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GAM · Vector Calculations — Strategy & Tricks

Every vector question is a small table: fill the rows, work the columns, check the bound, move on. · 13 min read

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📐 The rules the passage assumes you know

A vector is a list of numbers with a direction attached: (x, y) in the plane, (x, y, z) in space. dMAT passages spell out the vectors and restate the formula they want you to use — treat that restatement as a cue. The arithmetic is on you.

Exam facts About 60 questions in 90 minutes — roughly 90 seconds each — one passage shared by four or five questions, four options, no calculator, no negative marking. A blank scores exactly like a wrong answer, so you always answer.
a + b = (a₁+b₁, a₂+b₂, a₃+b₃) — add column by column
a − b = (a₁−b₁, a₂−b₂, a₃−b₃) — from A to B is B − A
k·a = (ka₁, ka₂, ka₃) — scaling; k < 0 flips the direction
|a| = √(a₁² + a₂² + a₃²) — length; never the component sum
a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ — a number, not a vector
|a × b| = |a||b| sin θ — parallelogram area; a × b ⊥ both
û = a/|a| — same direction, length 1
cos θ = a·b / (|a||b|)
a ⊥ b ⇔ a·b = 0;  a ∥ b ⇔ b = k·a — one ratio, every component
The entire toolbox: nine lines. Everything else is arithmetic.

Two facts do most of the heavy lifting. First, the dot product is a scalar: if the stem asks for it, every vector option is dead before you compute. Second, both products are bounded — neither |a·b| nor |a × b| can exceed |a||b|, because cos and sin never exceed 1. No other check kills as many distractors.

The passage is the law If the passage defines the dot product component-wise, use components. If it hands you |a|, |b| and θ, use |a||b| cos θ. Convert only when the question forces it — every conversion costs twenty seconds. Anything exotic will be defined in the passage; apply that definition as written.

👀 Name the variant in five seconds

Before you touch a number, classify the question. Six variants cover almost everything, each with exactly one operation attached.

If the stem says…VariantOperationBudget
net, resultant, total, ground velocityResultantAdd the columns (subtract for ‘correction’ or ‘return’)20–30 s
how far, length, distance, |v|, speedMagnitudeSquare, add, root — look for a triple first20–40 s
angle between, acute or obtuse, cos θAngleDot ÷ |a||b|, then read cos from memory45–60 s
work done, component along, projectionProjection / workDot with the direction (unit vector for ‘component along’)40–60 s
perpendicular, orthogonal, normal toPerpendicularityIs the dot product zero?15–25 s
doubled, halved, scaled by k, reversed, unit vectorScalingMultiply every component by k; magnitude scales by |k|10–20 s

Two extra labels: area, torque, ‘perpendicular to both’ means cross product, usually just |a||b| sin θ; ‘find k such that…’ means turning the stated condition into one equation. If a stem fits no row, re-read the passage — it has defined the operation for you.

🧠 The components table

Every question here is solved the same way: write the vectors as rows, one column per axis, and work down the columns. It looks childish. It is also why you stop dropping minus signs.

Rowxyz
r7−20
u−350
r + u430
r − u10−70
r·u (products)−21−100

Sum the products row: r·u = −31, negative, so the angle between r and u is obtuse. |r + u| = √(16 + 9) = 5, a 3-4-5 triple — no squaring needed. One table, four answers.

  1. Extract. Copy every vector from the passage into the table before reading the questions. One passage feeds four or five of them.
  2. Label the operation from the stem and write it as a new row.
  3. Work the columns one at a time — sign first, then size.
  4. Collapse. Magnitude squares-and-adds: |a| = √(x² + y² + z²). Dot multiplies-and-adds: a·b = a₁b₁ + a₂b₂ + a₃b₃, and cos θ = a·b ÷ (|a||b|). Each turns a row into one number.
  5. Bound check. At most |a||b|? At least the largest component? Then pick the option.
One table per passage, not per question The next question almost always reuses a row you already have: r + u, then |r + u|, then the correction back. Write the rows once and later questions are ten-second lookups.

✂️ Shortcuts that are actually faster

⚡ The 90-second budget

Ninety seconds a question, but a vector pack is four or five questions on one passage — think six or seven minutes for the pack. Spend the first 40 seconds on the table and each question then costs 20–60 seconds.

  1. 0–10 s: read the passage once, marking every vector and formula it states.
  2. 10–40 s: fill the components table, plus a spare row for the obvious next operation.
  3. Per question: 5 s to classify the variant, 10 s to bound-check the options, the rest computing down the columns, 5 s to sanity-check — then commit.
Where the seconds actually go Not into arithmetic. Into re-reading the passage a third time, and squaring numbers you never needed to because (6, 8) was sitting right there.

🪤 Seven ways to get it wrong fast

  1. Wrong subtraction order. ‘From A to B’ is B − A; the correction returning from r′ to r is r − r′ = −u. An answer with every sign flipped means you subtracted backwards — and that option is always on the list.
  2. Forgetting the square root. 6² + 8² = 100 is not a length. Watch for ‘100 m’ next to ‘10 m’.
  3. The component sum posing as a magnitude. (6, 8) is not 14 long, yet x + y appears in nearly every magnitude question.
  4. A product bigger than |a||b|. sin θ ≤ 1 and cos θ ≤ 1, so neither |a × b| nor |a·b| can exceed |a||b|.
  5. Adding magnitudes instead of vectors. |(9, 12)| + |(−3, −4)| = 20, but the sum (6, 8) is only 10 long. Lengths add only when both vectors point the same way.
  6. Degrees and radians. π/3 is 60°, cos = 1/2. Read the unit before you read the table.
  7. Sin for cos. Dot uses cos (the ‘along’ part), cross uses sin (the ‘across’ part). The 30° torque question offers √3/2 as bait.
The universal antidote After every answer, ask: right order? root taken? below the bound? Five seconds, and most careless errors die there.

🧩 Worked examples, easy to hard

Each example uses a mini passage in the exam’s style. Cover the steps, attempt it inside the time target, then compare.

easyNet force and its magnitudetarget 45s

What is the magnitude of the net force N?

Two harbour tugs push a barge. Along two perpendicular horizontal axes, the first applies P = (9, 12) kN and the second Q = (−3, −4) kN. The net force is N = P + Q, and the magnitude of (a, b) is √(a² + b²).
  1. 10 kN
  2. 14 kN
  3. 20 kN
  4. 100 kN
Show the answer and the steps
  1. Table: x → 9 − 3 = 6; y → 12 − 4 = 8. So N = (6, 8).
  2. Spot the triple: (6, 8) is 2 × (3, 4), so |N| = 10. No squaring needed.
  3. Distractors: 14 is the component sum, 100 forgot the root, 20 is |P| + |Q| = 15 + 5. Lengths add only when both point the same way; Q points backwards, so they subtract: 15 − 5 = 10, matching the columns.

Answer: 10 kN

easyPerpendicular to a displacementtarget 50s

Which vector is perpendicular to d?

A robot arm’s tip moves from A = (2, 5, −1) to B = (4, 1, 3), in centimetres, so the displacement is d = B − A. The dot product of (a₁, a₂, a₃) and (b₁, b₂, b₃) is a₁b₁ + a₂b₂ + a₃b₃, and two vectors are perpendicular exactly when it is zero.
  1. (2, 4, 4)
  2. (2, 1, 0)
  3. (1, 1, 1)
  4. (−2, 4, −4)
Show the answer and the steps
  1. Order first: d = B − A = (4 − 2, 1 − 5, 3 − (−1)) = (2, −4, 4). Mind the double negative in z.
  2. Products row per option: (2, 1, 0) → 4 − 4 + 0 = 0. That is the one.
  3. Check the rest: (2, 4, 4) → 4 − 16 + 16 = 4; (1, 1, 1) → 2; (−2, 4, −4) is −d — backwards is 180°, not 90°.

Answer: (2, 1, 0)

mediumAngle between two bondstarget 60s

What is the angle between the two bonds?

In a crystal model, two bonds leave the same atom along u = (1, 1, 0) and v = (1, 0, 1). For any two vectors, u·v = u₁v₁ + u₂v₂ + u₃v₃, and also u·v = |u||v| cos θ, where θ is the angle between them. |(x, y, z)| = √(x² + y² + z²).
  1. 30°
  2. 45°
  3. 60°
  4. 90°
Show the answer and the steps
  1. Products row: 1·1 + 1·0 + 0·1 = 1. Positive, so acute — 90° is gone.
  2. Magnitudes: |u| = |v| = √2. Keep them symbolic: |u||v| = √2 · √2 = 2.
  3. cos θ = 1/2, so θ = 60°. 45° baits ‘diagonal of a square’; 30° baits reading cos as sin.

Answer: 60°

mediumTorque with a bound checktarget 50s

What is the magnitude of the turning effect?

A technician pushes a valve handle. The handle’s position vector r from the pivot has magnitude |r| = 0.5 m, the push F has magnitude |F| = 20 N, and the angle between r and F is 30°. The turning effect is r × F, with magnitude |r||F| sin θ.
  1. 5 N·m
  2. 5√3 N·m
  3. 10 N·m
  4. 40 N·m
Show the answer and the steps
  1. Bound first: |r||F| = 0.5 × 20 = 10. Anything above 10 is impossible — 40 dies at once, and 10 itself would need θ = 90°.
  2. Cross product means sin, not cos: sin 30° = 1/2. The 5√3 option is what cos 30° = √3/2 produces.
  3. 10 × 1/2 = 5 N·m. Sanity: a push mostly along the handle turns it weakly — half the maximum fits.

Answer: 5 N·m

hardWork along a ramptarget 75s

How much work does the rope do?

A crate is hauled 5 m up a ramp whose direction is the unit vector û = (0.6, 0.8), so the displacement is d = 5û. The rope applies a constant force F = (24, 7) N. Work done is W = F·d with the dot product taken component-wise; the scalar projection of F onto û is F·û. The magnitude of (a, b) is √(a² + b²).
  1. 20 J
  2. 100 J
  3. 117 J
  4. 125 J
Show the answer and the steps
  1. Make d concrete: 5 × (0.6, 0.8) = (3, 4); |d| = 5 as promised.
  2. Bound: |F| = 25 (7-24-25), so |F||d| = 125 — but that needs F parallel to d, and 24/3 ≠ 7/4. Out.
  3. Columns: 24 × 3 = 72; 7 × 4 = 28; W = 100 J.
  4. The rest: 20 is F·û, the projection in newtons, still missing the 5 m; 117 crosses the columns (24 × 4 + 7 × 3).

Answer: 100 J

hardUnknown component under two conditionstarget 90s

What is the value of k?

A stage-rigging cable runs along w = (3, k, 6) in metres, with its length fixed at |w| = 7 m. It must also make an obtuse angle with the reference direction c = (1, 2, −1). Recall |(x, y, z)| = √(x² + y² + z²) and w·c = w₁c₁ + w₂c₂ + w₃c₃ = |w||c| cos θ, so w·c has the sign of cos θ.
  1. −4
  2. −2
  3. 2
  4. ±2
Show the answer and the steps
  1. Length: 9 + k² + 36 = 49, so k² = 4 and k = ±2. Don’t stop — ‘±2’ is the option for people who do.
  2. Obtuse means cos θ < 0, so w·c < 0. Products row: 3 + 2k − 6 = 2k − 3.
  3. k = 2 gives +1 (acute, rejected); k = −2 gives −7 (obtuse). So k = −2.
  4. Verify: w = (3, −2, 6), |w| = √(9 + 4 + 36) = 7. The −4 option skips the root on k² = 4 and borrows the minus from ‘obtuse’.

Answer: −2

Go Zero to dMAT

Strategy is the map; volume under the clock is the journey — 2,858 questions and 15 timed papers, the reasoning drawn on every one.

Start practising — free during launch (₹799 → ₹0)

🎯 Recognition drills

Say the operation out loud, then the answer. No paper on the first pass — the point is to make classification reflexive.

Which operation?

Read the stem, name the variant, answer inside the time.

  1. A walker goes (3, 4) km, then (−1, 2) km. Net displacement?12s
    Answer

    Resultant → add the columns: (2, 6) km

  2. How far from its start is a drone at (6, 8) m?8s
    Answer

    Magnitude → 3-4-5 doubled: 10 m

  3. Are (2, 1) and (1, −2) at right angles?10s
    Answer

    Dot product → 2 − 2 = 0: yes

  4. A rod along (1, −2, 2) is stretched to three times its length. New vector?12s
    Answer

    Scalar multiple → (3, −6, 6); length 3 → 9

  5. Work done by F = (5, 2) N over d = (2, 3) m?12s
    Answer

    Dot product → 10 + 6 = 16 J

  6. Area of the parallelogram spanned by lengths 3 and 4 at 90°?10s
    Answer

    Cross-product magnitude → 3 × 4 × sin 90° = 12

  7. Is (2, −4, 6) parallel to (−1, 2, −3)?12s
    Answer

    Ratio test → each component is −2 times the other: parallel, opposite

Numbers from memory

Magnitudes, dot products and trig values without squaring on paper.

  1. |(5, 12)|6s
    Answer

    13 (5-12-13)

  2. |(2, −1, 2)|8s
    Answer

    3 (4 + 1 + 4 = 9)

  3. (3, −2)·(4, 6)10s
    Answer

    12 − 12 = 0 → perpendicular

  4. |a| = 6, |b| = 5, θ = 30°: |a × b|10s
    Answer

    6 × 5 × 1/2 = 15

  5. r = (7, −2), u = (−3, 5): r − u10s
    Answer

    (10, −7) — not (−10, 7)

  6. Unit vector along (3, 4)8s
    Answer

    (0.6, 0.8): divide by 5

Bound check

Using only magnitudes, decide instantly whether the option is possible.

  1. |a| = 4, |b| = 5. Could a·b = 24?6s
    Answer

    No — the bound is 20

  2. |a| = 4, |b| = 5. Could |a × b| = 20?6s
    Answer

    Yes, only at θ = 90°

  3. w = (3, k, 6). Could |w| = 5?8s
    Answer

    No — shorter than its own component 6

  4. a·b = −8. Acute or obtuse?6s
    Answer

    Obtuse — negative

  5. 10 m, 14 m, 100 m for |(6, 8)|: which two are traps?8s
    Answer

    14 (component sum), 100 (no root)

  6. |a + b| with |a| = 3, |b| = 4, θ = 90°?8s
    Answer

    5 by Pythagoras; 7 only if they point the same way

🧭 Guided walk-through: one pack, five questions

Here is a full pack the way it lands in the exam, with the thinking that should run at each moment.

A warehouse robot has moved r = (7, −2) m from its dock. Its next scheduled move is u = (−3, 5) m, giving r′ = r + u. A correction c may be added afterwards. |(x, y)| = √(x² + y²), and for two vectors a·b = a₁b₁ + a₂b₂ = |a||b| cos θ.
One passage, five linked questions — the standard pack format.
  1. Seconds 0–30, the table. Rows r = (7, −2) and u = (−3, 5). Pre-compute r + u = (4, 3) and, because ‘correction’ is in the passage, −u = (3, −5).
  2. Q1: r′? Lookup: (4, 3). Five seconds.
  3. Q2: the correction returning the robot from r′ to r? r − r′ = −u = (3, −5). The (−3, 5) option is the wrong-order trap. Ten seconds.
  4. Q3: final displacement if c = (2, −1)? r′ + c = (6, 2). Ten seconds.
  5. Q4: |r′|? (4, 3) → 5 m on sight. A √-something option means the wrong row got squared: |r| = √53, |u| = √34. Five seconds.
  6. Q5: angle between r′ and u, acute or obtuse? 4·(−3) + 3·5 = 3 > 0, so acute. Fifteen seconds.

Five questions, about 75 seconds of computing, most of it in the table. Front-load, then harvest.

Glance at the options first In linked questions, read the options before computing. If three are vectors and one is a number, or the reverse, the stem’s operation says which kind can survive — a free elimination.

📈 Progressive practice

Untimed. The goal is zero careless errors, not speed. After each set, sort every mistake into one of the seven traps — the sorting is the practice.

Easy · resultants, magnitudes, perpendicularity → 6 real easy questions · free, no login

Medium · angles, cross-product magnitudes, unknown components → 6 real medium questions · free, no login

Hard · projections, two-condition constraints, 3D packs → 6 real hard questions · free, no login

Move up a tier when you score five of six with the three-question check applied.

⏱️ Speed training

Now the clock: three rounds, the target dropping each time. Write the table every round, even when you think you don’t need it — the habit is what gets faster, not your arithmetic.

  1. Round 1 — 75 s per question. Table, classify, bound check, compute, one written sanity check.
  2. Round 2 — 60 s. Same routine, sanity check in your head.
  3. Round 3 — 45 s. Skip the squaring whenever a triple is visible; guess and move at 45 s.

Timed · medium, 60 s each → 6 real medium questions, against the clock · free, no login

Timed · hard, 75 s each → 6 real hard questions, against the clock · free, no login

The time target is a ceiling At 90 seconds and not converging? Take the option that survives the bound and sign checks, and move. Vector packs are where you gain time for the rest of the GAM, not where you spend it.

🏆 On exam day

Vector packs arrive as one passage plus four or five linked questions — the most predictable points in the GAM, if you run the routine.

Exam simulation · hard passage questions, 90 s each → 6 real hard questions, against the clock · free, no login

Linked questions share a passage, not their answers. If Q3 feels wrong, don’t rebuild Q1 — re-check the single row you used.

📋 Cheat sheet

Add / subtract column by column. From A to B = B − A.
Scale: k·(x, y, z) = (kx, ky, kz); |k·a| = |k||a|.
Magnitude √(x² + y² + z²): take the root; never below the largest component.
Triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25; 3D (1,2,2)→3, (2,3,6)→7, (1,4,8)→9.
Dot: a·b = Σaᵢbᵢ = |a||b| cos θ — a number; |a·b| ≤ |a||b|.
Angle: cos θ = a·b/(|a||b|); sign → acute (+), right (0), obtuse (−).
cos 0°, 30°, 45°, 60°, 90° = 1, √3/2, 1/√2, 1/2, 0; sin is the list backwards.
Perpendicular ⇔ a·b = 0. Parallel ⇔ b = k·a.
Cross: |a × b| = |a||b| sin θ ≤ |a||b|; ⊥ both; parallelogram area.
Unit vector û = a/|a|. Component along û = a·û. Work = F·d.
Ten lines to reread the night before.

Go Zero to dMAT

Strategy is the map; volume under the clock is the journey — 2,858 questions and 15 timed papers, the reasoning drawn on every one.

Start practising — free during launch (₹799 → ₹0)

Every example here is original, written to the dMAT's mechanisms and designed with the official sample tasks in mind — nothing is copied from g.a.s.t.'s materials. dMAT.practice is an independent study aid by NinjaMinds Software Solutions, not connected to g.a.s.t., APS India or any university; use the official practice materials alongside it.