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GAM · Vector Calculations — Strategy & Tricks
⬇ Download this guide as a PDF📐 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.
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
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.
👀 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… | Variant | Operation | Budget |
|---|---|---|---|
| net, resultant, total, ground velocity | Resultant | Add the columns (subtract for ‘correction’ or ‘return’) | 20–30 s |
| how far, length, distance, |v|, speed | Magnitude | Square, add, root — look for a triple first | 20–40 s |
| angle between, acute or obtuse, cos θ | Angle | Dot ÷ |a||b|, then read cos from memory | 45–60 s |
| work done, component along, projection | Projection / work | Dot with the direction (unit vector for ‘component along’) | 40–60 s |
| perpendicular, orthogonal, normal to | Perpendicularity | Is the dot product zero? | 15–25 s |
| doubled, halved, scaled by k, reversed, unit vector | Scaling | Multiply 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.
| Row | x | y | z |
|---|---|---|---|
| r | 7 | −2 | 0 |
| u | −3 | 5 | 0 |
| r + u | 4 | 3 | 0 |
| r − u | 10 | −7 | 0 |
| r·u (products) | −21 | −10 | 0 |
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.
- Extract. Copy every vector from the passage into the table before reading the questions. One passage feeds four or five of them.
- Label the operation from the stem and write it as a new row.
- Work the columns one at a time — sign first, then size.
- 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.
- Bound check. At most |a||b|? At least the largest component? Then pick the option.
✂️ Shortcuts that are actually faster
- Pythagorean triples, memorised: 3-4-5, 5-12-13, 8-15-17, 7-24-25 and multiples (6-8-10, 9-12-15). In 3D: (1, 2, 2) → 3, (2, 3, 6) → 7, (1, 4, 8) → 9, (2, 6, 9) → 11.
- The sign of the dot product is free information: positive → acute, zero → right angle, negative → obtuse. Half the angle questions ask no more.
- Cos from memory: 0° → 1, 30° → √3/2, 45° → 1/√2, 60° → 1/2, 90° → 0; sin is the same list backwards. Read it in reverse too: cos θ = 1/2 means 60°.
- Bound before you compute: |a·b| and |a × b| cannot exceed |a||b|. With |a| = 4 and |b| = 5, any option above 20 is dead.
- A magnitude is never smaller than its biggest component: |(3, k, 6)| cannot be 5.
- Square roots multiply cleanly: √2 · √2 = 2, √3 · √12 = 6. Keep them symbolic; approximating √2 ≈ 1.41 mid-way loses the cancellation.
- Parallel by ratio, not by angle: (2, −4, 6) and (−1, 2, −3)? Every component is −2 times the other — parallel and opposite. No cos needed.
- Unit vectors usually hide a triple: (0.6, 0.8), (1/3)(1, 2, 2). If a ‘unit’ vector isn’t unit, make one: divide by the magnitude.
⚡ 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.
- 0–10 s: read the passage once, marking every vector and formula it states.
- 10–40 s: fill the components table, plus a spare row for the obvious next operation.
- 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.
🪤 Seven ways to get it wrong fast
- 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.
- Forgetting the square root. 6² + 8² = 100 is not a length. Watch for ‘100 m’ next to ‘10 m’.
- The component sum posing as a magnitude. (6, 8) is not 14 long, yet x + y appears in nearly every magnitude question.
- A product bigger than |a||b|. sin θ ≤ 1 and cos θ ≤ 1, so neither |a × b| nor |a·b| can exceed |a||b|.
- 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.
- Degrees and radians. π/3 is 60°, cos = 1/2. Read the unit before you read the table.
- Sin for cos. Dot uses cos (the ‘along’ part), cross uses sin (the ‘across’ part). The 30° torque question offers √3/2 as bait.
🧩 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.
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).
- 10 kN
- 14 kN
- 20 kN
- 100 kN
Show the answer and the steps
- Table: x → 9 − 3 = 6; y → 12 − 4 = 8. So N = (6, 8).
- Spot the triple: (6, 8) is 2 × (3, 4), so |N| = 10. No squaring needed.
- 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
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.
- (2, 4, 4)
- (2, 1, 0)
- (1, 1, 1)
- (−2, 4, −4)
Show the answer and the steps
- Order first: d = B − A = (4 − 2, 1 − 5, 3 − (−1)) = (2, −4, 4). Mind the double negative in z.
- Products row per option: (2, 1, 0) → 4 − 4 + 0 = 0. That is the one.
- 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)
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).
- 30°
- 45°
- 60°
- 90°
Show the answer and the steps
- Products row: 1·1 + 1·0 + 0·1 = 1. Positive, so acute — 90° is gone.
- Magnitudes: |u| = |v| = √2. Keep them symbolic: |u||v| = √2 · √2 = 2.
- cos θ = 1/2, so θ = 60°. 45° baits ‘diagonal of a square’; 30° baits reading cos as sin.
Answer: 60°
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.
- 5 N·m
- 5√3 N·m
- 10 N·m
- 40 N·m
Show the answer and the steps
- Bound first: |r||F| = 0.5 × 20 = 10. Anything above 10 is impossible — 40 dies at once, and 10 itself would need θ = 90°.
- Cross product means sin, not cos: sin 30° = 1/2. The 5√3 option is what cos 30° = √3/2 produces.
- 10 × 1/2 = 5 N·m. Sanity: a push mostly along the handle turns it weakly — half the maximum fits.
Answer: 5 N·m
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).
- 20 J
- 100 J
- 117 J
- 125 J
Show the answer and the steps
- Make d concrete: 5 × (0.6, 0.8) = (3, 4); |d| = 5 as promised.
- Bound: |F| = 25 (7-24-25), so |F||d| = 125 — but that needs F parallel to d, and 24/3 ≠ 7/4. Out.
- Columns: 24 × 3 = 72; 7 × 4 = 28; W = 100 J.
- 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
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.
- −4
- −2
- 2
- ±2
Show the answer and the steps
- Length: 9 + k² + 36 = 49, so k² = 4 and k = ±2. Don’t stop — ‘±2’ is the option for people who do.
- Obtuse means cos θ < 0, so w·c < 0. Products row: 3 + 2k − 6 = 2k − 3.
- k = 2 gives +1 (acute, rejected); k = −2 gives −7 (obtuse). So k = −2.
- 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.
Read the stem, name the variant, answer inside the time.
- A walker goes (3, 4) km, then (−1, 2) km. Net displacement?12s
Answer
Resultant → add the columns: (2, 6) km
- How far from its start is a drone at (6, 8) m?8s
Answer
Magnitude → 3-4-5 doubled: 10 m
- Are (2, 1) and (1, −2) at right angles?10s
Answer
Dot product → 2 − 2 = 0: yes
- 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
- Work done by F = (5, 2) N over d = (2, 3) m?12s
Answer
Dot product → 10 + 6 = 16 J
- Area of the parallelogram spanned by lengths 3 and 4 at 90°?10s
Answer
Cross-product magnitude → 3 × 4 × sin 90° = 12
- Is (2, −4, 6) parallel to (−1, 2, −3)?12s
Answer
Ratio test → each component is −2 times the other: parallel, opposite
Magnitudes, dot products and trig values without squaring on paper.
- |(5, 12)|6s
Answer
13 (5-12-13)
- |(2, −1, 2)|8s
Answer
3 (4 + 1 + 4 = 9)
- (3, −2)·(4, 6)10s
Answer
12 − 12 = 0 → perpendicular
- |a| = 6, |b| = 5, θ = 30°: |a × b|10s
Answer
6 × 5 × 1/2 = 15
- r = (7, −2), u = (−3, 5): r − u10s
Answer
(10, −7) — not (−10, 7)
- Unit vector along (3, 4)8s
Answer
(0.6, 0.8): divide by 5
Using only magnitudes, decide instantly whether the option is possible.
- |a| = 4, |b| = 5. Could a·b = 24?6s
Answer
No — the bound is 20
- |a| = 4, |b| = 5. Could |a × b| = 20?6s
Answer
Yes, only at θ = 90°
- w = (3, k, 6). Could |w| = 5?8s
Answer
No — shorter than its own component 6
- a·b = −8. Acute or obtuse?6s
Answer
Obtuse — negative
- 10 m, 14 m, 100 m for |(6, 8)|: which two are traps?8s
Answer
14 (component sum), 100 (no root)
- |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.
- 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).
- Q1: r′? Lookup: (4, 3). Five seconds.
- 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.
- Q3: final displacement if c = (2, −1)? r′ + c = (6, 2). Ten seconds.
- Q4: |r′|? (4, 3) → 5 m on sight. A √-something option means the wrong row got squared: |r| = √53, |u| = √34. Five seconds.
- 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.
📈 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.
- Round 1 — 75 s per question. Table, classify, bound check, compute, one written sanity check.
- Round 2 — 60 s. Same routine, sanity check in your head.
- 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
🏆 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.
- Read the passage once; build the table before question 1.
- Copy the passage’s formula exactly — it decides components versus |a||b| cos θ.
- Classify each stem in five seconds.
- Bound-check the options before computing.
- Subtraction order: ‘from A to B’ is B − A.
- Magnitude: root taken? At least the largest component?
- Sign of the dot product → acute or obtuse; sin for cross, cos for dot.
- Never leave a blank.
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.
- Table built before Q1
- Operation named from the stem
- Impossible options crossed out
- Order, root, bound checked
- Answered — never blank
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.