed are scalar quantities, not the defining properties of a vector. A vector is defined by its magnitude, or size, and its direction."},{"text":"

Position
Time

","correct":false,"explanation":"Position and time do not define vectors in general. The required properties are magnitude, meaning size, and direction."},{"text":"

Magnitude
Distance

","correct":false,"explanation":"Magnitude and distance are both measures of size, so this omits direction. Vectors require size, or magnitude, together with direction."}]},{"id":"fbf69eeb-26d1-4e35-8f3d-847e82d3dd79","cardType":"cloze","points":null,"front":"A [ ] quantity has size only, whereas a vector has both size and direction.","back":"A scalar quantity has size only, whereas a vector has both size and direction.","mcqOptions":[{"text":"scalar","correct":true,"explanation":"A scalar quantity is described by size only, whereas a vector quantity requires both size, or magnitude, and direction."},{"text":"vector","correct":false,"explanation":"A vector quantity has both magnitude and direction, so it cannot describe a quantity with size only. The correct term for size alone is scalar."},{"text":"displacement","correct":false,"explanation":"A displacement has magnitude and direction, making it a vector rather than a size-only quantity. The blank requires the term scalar."},{"text":"magnitude","correct":false,"explanation":"A magnitude describes the size of a quantity but is not the classification of a size-only quantity. The correct classification is scalar."}]},{"id":"cc896015-5e70-4247-b47b-fb8e2f100fa0","cardType":"cloze","points":null,"front":"The unit vectors $\\text{[...]}$, $\\mathbf{j}$, and $\\mathbf{k}$ represent the positive $x$, $y$, and $z$ directions.","back":"The unit vectors $\\boxed{\\mathbf{i}}$, $\\mathbf{j}$, and $\\mathbf{k}$ represent the positive $x$, $y$, and $z$ directions.","mcqOptions":[{"text":"$$\\mathbf{i}$","correct":true,"explanation":"The unit vector $\\mathbf{i}$ represents one unit in the positive $x$ direction; $\\mathbf{j}$ and $\\mathbf{k}$ represent the positive $y$ and $z$ directions."},{"text":"$$\\mathbf{j}$","correct":false,"explanation":"The unit vector $\\mathbf{j}$ represents the positive $y$ direction, not the positive $x$ direction. The $x$ direction is represented by $\\mathbf{i}$."},{"text":"$$\\mathbf{k}$","correct":false,"explanation":"The unit vector $\\mathbf{k}$ represents the positive $z$ direction, not the positive $x$ direction. The $x$ direction is represented by $\\mathbf{i}$."},{"text":"$$-\\mathbf{i}$","correct":false,"explanation":"The negative unit vector $-\\mathbf{i}$ points in the negative $x$ direction. The positive $x$ direction is represented by $\\mathbf{i}$."}]},{"id":"84535105-350e-483e-929e-818799fc6929","cardType":"cloze","points":null,"front":"The unit vectors $\\mathbf{i}$, $\\text{[...]}$, and $\\mathbf{k}$ represent the positive $x$, $y$, and $z$ directions.","back":"The unit vectors $\\mathbf{i}$, $\\boxed{\\mathbf{j}}$, and $\\mathbf{k}$ represent the positive $x$, $y$, and $z$ directions.","mcqOptions":[{"text":"$$\\mathbf{j}$","correct":true,"explanation":"The unit vector $\\mathbf{j}$ represents one unit in the positive $y$ direction; $\\mathbf{i}$ and $\\mathbf{k}$ represent the positive $x$ and $z$ directions."},{"text":"$$\\mathbf{i}$","correct":false,"explanation":"The unit vector $\\mathbf{i}$ represents the positive $x$ direction, not the positive $y$ direction. The $y$ direction is represented by $\\mathbf{j}$."},{"text":"$$\\mathbf{k}$","correct":false,"explanation":"The unit vector $\\mathbf{k}$ represents the positive $z$ direction, not the positive $y$ direction. The $y$ direction is represented by $\\mathbf{j}$."},{"text":"$$-\\mathbf{j}$","correct":false,"explanation":"The negative unit vector $-\\mathbf{j}$ points in the negative $y$ direction. The positive $y$ direction is represented by $\\mathbf{j}$."}]},{"id":"4777d49f-20f3-461f-b31c-d2fdf3c71c20","cardType":"cloze","points":null,"front":"The unit vectors $\\mathbf{i}$, $\\mathbf{j}$, and $\\text{[...]}$ represent the positive $x$, $y$, and $z$ directions.","back":"The unit vectors $\\mathbf{i}$, $\\mathbf{j}$, and $\\boxed{\\mathbf{k}}$ represent the positive $x$, $y$, and $z$ directions.","mcqOptions":[{"text":"$$\\mathbf{k}$","correct":true,"explanation":"The unit vector $\\mathbf{k}$ represents one unit in the positive $z$ direction; $\\mathbf{i}$ and $\\mathbf{j}$ represent the positive $x$ and $y$ directions."},{"text":"$$\\mathbf{i}$","correct":false,"explanation":"The unit vector $\\mathbf{i}$ represents the positive $x$ direction, not the positive $z$ direction. The $z$ direction is represented by $\\mathbf{k}$."},{"text":"$$\\mathbf{j}$","correct":false,"explanation":"The unit vector $\\mathbf{j}$ represents the positive $y$ direction, not the positive $z$ direction. The $z$ direction is represented by $\\mathbf{k}$."},{"text":"$$-\\mathbf{k}$","correct":false,"explanation":"The negative unit vector $-\\mathbf{k}$ points in the negative $z$ direction. The positive $z$ direction is represented by $\\mathbf{k}$."}]},{"id":"64be958a-2c09-46a4-8ba1-30890caf2a75","cardType":"basic","points":1,"front":"How do you calculate the resultant of several 3D vectors?","back":"Add the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components separately.","mcqOptions":[{"text":"Add the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components separately.","correct":true,"explanation":"Adding corresponding components separately preserves their directions: sum all the $\\mathbf{i}$ components, all the $\\mathbf{j}$ components, and all the $\\mathbf{k}$ components."},{"text":"Multiply the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components separately.","correct":false,"explanation":"Multiplying components combines magnitudes incorrectly and does not produce vector addition. The resultant is found by adding the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components separately."},{"text":"Add the vector magnitudes separately.","correct":false,"explanation":"Adding only magnitudes loses directional information and cannot determine a 3D resultant. Add the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components separately."},{"text":"Add the $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ components together.","correct":false,"explanation":"Adding vectors in a single combined step without matching components can mix perpendicular directions. The correct method is to add each $\\mathbf{i}$, $\\mathbf{j}$, and $\\mathbf{k}$ component separately."}]},{"id":"2a258de0-2941-4df1-8611-768de658a747","cardType":"basic","points":1,"front":"What is the physical meaning of a position vector?","back":"A vector that gives the position of a point relative to a fixed origin $O$.","mcqOptions":[{"text":"A vector that gives the position of a point relative to a fixed origin $O$.","correct":true,"explanation":"A position vector is drawn from the fixed origin $O$ to the point, so it specifies that point’s position relative to the chosen origin."},{"text":"A vector that gives the position of a point relative to a moving point $O$.","correct":false,"explanation":"A position vector is measured from a fixed origin, not from any moving point. It gives the position of a point relative to the fixed origin $O$."},{"text":"A vector that gives the distance travelled by a point relative to a fixed origin $O$.","correct":false,"explanation":"A position vector describes location, not the distance travelled over time. It gives a point’s position relative to a fixed origin $O$."},{"text":"A vector that gives the position of a point relative to the point from a fixed origin $O$.","correct":false,"explanation":"A position vector points from the origin to the point, rather than from the point to the origin. It gives the point’s position relative to the fixed origin $O$."}]},{"id":"112060e8-fdb5-4f3b-96e5-0256092ce485","cardType":"cloze","points":null,"front":"If point $A$ has position vector $\\mathbf{a}$ and point $B$ has position vector $\\mathbf{b}$, the displacement vector $\\vec{AB}$ is $\\text{[...]}$.","back":"If point $A$ has position vector $\\mathbf{a}$ and point $B$ has position vector $\\mathbf{b}$, the displacement vector $\\vec{AB}$ is $\\boxed{\\mathbf{b}-\\mathbf{a}}$.","mcqOptions":[{"text":"$$\\mathbf{b}-\\mathbf{a}$","correct":true,"explanation":"The displacement from $A$ to $B$ is found by subtracting the position vector of the starting point from that of the finishing point, giving $\\mathbf{b}-\\mathbf{a}$."},{"text":"$$\\mathbf{b}+\\mathbf{a}$","correct":false,"explanation":"Adding the two position vectors does not give the displacement between the points. Displacement from $A$ to $B$ is finishing position minus starting position: $\\mathbf{b}-\\mathbf{a}$."},{"text":"$$\\mathbf{a}-\\mathbf{b}$","correct":false,"explanation":"The expression $\\mathbf{a}-\\mathbf{b}$ gives the displacement from $B$ to $A$, reversing the required direction. From $A$ to $B$, the answer is $\\mathbf{b}-\\mathbf{a}$."},{"text":"$$-\\mathbf{b}-\\mathbf{a}$","correct":false,"explanation":"Subtracting both position vectors reverses and combines the wrong quantities. The displacement from $A$ to $B$ requires the final position minus the initial position, $\\mathbf{b}-\\mathbf{a}$."}]},{"id":"22821006-3323-4a08-8d55-c8986189ea2e","cardType":"basic","points":1,"front":"What is the formula for the magnitude of vector $\\mathbf{v}=x\\mathbf{i}+y\\mathbf{j}+z\\mathbf{k}$?","back":"$$|\\mathbf{v}|=\\sqrt{x^2+y^2+z^2}$","mcqOptions":[{"text":"$$|\\mathbf{v}|=\\sqrt{x^2+y^2+z^2}$","correct":true,"explanation":"The magnitude follows the three-dimensional Pythagorean theorem: square each Cartesian component, add the squares, and take the positive square root."},{"text":"$$|\\mathbf{v}|=\\sqrt{(x+y+z)^2}$","correct":false,"explanation":"Adding the components before squaring ignores their independent perpendicular directions. The magnitude is $|\\mathbf{v}|=\\sqrt{x^2+y^2+z^2}$ by the three-dimensional Pythagorean theorem."},{"text":"$$|\\mathbf{v}|=x^2+y^2+z^2$","correct":false,"explanation":"Omitting the square root gives the squared magnitude rather than the magnitude itself. The required formula is $|\\mathbf{v}|=\\sqrt{x^2+y^2+z^2}$."},{"text":"$$|\\mathbf{v}|=\\sqrt{x^2+y^2-z^2}$","correct":false,"explanation":"The signs of components do not remain after applying Pythagoras because every component is squared. The correct formula is $|\\mathbf{v}|=\\sqrt{x^2+y^2+z^2}$."}]},{"id":"85bed04f-ace7-4bc7-8e88-655fe9f4322c","cardType":"basic","points":2,"front":"How does the vector $\\vec{BA}$ relate to the vector $\\vec{AB}$?","back":"It has the same length but the opposite direction ($-\\vec{AB}$).","mcqOptions":[{"text":"It has the same length but the opposite direction ($-\\vec{AB}$).","correct":true,"explanation":"Reversing the order of the endpoints reverses a vector’s direction without changing its length. Therefore $\\vec{BA}=-\\vec{AB}$."},{"text":"It has the opposite length but the same direction ($\\vec{AB}$).","correct":false,"explanation":"Reversing the endpoints does not change the distance between them, so the length cannot change. It gives the same length and opposite direction: $\\vec{BA}=-\\vec{AB}$."},{"text":"It has the same length and the same direction ($\\vec{AB}$).","correct":false,"explanation":"The endpoint order determines direction, so $\\vec{BA}$ cannot point the same way as $\\vec{AB}$. It has the same length but opposite direction, giving $-\\vec{AB}$."},{"text":"It has a different length and the opposite direction ($-2\\vec{AB}$).","correct":false,"explanation":"Negating a vector reverses its direction but preserves its magnitude; it does not create a different length. Thus $\\vec{BA}$ has the same length and equals $-\\vec{AB}$."}]},{"id":"9129ff7c-6c50-4532-90bb-31f29a48aed0","cardType":"basic","points":2,"front":"When solving for an unknown component in $|2\\mathbf{i}+m\\mathbf{j}-3\\mathbf{k}|=7$, why might there be two possible values for $m$?","back":"The magnitude formula squares components, so both $m$ and $-m$ can produce the same result.","mcqOptions":[{"text":"The magnitude formula squares components, so both $m$ and $-m$ can produce the same result.","correct":true,"explanation":"The equation contains $m^2$, and positive and negative values with the same magnitude have equal squares. Consequently, both signs can satisfy the magnitude condition."},{"text":"The magnitude formula adds components, so both $m$ and $-m$ can produce the same result.","correct":false,"explanation":"The absolute value of the whole vector is not applied separately to each component. Two solutions arise because the magnitude equation contains $m^2$, so opposite values of $m$ have equal squares."},{"text":"The negative component $-3$ changes sign, so both $m$ and $-m$ can produce the same result.","correct":false,"explanation":"The fixed negative component becomes positive when squared, but that alone does not create two values of $m$. The pair arises because $m^2$ is unchanged when $m$ is negated."},{"text":"The magnitude formula has a square root, so both $m$ and $-m$ can produce the same result.","correct":false,"explanation":"A square root conventionally gives its non-negative principal value, not both signs. The two values occur earlier, when solving an equation involving $m^2$, since $m$ and $-m$ square equally."}]},{"id":"520de47e-bf90-42f3-9fd6-65a9cef0566c","cardType":"cloze","points":null,"front":"The journey rule states that the total journey $\\vec{AC}$ is equal to $\\text{[...]}$.","back":"The journey rule states that the total journey $\\vec{AC}$ is equal to $\\boxed{\\vec{AB}+\\vec{BC}}$.","mcqOptions":[{"text":"$$\\vec{AB}+\\vec{BC}$","correct":true,"explanation":"Travelling from $A$ to $B$ and then from $B$ to $C$ has the same overall displacement as travelling directly from $A$ to $C$."},{"text":"$$\\vec{AB}-\\vec{BC}$","correct":false,"explanation":"Subtracting the second stage does not represent consecutive travel from $A$ through $B$ to $C$. The journey rule is $\\vec{AC}=\\vec{AB}+\\vec{BC}$."},{"text":"$$\\vec{BA}+\\vec{CB}$","correct":false,"explanation":"The vectors $\\vec{BA}$ and $\\vec{CB}$ both point opposite to the required stages, so their sum gives $\\vec{CA}$. The correct journey is $\\vec{AB}+\\vec{BC}$."},{"text":"$$\\vec{AB}+\\vec{CB}$","correct":false,"explanation":"These vectors do not connect head-to-tail: both terminate at $B$. The route from $A$ to $C$ must be written as $\\vec{AB}+\\vec{BC}$."}]},{"id":"0d7b4b2b-ce07-4c67-aa83-49069ff44f52","cardType":"basic","points":1,"front":"If $A$ and $B$ have position vectors $\\mathbf{a}$ and $\\mathbf{b}$, what is the position vector of their midpoint?","back":"$$\\frac{\\mathbf{a}+\\mathbf{b}}{2}$","mcqOptions":[{"text":"$$\\frac{\\mathbf{a}+\\mathbf{b}}{2}$","correct":true,"explanation":"A midpoint lies halfway between the two position vectors, so its vector is their average: add $\\mathbf{a}$ and $\\mathbf{b}$, then divide by $2$."},{"text":"$$\\mathbf{a}+\\frac{\\mathbf{b}}{2}$","correct":false,"explanation":"Dividing only $\\mathbf{b}$ does not average the two endpoints and treats them asymmetrically. The midpoint position vector is $\\frac{\\mathbf{a}+\\mathbf{b}}{2}$."},{"text":"$$\\frac{\\mathbf{b}-\\mathbf{a}}{2}$","correct":false,"explanation":"The difference $\\mathbf{b}-\\mathbf{a}$ is the displacement from $A$ to $B$, so halving it is not generally a position vector. The midpoint is $\\frac{\\mathbf{a}+\\mathbf{b}}{2}$."},{"text":"$$\\mathbf{a}+\\mathbf{b}$","correct":false,"explanation":"The sum points to twice the midpoint position when measured from the origin. Averaging requires division by $2$, giving $\\frac{\\mathbf{a}+\\mathbf{b}}{2}$."}]},{"id":"3966334f-1219-46cd-8515-eaf5d274001b","cardType":"basic","points":1,"front":"What is the condition for two non-zero vectors $\\mathbf{u}$ and $\\mathbf{v}$ to be parallel?","back":"One vector is a scalar multiple of the other ($\\mathbf{u}=\\lambda\\mathbf{v}$).","mcqOptions":[{"text":"One vector is a scalar multiple of the other ($\\mathbf{u}=\\lambda\\mathbf{v}$).","correct":true,"explanation":"Parallel vectors have proportional corresponding components, so one is a scalar multiple of the other. A negative scalar permits them to point in opposite directions."},{"text":"The vectors have a zero scalar product with each other ($\\mathbf{u}\\boldsymbol{\\cdot}\\mathbf{v}=0$).","correct":false,"explanation":"A zero dot product is the condition for perpendicular vectors, not parallel vectors. Parallel non-zero vectors satisfy $\\mathbf{u}=\\lambda\\mathbf{v}$ for some scalar $\\lambda$."},{"text":"The vectors have the same magnitude as each other ($|\\mathbf{u}|=|\\mathbf{v}|$).","correct":false,"explanation":"Equal magnitudes do not ensure equal directions, so vectors of the same length can meet at any angle. Parallelism requires one vector to be a scalar multiple of the other."},{"text":"One vector is exactly equal to the other ($\\mathbf{u}=\\mathbf{v}$).","correct":false,"explanation":"Requiring equality is too restrictive because parallel vectors may have different lengths or opposite directions. The general condition is $\\mathbf{u}=\\lambda\\mathbf{v}$ for a scalar $\\lambda$."}]},{"id":"2bbfe283-3434-44c6-b3dc-b64797023df0","cardType":"cloze","points":null,"front":"A quadrilateral is a parallelogram if both pairs of opposite side vectors are [ ] (or equal).","back":"A quadrilateral is a parallelogram if both pairs of opposite side vectors are parallel (or equal).","mcqOptions":[{"text":"parallel","correct":true,"explanation":"In a parallelogram, each pair of opposite sides has the same direction line, so their vectors are parallel. Equivalently oriented side vectors may also be shown equal."},{"text":"perpendicular","correct":false,"explanation":"Perpendicular adjacent sides identify right angles but do not describe opposite-side relationships, and parallelograms need not contain right angles. Both pairs of opposite side vectors must be parallel."},{"text":"equal in magnitude","correct":false,"explanation":"Equal magnitudes alone establish only equal side lengths and do not ensure the opposite sides have matching directions. A parallelogram requires both pairs of opposite side vectors to be parallel."},{"text":"collinear","correct":false,"explanation":"Opposite sides of a non-degenerate parallelogram lie on distinct parallel lines rather than on one common line. The required vector relationship is parallelism for both opposite pairs."}]},{"id":"19595778-d750-4d46-bc43-3e25857eb8f1","cardType":"basic","points":2,"front":"How do you prove a quadrilateral $OABC$ is a trapezium using vectors?","back":"Show that exactly one pair of opposite sides are scalar multiples of each other.","mcqOptions":[{"text":"Show that exactly one pair of opposite sides are scalar multiples of each other.","correct":true,"explanation":"Under the UK definition used here, a trapezium has exactly one pair of parallel opposite sides. Parallel vectors are scalar multiples, so demonstrating exactly one such pair proves the classification."},{"text":"Show that at least one pair of opposite sides are scalar multiples of each other.","correct":false,"explanation":"Showing at least one parallel pair does not exclude a parallelogram, which has two parallel pairs. To prove a trapezium, show exactly one pair of opposite sides are scalar multiples."},{"text":"Show that both pairs of opposite sides are scalar multiples of each other.","correct":false,"explanation":"Two pairs of opposite sides that are scalar multiples establish two pairs of parallel sides, which characterises a parallelogram. A trapezium requires exactly one parallel pair."},{"text":"Show that exactly one pair of opposite sides have equal magnitudes to each other.","correct":false,"explanation":"Equal lengths do not establish parallel directions, so this test cannot prove a trapezium. The vector test is that exactly one pair of opposite sides are scalar multiples."}]},{"id":"123b03d7-28c4-42a5-b03c-775e3af8a942","cardType":"basic","points":2,"front":"In an isosceles triangle where $AB=AC$, how is the perpendicular height from $A$ found using vectors?","back":"Calculate the vector from $A$ to the midpoint of the base $BC$.","mcqOptions":[{"text":"Calculate the vector from $A$ to the midpoint of the base $BC$.","correct":true,"explanation":"In an isosceles triangle, the perpendicular from the apex $A$ bisects the base $BC$. Therefore, its vector runs from $A$ to the midpoint of $BC$."},{"text":"Calculate the vector from $A$ to the endpoint $B$ of the base $BC$.","correct":false,"explanation":"The vector from $A$ to point $B$ follows an equal side rather than the perpendicular height. The height runs from $A$ to the midpoint of $BC$."},{"text":"Calculate the vector from $A$ to the midpoint of the side $AB$.","correct":false,"explanation":"The midpoint of $AB$ lies on one of the equal sides, so this vector is not the height. The required vector goes from $A$ to the midpoint of $BC$."},{"text":"Calculate the vector from $B$ to the endpoint $C$ of the base $BC$.","correct":false,"explanation":"The vector between the endpoints of $BC$ lies along the base, not perpendicular to it. The height is the vector from $A$ to the midpoint of $BC$."}]},{"id":"69a57f6f-40f7-4dac-9511-b601532fcf94","cardType":"cloze","points":null,"front":"To solve problems involving unknown components and magnitudes, apply the formula then [ ] both sides to eliminate the square root.","back":"To solve problems involving unknown components and magnitudes, apply the formula then square both sides to eliminate the square root.","mcqOptions":[{"text":"square","correct":true,"explanation":"A magnitude formula contains a square root, so squaring both sides removes that root and produces an equation that can be solved for the unknown component."},{"text":"square-root","correct":false,"explanation":"Square-rooting introduces another root rather than eliminating the existing one. The correct operation is to square both sides of the magnitude equation."},{"text":"divide","correct":false,"explanation":"Dividing both sides does not generally remove a square root surrounding a sum of squared components. Squaring both sides eliminates the root correctly."},{"text":"expand","correct":false,"explanation":"Expanding may simplify expressions inside brackets, but it does not undo an outer square root. The required inverse operation is to square both sides."}]},{"id":"bc6c3272-da6b-45f2-9b17-8d136dc963f3","cardType":"basic","points":1,"front":"Express the vector $3\\mathbf{i}-2\\mathbf{j}+5\\mathbf{k}$ as a column vector.","back":"$$$\\begin{pmatrix}3\\\\-2\\\\5\\end{pmatrix}$$","mcqOptions":[{"text":"$$$\\begin{pmatrix}3\\\\-2\\\\5\\end{pmatrix}$$","correct":true,"explanation":"The coefficients of $mathbf{i}$, $mathbf{j}$ and $mathbf{k}$ become the first, second and third entries respectively, including the negative sign on the second component."},{"text":"$$$begin{pmatrix}325end{pmatrix}$$","correct":false,"explanation":"The second entry must retain the negative sign from $-2mathbf{j}$. The correct column vector is $begin{pmatrix}3-25end{pmatrix}$."},{"text":"$$$begin{pmatrix}35-2end{pmatrix}$$","correct":false,"explanation":"The entries must follow the order $mathbf{i}$, $mathbf{j}$, $mathbf{k}$, not place the $mathbf{k}$-component second. The correct entries are $3,-2,5$."},{"text":"$$$begin{pmatrix}-32-5end{pmatrix}$$","correct":false,"explanation":"Negating every component gives the opposite vector, not the given vector. The coefficients must be copied directly, producing entries $3,-2,5$."}]},{"id":"a520b172-07b0-4f81-80e3-3e7de71e0165","cardType":"basic","points":1,"front":"If $\\vec{AB}=\\vec{BD}$, how is the position vector $\\mathbf{d}$ calculated from point $B$?","back":"Add vector $\\vec{AB}$ to the position vector of $B$ ($\\mathbf{d}=\\mathbf{b}+\\vec{AB}$).","mcqOptions":[{"text":"Add vector $\\vec{AB}$ to the position vector of $B$ ($\\mathbf{d}=\\mathbf{b}+\\vec{AB}$).","correct":true,"explanation":"Starting at the origin, travel to $B$ using $mathbf{b}$ and then from $B$ to $D$ using $vec{AB}$. Vector addition therefore gives $mathbf{d}=mathbf{b}+vec{AB}$."},{"text":"Subtract vector $vec{AB}$ from the position vector of $B$ ($mathbf{d}=mathbf{b}-vec{AB}$).","correct":false,"explanation":"Subtracting $vec{AB}$ moves from $B$ in the direction opposite to $vec{BD}$. Since $vec{BD}=vec{AB}$, the correct calculation is $mathbf{d}=mathbf{b}+vec{AB}$."},{"text":"Add the position vector of $A$ to the position vector of $B$ ($mathbf{d}=mathbf{b}+mathbf{a}$).","correct":false,"explanation":"Adding the position vectors of $A$ and $B$ does not represent the displacement from $B$ to $D$. The required relation is $mathbf{d}=mathbf{b}+vec{AB}$."},{"text":"Add the position vector of $B$ to itself ($mathbf{d}=mathbf{b}+mathbf{b}$).","correct":false,"explanation":"Doubling $mathbf{b}$ incorrectly treats $vec{BD}$ as equal to the position vector of $B$. Instead, add the given displacement: $mathbf{d}=mathbf{b}+vec{AB}$."}]}]}],"examCountdown":{"examDate":"2027-05-26","paperLabel":"Paper 1"},"heading":"Vectors","title":"Vectors Flashcards","topbar":{"backHref":"/a-level/maths/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"53f28e45-68bc-4049-aa25-45ef430b2cc3","walled":"$L38"}],"$L39","$L3a"]