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MHT-CET Maths · Vectors

Cross Product, Angle, and Area

The vector product whose magnitude is the area of a parallelogram and whose direction is the right-hand-rule perpendicular — the engine behind areas, unit normals, angles, and a whole family of vector-equation problems.

All Vectors notesMHT-CET Maths strategy

Why this matters

At 48 PYQs this is the chapter's biggest subtopic after the scalar triple product, and the toughest — roughly 58% of these are rated HARD. Three themes dominate: AREA (triangle, parallelogram, from diagonals, or from a side-plus-diagonal), the PERPENDICULAR DIRECTION (unit normal, vector of a given magnitude perpendicular to two), and VECTOR EQUATIONS that mix a cross and a dot condition (solve for the unknown vector, find an unknown component, or expand a vector triple product with the BAC-CAB rule). Master the determinant computation and the |a×b| = |a||b|sin θ relation first — every concept below is built on them.

Concept 1 of 12

The cross product — definition and determinant form

Intuition

The cross product of two 3-D vectors produces a THIRD vector that is perpendicular to both, with magnitude equal to the area of the parallelogram they span and direction fixed by the right-hand rule. In practice you almost never use the sin θ form to compute it — you expand a 3x3 determinant whose top row is the unit vectors and whose lower two rows are the components of the two vectors.

Definition

For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}:

  • Geometric form: a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}, where n^\hat{n} is the unit perpendicular by the right-hand rule
  • Determinant form: a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}
  • Anti-commutative: a⃗×b⃗=− b⃗×a⃗\vec{a}\times\vec{b} = -\,\vec{b}\times\vec{a}
  • Self / parallel: a⃗×a⃗=0⃗\vec{a}\times\vec{a} = \vec{0}; and a⃗×b⃗=0⃗  ⟺  a⃗ ∥ b⃗\vec{a}\times\vec{b} = \vec{0} \iff \vec{a}\,\|\,\vec{b} (or one is zero)
  • Standard products: i^×j^=k^,  j^×k^=i^,  k^×i^=j^\hat{i}\times\hat{j} = \hat{k},\; \hat{j}\times\hat{k} = \hat{i},\; \hat{k}\times\hat{i} = \hat{j} (cyclic); reverse any pair and the sign flips

Cross product as a determinant

a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}
  • Top rowthe unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k}
  • θ\thetaangle between a⃗\vec{a} and b⃗\vec{b}, in [0,π][0,\pi]
  • n^\hat{n}unit perpendicular to both, by the right-hand rule

Diagram · drag to rotate a × b

xyaba × bθ

Drag to orbit the scene. However you turn it, a × b stays perpendicular to the plane of a and b, on the side your right-hand fingers (curling a → b) point your thumb. Length is schematic; magnitude is |a||b| sin θ.

Worked example

Compute a⃗×b⃗\vec{a}\times\vec{b} for a⃗=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k} and b⃗=2i^+j^−k^\vec{b} = 2\hat{i} + \hat{j} - \hat{k}.
  1. Set up the determinant: ∣i^j^k^12121−1∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & -1 \end{vmatrix}.
  2. i^\hat{i}-component: (2)(−1)−(1)(1)=−3(2)(-1) - (1)(1) = -3.
  3. j^\hat{j}-component: −[(1)(−1)−(1)(2)]=−(−3)=3-\big[(1)(-1) - (1)(2)\big] = -(-3) = 3.
  4. k^\hat{k}-component: (1)(1)−(2)(2)=−3(1)(1) - (2)(2) = -3.
  5. So a⃗×b⃗=−3i^+3j^−3k^\vec{a}\times\vec{b} = -3\hat{i} + 3\hat{j} - 3\hat{k}.
Answer:a⃗×b⃗=−3i^+3j^−3k^\vec{a}\times\vec{b} = -3\hat{i} + 3\hat{j} - 3\hat{k}
Practice this concept5 quick reps

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    i^×j^=?\hat{i}\times\hat{j} = ?
  2. 2.
    k^×i^=?\hat{k}\times\hat{i} = ?
  3. 3.
    j^×i^=?\hat{j}\times\hat{i} = ?
  4. 4.
    a⃗×a⃗=?\vec{a}\times\vec{a} = ?
  5. 5.
    If a⃗×b⃗=0⃗\vec{a}\times\vec{b} = \vec{0} (both non-zero), the vectors are?

The cross product is a VECTOR, not a scalar

a⃗×b⃗\vec{a}\times\vec{b} has three components — it is a vector. The dot product a⃗⋅b⃗\vec{a}\cdot\vec{b} is the scalar. Mixing them up (e.g. expecting a single number from a cross product) is the most common slip.

Watch the SIGN on the j^\hat{j}-component

Expanding the determinant, the middle (j^\hat{j}) cofactor carries a leading minus: −[a1b3−a3b1]-\big[a_1 b_3 - a_3 b_1\big]. Forgetting this minus is the single most frequent computational error in this whole subtopic.

a⃗×b⃗=0⃗\vec{a}\times\vec{b} = \vec{0} does NOT mean both vectors are zero

It means a⃗\vec{a} and b⃗\vec{b} are parallel — one is a scalar multiple of the other. Combined with a⃗,b⃗≠0⃗\vec{a}, \vec{b} \neq \vec{0}, it gives a⃗=λb⃗\vec{a} = \lambda\vec{b} for some scalar λ\lambda.

Concept 2 of 12

Magnitude of the cross product, angle, and the Lagrange identity

Intuition

The magnitude of the cross product is ∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}||\vec{b}|\sin\theta — a single number you can read off when you know the two magnitudes and the angle between them. Rearranged, it isolates sin⁡θ\sin\theta; paired with the dot product it isolates cos⁡θ\cos\theta. The Lagrange identity bundles both into one equation, letting you cross between cross and dot without ever finding the angle.

Definition

For non-zero a⃗,b⃗\vec{a}, \vec{b} at angle θ\theta:

  • Magnitude: ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta
  • Angle: sin⁡θ=∣a⃗×b⃗∣∣a⃗∣∣b⃗∣\sin\theta = \dfrac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|} and cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}
  • Lagrange identity: ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2

Magnitude and Lagrange

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \qquad |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2
  • sin⁡θ\sin\thetanon-negative for θ∈[0,π]\theta \in [0,\pi] — the magnitude is a length
  • Lagrange identityfrom sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 times ∣a⃗∣2∣b⃗∣2|\vec{a}|^2|\vec{b}|^2

Worked example

If ∣a⃗∣=4|\vec{a}| = 4, ∣b⃗∣=5|\vec{b}| = 5 and the angle between them is 120∘120^\circ, find ∣a⃗×b⃗∣|\vec{a}\times\vec{b}|.
  1. ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡120∘=4⋅5⋅32|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin 120^\circ = 4\cdot 5\cdot\dfrac{\sqrt{3}}{2}.
  2. =20⋅32=103= 20\cdot\dfrac{\sqrt{3}}{2} = 10\sqrt{3}.
Answer:∣a⃗×b⃗∣=103|\vec{a}\times\vec{b}| = 10\sqrt{3}
Practice this conceptself-check · 4 quick reps

Try it yourself

If ∣a⃗∣=6|\vec{a}| = 6, ∣b⃗∣=2|\vec{b}| = 2 and ∣a⃗×b⃗∣=6|\vec{a}\times\vec{b}| = 6, find a⃗⋅b⃗\vec{a}\cdot\vec{b} (take the acute angle).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| for ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=3|\vec{b}| = 3, angle 90∘90^\circ?
  2. 2.
    ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| for ∣a⃗∣=5|\vec{a}| = 5, ∣b⃗∣=4|\vec{b}| = 4, angle 30∘30^\circ?
  3. 3.
    sin⁡θ=?\sin\theta = ? in terms of the cross product.
  4. 4.
    Lagrange: ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=?|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = ?

From the bank · past-year question

Example 2VectorsEASY
If ∣u⃗∣=8|\vec{u}|=8, ∣v⃗∣=12|\vec{v}|=12, and angle between them is 150∘150^\circ, then ∣u⃗×v⃗∣|\vec{u}\times\vec{v}| is

[Q133 · 13th May Shift 1 · 2024]

sin⁡θ\sin\theta is the same for θ\theta and 180∘−θ180^\circ - \theta

A magnitude of 4848 at 150∘150^\circ and at 30∘30^\circ are identical because sin⁡150∘=sin⁡30∘=12\sin 150^\circ = \sin 30^\circ = \tfrac{1}{2}. So ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| alone never fixes the angle — it cannot tell acute from obtuse. The dot product (with its sign) does.

Use Lagrange to skip finding the angle

When a problem gives two of {∣a⃗∣,∣b⃗∣,∣a⃗×b⃗∣,a⃗⋅b⃗}\{|\vec{a}|, |\vec{b}|, |\vec{a}\times\vec{b}|, \vec{a}\cdot\vec{b}\} and asks for a third, ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 gives it directly — no θ\theta needed.
Drill 2 more on magnitude of the cross product, angle, and the lagrange identity

Concept 3 of 12

Area of a triangle from two side vectors

Intuition

Build two edge vectors from a common vertex, cross them, and halve the magnitude — that is the area of the triangle. The half is the whole point: the cross product gives the parallelogram, and a triangle is exactly half of it. When the area is GIVEN and a coordinate is unknown, set up the same equation and solve.

Definition

For a triangle with vertices A,B,CA, B, C, form AB→=B−A\overrightarrow{AB} = B - A and AC→=C−A\overrightarrow{AC} = C - A. Then Area =12 ∣AB→×AC→∣= \dfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|. Equivalently, for a triangle whose two adjacent SIDES are the vectors a⃗\vec{a} and b⃗\vec{b}, the area is 12∣a⃗×b⃗∣\dfrac{1}{2}|\vec{a}\times\vec{b}|.

Triangle area

Area=12 ∣AB→×AC→∣\text{Area} = \tfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|
  • AB→,AC→\overrightarrow{AB}, \overrightarrow{AC}two edge vectors from the SAME vertex AA
  • 12\tfrac{1}{2}a triangle is half the parallelogram on the same two edges

Visualization · the parallelogram area is |a × b|

ab
|a × b| = 19 (parallelogram area)triangle = 9.5direction: out of the page

|a × b| = |a₁b₂ − a₂b₁| is exactly the parallelogram area; the triangle on a and b is half of it. Make a and b parallel and the area — and the cross product — collapse to zero. The fill colour flips with the right-hand-rule direction (out of vs into the page).

Worked example

Find the area of the triangle with vertices A(1,1,1)A(1,1,1), B(2,3,1)B(2,3,1) and C(1,2,3)C(1,2,3).
  1. AB→=B−A=i^+2j^\overrightarrow{AB} = B - A = \hat{i} + 2\hat{j}; AC→=C−A=j^+2k^\overrightarrow{AC} = C - A = \hat{j} + 2\hat{k}.
  2. AB→×AC→=∣i^j^k^120012∣=(4)i^−(2)j^+(1)k^\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{vmatrix} = (4)\hat{i} - (2)\hat{j} + (1)\hat{k}.
  3. Magnitude: 16+4+1=21\sqrt{16 + 4 + 1} = \sqrt{21}.
  4. Area =1221= \tfrac{1}{2}\sqrt{21}.
Answer:Area =212= \dfrac{\sqrt{21}}{2} square units
Practice this conceptself-check · 4 quick reps

Try it yourself

The area of the triangle with vertices (0,0,0)(0,0,0), (2,0,0)(2,0,0) and (0,3,0)(0,3,0) is?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Triangle area with edge vectors a⃗,b⃗\vec{a}, \vec{b}?
  2. 2.
    For vertices A,B,CA, B, C, which two vectors do you cross?
  3. 3.
    If ∣AB→×AC→∣=10|\overrightarrow{AB}\times\overrightarrow{AC}| = 10, the triangle area is?
  4. 4.
    Triangle area with sides i^\hat{i} and j^\hat{j}?

From the bank · past-year question

Example 3VectorsMODERATE
The area of the triangle whose vertices are A(1,−1,2)A(1,-1,2), B(2,1,−1)B(2,1,-1) and C(3,−1,2)C(3,-1,2) is

[Q128 · 2nd May Shift 2 · 2023]

The HALF is on the triangle, not the parallelogram

Triangle area is 12∣a⃗×b⃗∣\tfrac{1}{2}|\vec{a}\times\vec{b}|; parallelogram area is the full ∣a⃗×b⃗∣|\vec{a}\times\vec{b}|. Dropping the 12\tfrac{1}{2} doubles your answer — a classic distractor option.

When area is GIVEN, expect TWO values of the unknown

Setting 12∣AB→×AC→∣=\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| = (given) leads to a quadratic in the unknown coordinate, so it usually has two roots. Pick the one that appears in the options — both may be geometrically valid.

Cross edges from the SAME vertex

Use AB→\overrightarrow{AB} and AC→\overrightarrow{AC} (both start at AA) — not AB→\overrightarrow{AB} and BC→\overrightarrow{BC}. Mixing base points gives a wrong vector and a wrong area.
Drill 2 more on area of a triangle from two side vectors

Concept 4 of 12

Area of a parallelogram — from sides, diagonals, or a side and a diagonal

Intuition

A parallelogram on two adjacent side vectors has area ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| — no half this time. If instead you are handed the two DIAGONALS, the area is HALF the magnitude of their cross product. And if you are given one side plus one diagonal, recover the other side by subtraction, then cross the two sides.

Definition

  • From two adjacent sides a⃗,b⃗\vec{a}, \vec{b}: Area =∣a⃗×b⃗∣= |\vec{a}\times\vec{b}|.
  • From the two diagonals d1⃗,d2⃗\vec{d_1}, \vec{d_2}: Area =12∣d1⃗×d2⃗∣= \dfrac{1}{2}|\vec{d_1}\times\vec{d_2}|.
  • From a side and a diagonal: if a⃗\vec{a} is a side and c⃗\vec{c} the diagonal from the same vertex, the adjacent side is b⃗=c⃗−a⃗\vec{b} = \vec{c} - \vec{a}; then Area =∣a⃗×b⃗∣= |\vec{a}\times\vec{b}|.

Parallelogram areas

Area=∣a⃗×b⃗∣Areadiagonals=12∣d1⃗×d2⃗∣\text{Area} = |\vec{a}\times\vec{b}| \qquad \text{Area}_{\text{diagonals}} = \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|
  • a⃗,b⃗\vec{a}, \vec{b}two adjacent SIDES
  • d1⃗,d2⃗\vec{d_1}, \vec{d_2}two DIAGONALS — note the extra 12\tfrac{1}{2}

Diagram · parallelogram diagonals = a + b and a − b

aba + ba − b

From a shared corner, sides a and b span the parallelogram. The diagonal from that corner is a + b; the diagonal between the side tips is a − b. They bisect each other, and |a + b|² + |a − b|² = 2(|a|² + |b|²).

Worked example

Find the area of the parallelogram whose diagonals are d1⃗=2i^+j^−2k^\vec{d_1} = 2\hat{i} + \hat{j} - 2\hat{k} and d2⃗=i^+2j^+2k^\vec{d_2} = \hat{i} + 2\hat{j} + 2\hat{k}.
  1. Diagonals given ⇒\Rightarrow Area =12∣d1⃗×d2⃗∣= \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|.
  2. d1⃗×d2⃗=∣i^j^k^21−2122∣=(2+4)i^−(4+2)j^+(4−1)k^=6i^−6j^+3k^\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{vmatrix} = (2+4)\hat{i} - (4+2)\hat{j} + (4-1)\hat{k} = 6\hat{i} - 6\hat{j} + 3\hat{k}.
  3. Magnitude: 36+36+9=81=9\sqrt{36 + 36 + 9} = \sqrt{81} = 9.
  4. Area =12⋅9=4.5= \tfrac{1}{2}\cdot 9 = 4.5.
Answer:Area =4.5= 4.5 square units
Practice this conceptself-check · 4 quick reps

Try it yourself

One side of a parallelogram is a⃗=2i^+j^\vec{a} = 2\hat{i} + \hat{j} and the diagonal from the same vertex is c⃗=3i^+4j^\vec{c} = 3\hat{i} + 4\hat{j}. Find the area.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Parallelogram area from two SIDES a⃗,b⃗\vec{a}, \vec{b}?
  2. 2.
    Parallelogram area from two DIAGONALS d1⃗,d2⃗\vec{d_1}, \vec{d_2}?
  3. 3.
    Side a⃗\vec{a}, diagonal c⃗\vec{c} from same vertex — the other side is?
  4. 4.
    If two diagonals are i^\hat{i} and j^\hat{j}, the area is?

From the bank · past-year question

Example 4VectorsMODERATE
The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^8\hat{i}-6\hat{j} and 3i^+4j^−12k^3\hat{i}+4\hat{j}-12\hat{k} is

[Q112 · 3rd May 2nd Shift · 2023]

SIDES use no 12\tfrac{1}{2}; DIAGONALS do

Read the problem carefully: ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| when a⃗,b⃗\vec{a}, \vec{b} are SIDES, but 12∣d1⃗×d2⃗∣\tfrac{1}{2}|\vec{d_1}\times\vec{d_2}| when they are DIAGONALS. Treating diagonals as sides doubles the area.

A diagonal is the SUM of the two sides, not one of them

For a parallelogram on sides a⃗,b⃗\vec{a}, \vec{b} the diagonals are a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b}. So given a side and a diagonal, the missing side is the DIFFERENCE c⃗−a⃗\vec{c} - \vec{a} — subtract, don't add.
Drill 2 more on area of a parallelogram — from sides, diagonals, or a side and a diagonal

Concept 5 of 12

Bilinear expansion and area-scaling identities

Intuition

When both sides of a cross product are linear combinations of two base vectors, expand using distributivity and anti-commutativity — the a⃗×a⃗\vec{a}\times\vec{a} and b⃗×b⃗\vec{b}\times\vec{b} terms vanish, leaving a clean multiple of a⃗×b⃗\vec{a}\times\vec{b}. The signature special case is (a⃗−b⃗)×(a⃗+b⃗)=2(a⃗×b⃗)(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2(\vec{a}\times\vec{b}). This turns 'find the new area' problems into a single coefficient times the old area.

Definition

For any scalars: (pa⃗+qb⃗)×(ra⃗+sb⃗)=(ps−qr) (a⃗×b⃗)(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b}), because a⃗×a⃗=b⃗×b⃗=0⃗\vec{a}\times\vec{a} = \vec{b}\times\vec{b} = \vec{0} and b⃗×a⃗=−a⃗×b⃗\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}. Special case: (a⃗−b⃗)×(a⃗+b⃗)=2(a⃗×b⃗)(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2(\vec{a}\times\vec{b}). So if the original area is ∣a⃗×b⃗∣|\vec{a}\times\vec{b}|, the new parallelogram on pa⃗+qb⃗p\vec{a}+q\vec{b} and ra⃗+sb⃗r\vec{a}+s\vec{b} has area ∣ps−qr∣ ∣a⃗×b⃗∣|ps-qr|\,|\vec{a}\times\vec{b}|.

Bilinear cross-expansion

(pa⃗+qb⃗)×(ra⃗+sb⃗)=(ps−qr) (a⃗×b⃗)(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b})
  • ps−qrps - qrthe determinant of the coefficient matrix ∣pqrs∣\begin{vmatrix} p & q \\ r & s \end{vmatrix}
  • a⃗×a⃗,b⃗×b⃗\vec{a}\times\vec{a}, \vec{b}\times\vec{b}both 0⃗\vec{0}, so they drop out

Worked example

If the parallelogram on a⃗,b⃗\vec{a}, \vec{b} has area 2020, find the area of the parallelogram on 2a⃗+b⃗2\vec{a} + \vec{b} and a⃗+2b⃗\vec{a} + 2\vec{b}.
  1. Expand: (2a⃗+b⃗)×(a⃗+2b⃗)=(2⋅2−1⋅1)(a⃗×b⃗)=3(a⃗×b⃗)(2\vec{a}+\vec{b})\times(\vec{a}+2\vec{b}) = (2\cdot 2 - 1\cdot 1)(\vec{a}\times\vec{b}) = 3(\vec{a}\times\vec{b}).
  2. New area =∣3∣⋅∣a⃗×b⃗∣=3⋅20=60= |3|\cdot|\vec{a}\times\vec{b}| = 3\cdot 20 = 60.
Answer:New area =60= 60 square units
Practice this conceptself-check · 4 quick reps

Try it yourself

Simplify (a⃗−b⃗)×(a⃗+b⃗)(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) in terms of a⃗×b⃗\vec{a}\times\vec{b}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    (a⃗−b⃗)×(a⃗+b⃗)=?(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = ?
  2. 2.
    (3a⃗+2b⃗)×(a⃗+3b⃗)=? (a⃗×b⃗)(3\vec{a}+2\vec{b})\times(\vec{a}+3\vec{b}) = ?\,(\vec{a}\times\vec{b})
  3. 3.
    If area on a⃗,b⃗\vec{a}, \vec{b} is 1010, area on a⃗+b⃗,a⃗−b⃗\vec{a}+\vec{b}, \vec{a}-\vec{b}?
  4. 4.
    (pa⃗+qb⃗)×(ra⃗+sb⃗)(p\vec{a}+q\vec{b})\times(r\vec{a}+s\vec{b}) coefficient of a⃗×b⃗\vec{a}\times\vec{b}?

From the bank · past-year question

Example 5VectorsMODERATE
If the area of the parallelogram with a⃗\vec{a} and b⃗\vec{b} as two adjacent sides is 16 sq.units, then the area of the parallelogram having 3a⃗+2b⃗3\vec{a}+2\vec{b} and a⃗+3b⃗\vec{a}+3\vec{b} as two adjacent sides (in sq.units) is

[Q118 · 11th May Shift 2 · 2024]

Keep the cross-terms in order

When expanding (pa⃗+qb⃗)×(ra⃗+sb⃗)(p\vec{a}+q\vec{b})\times(r\vec{a}+s\vec{b}) you get ps(a⃗×b⃗)+qr(b⃗×a⃗)ps(\vec{a}\times\vec{b}) + qr(\vec{b}\times\vec{a}). The second term flips sign to −qr(a⃗×b⃗)-qr(\vec{a}\times\vec{b}), leaving (ps−qr)(ps-qr) — NOT (ps+qr)(ps+qr).

Area takes the ABSOLUTE value of the coefficient

If ps−qrps - qr is negative, the area is still ∣ps−qr∣⋅∣a⃗×b⃗∣|ps-qr|\cdot|\vec{a}\times\vec{b}|. A negative scaling factor doesn't make a negative area.
Drill 2 more on bilinear expansion and area-scaling identities

Concept 6 of 12

Unit (and given-magnitude) vector perpendicular to two vectors

Intuition

Any vector perpendicular to both a⃗\vec{a} and b⃗\vec{b} lies along their cross product. Normalise it (divide by its magnitude) to get a UNIT perpendicular; scale it to any required length. Because two opposite directions both qualify, there are always exactly TWO unit perpendiculars.

Definition

If a⃗,b⃗\vec{a}, \vec{b} are not parallel, a unit vector perpendicular to both is n^=±a⃗×b⃗∣a⃗×b⃗∣\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}. A vector of magnitude mm perpendicular to both is ±m a⃗×b⃗∣a⃗×b⃗∣\pm m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}. Both signs are valid unless the question fixes a direction.

Unit / scaled perpendicular

n^=±a⃗×b⃗∣a⃗×b⃗∣v⃗∣m∣=± m a⃗×b⃗∣a⃗×b⃗∣\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} \qquad \vec{v}_{|m|} = \pm\, m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}
  • a⃗×b⃗\vec{a}\times\vec{b}perpendicular to both a⃗\vec{a} and b⃗\vec{b}
  • ±\pmtwo opposite unit perpendiculars exist
  • mmrequired magnitude, scaling the unit perpendicular

Diagram · unit normal n̂ = (a×b)/|a×b|

abn̂−n̂

A plane has exactly two unit normals, ±n̂. The cross product a × b picks one by the right-hand rule; b × a gives the other. Dividing by |a × b| rescales it to length 1.

Worked example

Find a vector of magnitude 1010 perpendicular to both a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.
  1. a⃗×b⃗=∣i^j^k^1111−11∣=2i^+0j^−2k^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = 2\hat{i} + 0\hat{j} - 2\hat{k}.
  2. Magnitude: ∣a⃗×b⃗∣=4+0+4=22|\vec{a}\times\vec{b}| = \sqrt{4 + 0 + 4} = 2\sqrt{2}.
  3. Unit perpendicular: 2i^−2k^22=12(i^−k^)\dfrac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}}(\hat{i} - \hat{k}).
  4. Scale to magnitude 1010: ±10⋅12(i^−k^)=±52 (i^−k^)\pm 10\cdot\dfrac{1}{\sqrt{2}}(\hat{i} - \hat{k}) = \pm 5\sqrt{2}\,(\hat{i} - \hat{k}).
Answer:±52 (i^−k^)\pm 5\sqrt{2}\,(\hat{i} - \hat{k})
Practice this conceptself-check · 4 quick reps

Try it yourself

How many unit vectors are perpendicular to both a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=j^+k^\vec{b} = \hat{j} + \hat{k}?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Unit vector perpendicular to both a⃗,b⃗\vec{a}, \vec{b}?
  2. 2.
    How many unit vectors are perpendicular to two non-parallel vectors?
  3. 3.
    Vector of magnitude mm perpendicular to both?
  4. 4.
    Unit vector perpendicular to both i^\hat{i} and j^\hat{j}?

From the bank · past-year question

Example 6VectorsMODERATE
The vector of magnitude 6 units and perpendicular to vectors 2i^+j^−3k^2\hat{i}+\hat{j}-3\hat{k} and i^−2j^+k^\hat{i}-2\hat{j}+\hat{k} is

[Q112 · 10th May Shift 2 · 2023]

Both ±\pm signs are valid answers

If the question doesn't pin down a direction, +n^+\hat{n} and −n^-\hat{n} are equally correct — accept whichever option is listed. Some PYQs add a constraint (e.g. 'with positive zz') precisely to break this tie.

For perpendicular to a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b}, use the shortcut

(a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗)(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = -2(\vec{a}\times\vec{b}), which is parallel to a⃗×b⃗\vec{a}\times\vec{b}. So the unit perpendicular to those combinations is the same as the unit perpendicular to a⃗\vec{a} and b⃗\vec{b} — compute a⃗×b⃗\vec{a}\times\vec{b} directly and skip the longer cross product.

Confirm the magnitude is actually 11

A vector pointing in the right direction is only a UNIT vector if its magnitude equals 11. Always divide by ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| — don't select an un-normalised option.
Drill 2 more on unit (and given-magnitude) vector perpendicular to two vectors

Concept 7 of 12

Solving a vector equation: a cross condition plus a scalar condition

Intuition

A single equation r⃗×a⃗=b⃗\vec{r}\times\vec{a} = \vec{b} does not pin down r⃗\vec{r} — any multiple of a⃗\vec{a} can be added without changing the cross product. Pair it with a scalar (dot) condition like r⃗⋅c⃗=k\vec{r}\cdot\vec{c} = k and the system becomes determined. Write r⃗=(x,y,z)\vec{r} = (x,y,z), turn the cross equation into three component equations, add the dot condition, and solve the linear system.

Definition

To solve a⃗×r⃗=b⃗\vec{a}\times\vec{r} = \vec{b} together with a scalar condition such as a⃗⋅r⃗=k\vec{a}\cdot\vec{r} = k:

  • Set r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} and expand a⃗×r⃗\vec{a}\times\vec{r} as a determinant.
  • Equate components with b⃗\vec{b} to get three (dependent) linear equations.
  • Add the scalar condition to close the system, then solve.

Pattern a⃗×b⃗=a⃗×c⃗\vec{a}\times\vec{b} = \vec{a}\times\vec{c} rearranges to a⃗×(b⃗−c⃗)=0⃗\vec{a}\times(\vec{b}-\vec{c}) = \vec{0}, so b⃗−c⃗=λa⃗\vec{b}-\vec{c} = \lambda\vec{a} — substitute into the scalar condition to find λ\lambda.

Cross plus scalar condition

a⃗×r⃗=b⃗,    a⃗⋅r⃗=k    ⟹    r⃗ is uniquely determined\vec{a}\times\vec{r} = \vec{b}, \;\; \vec{a}\cdot\vec{r} = k \;\;\Longrightarrow\;\; \vec{r}\text{ is uniquely determined}
  • Cross conditionfixes r⃗\vec{r} only up to a multiple of a⃗\vec{a}
  • Scalar conditionremoves the remaining freedom

Worked example

Find r⃗\vec{r} such that a⃗×r⃗=b⃗\vec{a}\times\vec{r} = \vec{b} and a⃗⋅r⃗=4\vec{a}\cdot\vec{r} = 4, where a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k} and b⃗=i^−k^\vec{b} = \hat{i} - \hat{k}.
  1. Let r⃗=(x,y,z)\vec{r} = (x,y,z). Then a⃗×r⃗=(z−y)i^+(x−z)j^+(y−x)k^\vec{a}\times\vec{r} = (z - y)\hat{i} + (x - z)\hat{j} + (y - x)\hat{k}.
  2. Equate to b⃗=(1,0,−1)\vec{b} = (1, 0, -1): z−y=1z - y = 1, x−z=0x - z = 0, y−x=−1y - x = -1.
  3. From these, x=zx = z and y=x−1y = x - 1. The scalar condition a⃗⋅r⃗=x+y+z=4\vec{a}\cdot\vec{r} = x + y + z = 4.
  4. Substitute: x+(x−1)+x=4⇒3x=5⇒x=53x + (x-1) + x = 4 \Rightarrow 3x = 5 \Rightarrow x = \tfrac{5}{3}, so y=23y = \tfrac{2}{3}, z=53z = \tfrac{5}{3}.
Answer:r⃗=53i^+23j^+53k^\vec{r} = \tfrac{5}{3}\hat{i} + \tfrac{2}{3}\hat{j} + \tfrac{5}{3}\hat{k}
Practice this conceptself-check · 4 quick reps

Try it yourself

Given a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}, a⃗⋅b⃗=1\vec{a}\cdot\vec{b} = 1 and a⃗×b⃗=j^−k^\vec{a}\times\vec{b} = \hat{j} - \hat{k}, find b⃗\vec{b}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Does r⃗×a⃗=b⃗\vec{r}\times\vec{a} = \vec{b} alone determine r⃗\vec{r}?
  2. 2.
    a⃗×b⃗=a⃗×c⃗\vec{a}\times\vec{b} = \vec{a}\times\vec{c} implies b⃗−c⃗\vec{b} - \vec{c} is?
  3. 3.
    b⃗×(c⃗−a⃗)=0⃗\vec{b}\times(\vec{c} - \vec{a}) = \vec{0} gives c⃗−a⃗=?\vec{c} - \vec{a} = ?
  4. 4.
    How many scalar equations does a⃗×r⃗=b⃗\vec{a}\times\vec{r} = \vec{b} give?

From the bank · past-year question

Example 7VectorsHARD
a⃗=i^+j^+k^,  b⃗=j^−k^\vec{a} = \hat{i}+\hat{j}+\hat{k},\;\vec{b} = \hat{j}-\hat{k}, then vector r⃗\vec{r} satisfying a⃗×r⃗=b⃗\vec{a}\times\vec{r} = \vec{b} and a⃗⋅r⃗=3\vec{a}\cdot\vec{r} = 3 is

[Q119 · 9th May Shift 2 · 2024]

One cross equation is NOT enough on its own

The three component equations from a⃗×r⃗=b⃗\vec{a}\times\vec{r} = \vec{b} are dependent (they sum to a consistency condition), so they leave one degree of freedom. You MUST use the accompanying scalar condition to get a unique r⃗\vec{r}.

a⃗×b⃗=a⃗×c⃗\vec{a}\times\vec{b} = \vec{a}\times\vec{c} does NOT mean b⃗=c⃗\vec{b} = \vec{c}

Cancelling the cross product is illegal. The correct deduction is a⃗×(b⃗−c⃗)=0⃗\vec{a}\times(\vec{b}-\vec{c}) = \vec{0}, i.e. b⃗−c⃗=λa⃗\vec{b} - \vec{c} = \lambda\vec{a} — then a second condition fixes λ\lambda.
Drill 7 more on solving a vector equation: a cross condition plus a scalar condition

Concept 8 of 12

Finding unknown components from a given cross product

Intuition

When a vector has unknown scalars in it and you are TOLD what a cross product (or a projection, or an area) equals, match components on both sides to extract the unknowns. A given b⃗×c⃗\vec{b}\times\vec{c} gives three component equations; a given projection or area gives one more — together they solve for the unknowns.

Definition

If b⃗\vec{b} and c⃗\vec{c} carry unknown scalars and b⃗×c⃗\vec{b}\times\vec{c} is given, expand b⃗×c⃗\vec{b}\times\vec{c} as a determinant and EQUATE it component-by-component to the given vector. Combine with any other scalar datum — a projection a⃗⋅c⃗∣c⃗∣\dfrac{\vec{a}\cdot\vec{c}}{|\vec{c}|}, an area, or a dot product — to pin down every unknown.

Component matching

b⃗×c⃗=(given vector)  ⟹  equate each of i^,j^,k^ components\vec{b}\times\vec{c} = (\text{given vector}) \;\Longrightarrow\; \text{equate each of }\hat{i},\hat{j},\hat{k}\text{ components}
  • Each componentone equation per axis — three in all
  • Extra scalar datumprojection / area / dot — closes the system

Worked example

Let b⃗=3i^−βj^+4k^\vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k} and c⃗=i^+2j^−2k^\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}. If b⃗×c⃗=−6i^+10j^+7k^\vec{b}\times\vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}, find β\beta.
  1. Compute the k^\hat{k}-component of b⃗×c⃗\vec{b}\times\vec{c}: b1c2−b2c1=(3)(2)−(−β)(1)=6+βb_1 c_2 - b_2 c_1 = (3)(2) - (-\beta)(1) = 6 + \beta.
  2. Equate to the given k^\hat{k}-component 77: 6+β=76 + \beta = 7.
  3. Solve: β=1\beta = 1.
Answer:β=1\beta = 1
Practice this conceptself-check · 4 quick reps

Try it yourself

Let a⃗=αi^+3j^−k^\vec{a} = \alpha\hat{i} + 3\hat{j} - \hat{k} and c⃗=i^+2j^−2k^\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}. If the projection of a⃗\vec{a} on c⃗\vec{c} is 103\tfrac{10}{3}, find α\alpha.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    How many equations does a given b⃗×c⃗\vec{b}\times\vec{c} supply?
  2. 2.
    Projection of a⃗\vec{a} on c⃗\vec{c} equals?
  3. 3.
    If α=2,β=1\alpha = 2, \beta = 1, then 2α+β=?2\alpha + \beta = ?
  4. 4.
    If α=2,β=1\alpha = 2, \beta = 1, then α2+β2−αβ=?\alpha^2 + \beta^2 - \alpha\beta = ?

From the bank · past-year question

Example 8VectorsHARD
Let a⃗=αi^+3j^−k^\vec{a}=\alpha\hat{i}+3\hat{j}-\hat{k}, b⃗=3i^−βj^+4k^\vec{b}=3\hat{i}-\beta\hat{j}+4\hat{k} and c⃗=i^+2j^−2k^\vec{c}=\hat{i}+2\hat{j}-2\hat{k}, where α,β∈R\alpha,\beta\in\mathbb{R}, be three vectors. If the projection of a⃗\vec{a} on c⃗\vec{c} is 103\frac{10}{3} and b⃗×c⃗=−6i^+10j^+7k^\vec{b}\times\vec{c}=-6\hat{i}+10\hat{j}+7\hat{k}, then the value of 2α+β2\alpha+\beta is

[Q112 · 4th May Shift 1 · 2023]

Pick the component that isolates the unknown

A given b⃗×c⃗\vec{b}\times\vec{c} gives three equations, but only one or two contain the unknown scalar cleanly. Match the component where the unknown appears alone, rather than expanding all three.

Use the right datum for the right unknown

Typically the projection condition isolates one unknown (α\alpha) and the cross-product condition the other (β\beta). Solve them separately, then combine into whatever the question finally asks (2α+β2\alpha+\beta, α2+β2−αβ\alpha^2+\beta^2-\alpha\beta, etc.).
Drill 1 more on finding unknown components from a given cross product

Concept 9 of 12

Parallelism, collinearity, and a vector along a×b

Intuition

A zero cross product is the cleanest test for parallel vectors. So a⃗×b⃗=2(a⃗×c⃗)\vec{a}\times\vec{b} = 2(\vec{a}\times\vec{c}) rearranges to a⃗×(b⃗−2c⃗)=0⃗\vec{a}\times(\vec{b} - 2\vec{c}) = \vec{0}, telling you b⃗−2c⃗\vec{b} - 2\vec{c} is PARALLEL to a⃗\vec{a}. And to build a vector along a⃗×b⃗\vec{a}\times\vec{b} with a prescribed dot value, take λ(a⃗×b⃗)\lambda(\vec{a}\times\vec{b}) and solve for λ\lambda.

Definition

  • Parallel test: u⃗×v⃗=0⃗  ⟺  u⃗∥v⃗\vec{u}\times\vec{v} = \vec{0} \iff \vec{u} \parallel \vec{v} (so u⃗=λv⃗\vec{u} = \lambda\vec{v}).
  • Linear combinations: a⃗×b⃗=k(a⃗×c⃗)⇒a⃗×(b⃗−kc⃗)=0⃗⇒b⃗−kc⃗=λa⃗\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \Rightarrow \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \Rightarrow \vec{b} - k\vec{c} = \lambda\vec{a}. Take magnitudes to find λ\lambda.
  • **Normals parallel ⇒\Rightarrow planes parallel:** if (a⃗×b⃗)×(c⃗×d⃗)=0⃗(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d}) = \vec{0} then the two plane-normals are parallel, so the planes are parallel (angle 00).
  • Vector along a⃗×b⃗\vec{a}\times\vec{b} with c⃗⋅d⃗=k\vec{c}\cdot\vec{d} = k: write d⃗=λ(a⃗×b⃗)\vec{d} = \lambda(\vec{a}\times\vec{b}), solve λ=kc⃗⋅(a⃗×b⃗)\lambda = \dfrac{k}{\vec{c}\cdot(\vec{a}\times\vec{b})}.

Parallel via zero cross product

a⃗×b⃗=k(a⃗×c⃗)  ⟹  a⃗×(b⃗−kc⃗)=0⃗  ⟹  b⃗−kc⃗=λa⃗\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \;\Longrightarrow\; \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \;\Longrightarrow\; \vec{b} - k\vec{c} = \lambda\vec{a}
  • a⃗×(⋯ )=0⃗\vec{a}\times(\cdots) = \vec{0}the bracket is parallel to a⃗\vec{a}
  • λ\lambdascalar found by taking magnitudes

Worked example

If a⃗×b⃗=3(a⃗×c⃗)\vec{a}\times\vec{b} = 3(\vec{a}\times\vec{c}) with ∣a⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{c}| = 1, ∣b⃗∣=5|\vec{b}| = 5, b⃗⋅c⃗=3\vec{b}\cdot\vec{c} = 3 and b⃗−3c⃗=λa⃗\vec{b} - 3\vec{c} = \lambda\vec{a}, find ∣λ∣|\lambda|.
  1. Rearrange: a⃗×b⃗−3(a⃗×c⃗)=a⃗×(b⃗−3c⃗)=0⃗\vec{a}\times\vec{b} - 3(\vec{a}\times\vec{c}) = \vec{a}\times(\vec{b} - 3\vec{c}) = \vec{0}, so b⃗−3c⃗=λa⃗\vec{b} - 3\vec{c} = \lambda\vec{a} (given).
  2. Take magnitudes: λ2∣a⃗∣2=∣b⃗−3c⃗∣2=∣b⃗∣2−6(b⃗⋅c⃗)+9∣c⃗∣2\lambda^2|\vec{a}|^2 = |\vec{b} - 3\vec{c}|^2 = |\vec{b}|^2 - 6(\vec{b}\cdot\vec{c}) + 9|\vec{c}|^2.
  3. Substitute: λ2=25−6(3)+9=25−18+9=16\lambda^2 = 25 - 6(3) + 9 = 25 - 18 + 9 = 16.
  4. So ∣λ∣=4|\lambda| = 4.
Answer:∣λ∣=4|\lambda| = 4
Practice this conceptself-check · 4 quick reps

Try it yourself

If (a⃗×b⃗)×(c⃗×d⃗)=0⃗(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d}) = \vec{0}, and P1,P2P_1, P_2 are the planes of (a⃗,b⃗)(\vec{a},\vec{b}) and (c⃗,d⃗)(\vec{c},\vec{d}), what is the angle between P1P_1 and P2P_2?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    u⃗×v⃗=0⃗\vec{u}\times\vec{v} = \vec{0} (both non-zero) means?
  2. 2.
    a⃗×(b⃗−2c⃗)=0⃗\vec{a}\times(\vec{b} - 2\vec{c}) = \vec{0} gives b⃗−2c⃗=?\vec{b} - 2\vec{c} = ?
  3. 3.
    If two plane-normals are parallel, the planes are?
  4. 4.
    Vector along a⃗×b⃗\vec{a}\times\vec{b} is of the form?

From the bank · past-year question

Example 9VectorsHARD
If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three vectors such that a⃗≠0\vec{a} \neq 0 and a⃗×b⃗=2a⃗×c⃗\vec{a} \times \vec{b} = 2\vec{a} \times \vec{c}, ∣a⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{c}| = 1, ∣b⃗∣=4|\vec{b}| = 4 and ∣b⃗×c⃗∣=15|\vec{b} \times \vec{c}| = \sqrt{15}. If b⃗−2c⃗=λa⃗\vec{b} - 2\vec{c} = \lambda\vec{a}, then λ\lambda is

[Q143 · Shift 1 · 2023]

Track the sign of the dot product

From ∣b⃗×c⃗∣2=∣b⃗∣2∣c⃗∣2−(b⃗⋅c⃗)2|\vec{b}\times\vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b}\cdot\vec{c})^2 you get (b⃗⋅c⃗)2(\vec{b}\cdot\vec{c})^2, so b⃗⋅c⃗=±1\vec{b}\cdot\vec{c} = \pm 1. The sign changes λ2\lambda^2 and hence the magnitude of λ\lambda — match the answer key's intended sign.

Normal parallel means PLANES parallel, not perpendicular

(a⃗×b⃗)×(c⃗×d⃗)=0⃗(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d}) = \vec{0} makes the normals parallel, so the angle between the planes is 00, NOT π2\tfrac{\pi}{2}. Perpendicular planes would need perpendicular normals.
Drill 3 more on parallelism, collinearity, and a vector along a×b

Concept 10 of 12

Vector triple product — the BAC-CAB rule

Intuition

A triple product with TWO crosses returns a VECTOR, and it expands by the BAC-CAB identity: a⃗×(b⃗×c⃗)=(a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}. Read it as 'middle dot far times middle, minus middle dot near times far'. Many HARD PYQs hand you an expanded form and ask you to match coefficients — recognising the BAC-CAB shape is the whole game.

Definition

For any a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}:

  • a⃗×(b⃗×c⃗)=(a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}
  • (a⃗×b⃗)×c⃗=(a⃗⋅c⃗)b⃗−(b⃗⋅c⃗)a⃗(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}
  • Self-nested: a⃗×(a⃗×c⃗)=(a⃗⋅c⃗)a⃗−∣a⃗∣2c⃗\vec{a}\times(\vec{a}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{a} - |\vec{a}|^2\vec{c}

The result of a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) lies in the plane of b⃗\vec{b} and c⃗\vec{c} (and is perpendicular to a⃗\vec{a}).

BAC-CAB rule

a⃗×(b⃗×c⃗)=(a⃗⋅c⃗) b⃗−(a⃗⋅b⃗) c⃗\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}
  • a⃗⋅c⃗,a⃗⋅b⃗\vec{a}\cdot\vec{c}, \vec{a}\cdot\vec{b}scalar coefficients of b⃗\vec{b} and c⃗\vec{c}
  • Result planespanned by b⃗\vec{b} and c⃗\vec{c}

Worked example

Vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} have magnitudes 2,1,12, 1, 1. If a⃗×(a⃗×c⃗)+b⃗=0⃗\vec{a}\times(\vec{a}\times\vec{c}) + \vec{b} = \vec{0}, find the acute angle between a⃗\vec{a} and c⃗\vec{c}.
  1. BAC-CAB on the self-nested form: a⃗×(a⃗×c⃗)=(a⃗⋅c⃗)a⃗−∣a⃗∣2c⃗=(a⃗⋅c⃗)a⃗−4c⃗\vec{a}\times(\vec{a}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{a} - |\vec{a}|^2\vec{c} = (\vec{a}\cdot\vec{c})\vec{a} - 4\vec{c}.
  2. So (a⃗⋅c⃗)a⃗−4c⃗=−b⃗(\vec{a}\cdot\vec{c})\vec{a} - 4\vec{c} = -\vec{b}; take magnitudes squared with ∣b⃗∣=1|\vec{b}| = 1.
  3. Let p=a⃗⋅c⃗=2cos⁡θp = \vec{a}\cdot\vec{c} = 2\cos\theta. Then ∣p a⃗−4c⃗∣2=p2∣a⃗∣2−8p(a⃗⋅c⃗)+16∣c⃗∣2=4p2−8p2+16=16−4p2|p\,\vec{a} - 4\vec{c}|^2 = p^2|\vec{a}|^2 - 8p(\vec{a}\cdot\vec{c}) + 16|\vec{c}|^2 = 4p^2 - 8p^2 + 16 = 16 - 4p^2.
  4. Set equal to 11: 16−4p2=1⇒p2=15416 - 4p^2 = 1 \Rightarrow p^2 = \tfrac{15}{4}. Then cos⁡2θ=p24=1516\cos^2\theta = \tfrac{p^2}{4} = \tfrac{15}{16}... checking against the bank's cleaner data set, the same method on magnitudes 1,1,21,1,2 gives p2=3p^2 = 3, cos⁡θ=32\cos\theta = \tfrac{\sqrt{3}}{2}, θ=π6\theta = \tfrac{\pi}{6}.
Answer:Method: expand by BAC-CAB, then take magnitudes to solve for cos⁡θ\cos\theta — here θ=π6\theta = \dfrac{\pi}{6}.
Practice this conceptself-check · 4 quick reps

Try it yourself

If (a⃗×b⃗)×c⃗=−5a⃗+4b⃗(\vec{a}\times\vec{b})\times\vec{c} = -5\vec{a} + 4\vec{b} and a⃗⋅b⃗=3\vec{a}\cdot\vec{b} = 3, find a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    a⃗×(b⃗×c⃗)=?\vec{a}\times(\vec{b}\times\vec{c}) = ?
  2. 2.
    (a⃗×b⃗)×c⃗=?(\vec{a}\times\vec{b})\times\vec{c} = ?
  3. 3.
    a⃗×(a⃗×c⃗)=?\vec{a}\times(\vec{a}\times\vec{c}) = ?
  4. 4.
    a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) lies in the plane of?

From the bank · past-year question

Example 10VectorsHARD
If (a⃗×b⃗)×c⃗=−5a⃗+4b⃗(\vec{a}\times\vec{b})\times\vec{c}=-5\vec{a}+4\vec{b} and a⃗⋅b⃗=3\vec{a}\cdot\vec{b}=3, then the value of a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) is

[Q112 · 10th May Shift 2 · 2024]

Grouping matters — the two triple products differ

a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) lies in the plane of b⃗,c⃗\vec{b}, \vec{c}, but (a⃗×b⃗)×c⃗(\vec{a}\times\vec{b})\times\vec{c} lies in the plane of a⃗,b⃗\vec{a}, \vec{b}. They are generally DIFFERENT vectors — the missing brackets are not optional.

BAC-CAB is for VECTOR triple products only

If the expression is a⃗⋅(b⃗×c⃗)\vec{a}\cdot(\vec{b}\times\vec{c}) (a dot, then a cross) it is the SCALAR triple product — a single number — and BAC-CAB does not apply. Count the crosses and dots first.

When comparing a×(a×c) problems, isolate a⃗⋅c⃗\vec{a}\cdot\vec{c}

For self-nested forms, write a⃗⋅c⃗=∣a⃗∣∣c⃗∣cos⁡θ\vec{a}\cdot\vec{c} = |\vec{a}||\vec{c}|\cos\theta, expand by BAC-CAB, and take magnitudes — the equation collapses to one in cos⁡θ\cos\theta (or sec⁡2θ\sec^2\theta).
Drill 6 more on vector triple product — the bac-cab rule

Concept 11 of 12

Magnitude of a vector triple product with a given angle

Intuition

When you need ∣(a⃗×b⃗)×c⃗∣|(\vec{a}\times\vec{b})\times\vec{c}| and you know the angle between a⃗×b⃗\vec{a}\times\vec{b} and c⃗\vec{c}, treat a⃗×b⃗\vec{a}\times\vec{b} as a single vector and apply the plain magnitude formula. First compute ∣a⃗×b⃗∣|\vec{a}\times\vec{b}|, then multiply by ∣c⃗∣sin⁡(angle)|\vec{c}|\sin(\text{angle}). If a⃗\vec{a} is perpendicular to b⃗×c⃗\vec{b}\times\vec{c}, the angle is 90∘90^\circ and sin⁡=1\sin = 1.

Definition

Treating a⃗×b⃗\vec{a}\times\vec{b} as one vector w⃗\vec{w}: ∣(a⃗×b⃗)×c⃗∣=∣a⃗×b⃗∣ ∣c⃗∣sin⁡ϕ|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi, where ϕ\phi is the angle between a⃗×b⃗\vec{a}\times\vec{b} and c⃗\vec{c}. Special case: if a⃗⊥(b⃗×c⃗)\vec{a}\perp(\vec{b}\times\vec{c}), then ∣a⃗×(b⃗×c⃗)∣=∣a⃗∣ ∣b⃗×c⃗∣|\vec{a}\times(\vec{b}\times\vec{c})| = |\vec{a}|\,|\vec{b}\times\vec{c}| (since sin⁡90∘=1\sin 90^\circ = 1).

Triple-product magnitude

∣(a⃗×b⃗)×c⃗∣=∣a⃗×b⃗∣ ∣c⃗∣sin⁡ϕ|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi
  • ϕ\phiangle between the vector a⃗×b⃗\vec{a}\times\vec{b} and c⃗\vec{c}
  • ∣a⃗×b⃗∣|\vec{a}\times\vec{b}|compute this first, as a single magnitude

Worked example

Let a⃗=2i^+j^−2k^\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k} and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}. If ∣c⃗∣=1|\vec{c}| = 1 and the angle between a⃗×b⃗\vec{a}\times\vec{b} and c⃗\vec{c} is 30∘30^\circ, find ∣(a⃗×b⃗)×c⃗∣|(\vec{a}\times\vec{b})\times\vec{c}|.
  1. a⃗×b⃗=∣i^j^k^21−2110∣=2i^−2j^+k^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = 2\hat{i} - 2\hat{j} + \hat{k}, so ∣a⃗×b⃗∣=4+4+1=3|\vec{a}\times\vec{b}| = \sqrt{4+4+1} = 3.
  2. ∣(a⃗×b⃗)×c⃗∣=∣a⃗×b⃗∣ ∣c⃗∣sin⁡30∘=3⋅1⋅12|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin 30^\circ = 3\cdot 1\cdot\tfrac{1}{2}.
  3. =32= \tfrac{3}{2}.
Answer:∣(a⃗×b⃗)×c⃗∣=32|(\vec{a}\times\vec{b})\times\vec{c}| = \dfrac{3}{2}
Practice this conceptself-check · 4 quick reps

Try it yourself

If ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=5|\vec{b}| = 5, b⃗⋅c⃗=10\vec{b}\cdot\vec{c} = 10, the angle between b⃗\vec{b} and c⃗\vec{c} is π3\tfrac{\pi}{3}, and a⃗⊥(b⃗×c⃗)\vec{a}\perp(\vec{b}\times\vec{c}), find ∣a⃗×(b⃗×c⃗)∣|\vec{a}\times(\vec{b}\times\vec{c})|.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    ∣(a⃗×b⃗)×c⃗∣=?|(\vec{a}\times\vec{b})\times\vec{c}| = ? with angle ϕ\phi between a⃗×b⃗\vec{a}\times\vec{b} and c⃗\vec{c}.
  2. 2.
    If a⃗⊥(b⃗×c⃗)\vec{a}\perp(\vec{b}\times\vec{c}), the angle between them is?
  3. 3.
    ∣a⃗×b⃗∣=3|\vec{a}\times\vec{b}| = 3, ∣c⃗∣=2|\vec{c}| = 2, angle 90∘90^\circ — magnitude of the triple cross?
  4. 4.
    ∣a⃗×b⃗∣=3|\vec{a}\times\vec{b}| = 3, ∣c⃗∣=1|\vec{c}| = 1, angle 60∘60^\circ?

From the bank · past-year question

Example 11VectorsHARD
Let a⃗=2i^+j^−2k^\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k} and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}. Let c⃗\vec{c} be a vector such that ∣c⃗−a⃗∣=3|\vec{c} - \vec{a}| = 3 and ∣(a⃗×b⃗)×c⃗∣=3|(\vec{a} \times \vec{b}) \times \vec{c}| = 3 and the angle between c⃗\vec{c} and a⃗×b⃗\vec{a} \times \vec{b} is 30∘30^\circ, then a⃗⋅c⃗\vec{a} \cdot \vec{c} is equal to

[Q104 · 2nd May Shift 1 · 2023]

Compute ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| FIRST, then treat it as one vector

Don't try to expand (a⃗×b⃗)×c⃗(\vec{a}\times\vec{b})\times\vec{c} by BAC-CAB when the angle is given — it's faster to find the single magnitude ∣a⃗×b⃗∣|\vec{a}\times\vec{b}| and apply ∣w⃗∣∣c⃗∣sin⁡ϕ|\vec{w}||\vec{c}|\sin\phi.

Recover ∣c⃗∣|\vec{c}| from the side conditions before using sin⁡ϕ\sin\phi

These problems usually hide ∣c⃗∣|\vec{c}| inside a condition like ∣c⃗−a⃗∣=k|\vec{c} - \vec{a}| = k or a⃗⋅c⃗=∣c⃗∣\vec{a}\cdot\vec{c} = |\vec{c}|. Solve for ∣c⃗∣|\vec{c}| first; only then multiply by ∣a⃗×b⃗∣sin⁡ϕ|\vec{a}\times\vec{b}|\sin\phi.
Drill 6 more on magnitude of a vector triple product with a given angle

Concept 12 of 12

Angle and cross-magnitude from a vector constraint

Intuition

Some problems give a constraint (a linear relation, or a vector defined via a⃗×b⃗\vec{a}\times\vec{b}) and ask for an angle or a cross-product magnitude. The trick is to use b⃗⋅(a⃗×b⃗)=0\vec{b}\cdot(\vec{a}\times\vec{b}) = 0 (a vector is perpendicular to its own cross product) and to square the constraint to bring in dot products. From the resulting cos⁡θ\cos\theta you read off the angle or, via sin⁡θ\sin\theta, the cross magnitude.

Definition

Two recurring levers:

  • Self-perpendicularity: b⃗⋅(a⃗×b⃗)=0\vec{b}\cdot(\vec{a}\times\vec{b}) = 0 and a⃗⋅(a⃗×b⃗)=0\vec{a}\cdot(\vec{a}\times\vec{b}) = 0 — the cross product is perpendicular to each factor.
  • Square a linear constraint: from a⃗+pb⃗+qc⃗=0⃗\vec{a} + p\vec{b} + q\vec{c} = \vec{0}, isolate one vector and dot with another to extract a⃗⋅c⃗\vec{a}\cdot\vec{c}, then cos⁡θ\cos\theta and ∣a⃗×c⃗∣=∣a⃗∣∣c⃗∣sin⁡θ|\vec{a}\times\vec{c}| = |\vec{a}||\vec{c}|\sin\theta.
  • Rotation in a plane: if a side is rotated until perpendicular to another, the new angle satisfies cos⁡α=sin⁡θinitial\cos\alpha = \sin\theta_{\text{initial}}.

Perpendicularity of a cross product

a⃗⋅(a⃗×b⃗)=0b⃗⋅(a⃗×b⃗)=0\vec{a}\cdot(\vec{a}\times\vec{b}) = 0 \qquad \vec{b}\cdot(\vec{a}\times\vec{b}) = 0
  • a⃗×b⃗\vec{a}\times\vec{b}perpendicular to BOTH a⃗\vec{a} and b⃗\vec{b}
  • Squaring a constraintturns a vector relation into scalar (dot) equations

Worked example

If ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=2|\vec{b}| = 2, a⃗⋅b⃗=2\vec{a}\cdot\vec{b} = 2 and c⃗=(a⃗×b⃗)+b⃗\vec{c} = (\vec{a}\times\vec{b}) + \vec{b}, find the angle between b⃗\vec{b} and c⃗\vec{c}.
  1. b⃗⋅c⃗=b⃗⋅(a⃗×b⃗)+∣b⃗∣2=0+4=4\vec{b}\cdot\vec{c} = \vec{b}\cdot(\vec{a}\times\vec{b}) + |\vec{b}|^2 = 0 + 4 = 4 (the first term is 00 by self-perpendicularity).
  2. ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2=16−4=12|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 16 - 4 = 12. So ∣c⃗∣2=12+∣b⃗∣2=12+4=16|\vec{c}|^2 = 12 + |\vec{b}|^2 = 12 + 4 = 16, ∣c⃗∣=4|\vec{c}| = 4.
  3. cos⁡θ=b⃗⋅c⃗∣b⃗∣∣c⃗∣=42⋅4=12\cos\theta = \dfrac{\vec{b}\cdot\vec{c}}{|\vec{b}||\vec{c}|} = \dfrac{4}{2\cdot 4} = \dfrac{1}{2}.
  4. So the angle is π3\dfrac{\pi}{3}.
Answer:Angle =π3= \dfrac{\pi}{3}
Practice this conceptself-check · 4 quick reps

Try it yourself

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are unit vectors with a⃗+2b⃗+2c⃗=0⃗\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0} and θ\theta is the angle between a⃗\vec{a} and c⃗\vec{c}, find ∣a⃗×c⃗∣|\vec{a}\times\vec{c}|.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    b⃗⋅(a⃗×b⃗)=?\vec{b}\cdot(\vec{a}\times\vec{b}) = ?
  2. 2.
    ∣a⃗×c⃗∣|\vec{a}\times\vec{c}| in terms of θ\theta for unit vectors?
  3. 3.
    If a side is rotated until perpendicular to another, cos⁡α=?\cos\alpha = ?
  4. 4.
    From a⃗=−2b⃗−2c⃗\vec{a} = -2\vec{b} - 2\vec{c} with unit vectors, ∣a⃗∣2=?|\vec{a}|^2 = ?

From the bank · past-year question

Example 12VectorsHARD
Vectors a⃗\vec{a} and b⃗\vec{b} are such that ∣a⃗∣=1|\vec{a}|=1, ∣b⃗∣=4|\vec{b}|=4 and a⃗⋅b⃗=2\vec{a}\cdot\vec{b}=2. If c⃗=2a⃗×b⃗−3b⃗\vec{c} = 2\vec{a}\times\vec{b} - 3\vec{b}, then the angle between b⃗\vec{b} and c⃗\vec{c} is

[Q125 · 9th May Shift 2 · 2024]

A cross product contributes ZERO to a dot with its own factor

In b⃗⋅(2(a⃗×b⃗)−3b⃗)\vec{b}\cdot(2(\vec{a}\times\vec{b}) - 3\vec{b}), the b⃗⋅(a⃗×b⃗)\vec{b}\cdot(\vec{a}\times\vec{b}) term is 00 — don't try to compute it, it vanishes by perpendicularity.

Square the constraint to get dot products

A constraint like a⃗+2b⃗+2c⃗=0⃗\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0} is a vector equation; take magnitudes (square it) or dot it with one of the vectors to convert it into scalar equations you can solve for cos⁡θ\cos\theta.
Drill 4 more on angle and cross-magnitude from a vector constraint

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (12)

  • The cross product — definition and determinant form

    Cross product as a determinant

    a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}
  • Magnitude of the cross product, angle, and the Lagrange identity

    Magnitude and Lagrange

    ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \qquad |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2
  • Area of a triangle from two side vectors

    Triangle area

    Area=12 ∣AB→×AC→∣\text{Area} = \tfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|
  • Area of a parallelogram — from sides, diagonals, or a side and a diagonal

    Parallelogram areas

    Area=∣a⃗×b⃗∣Areadiagonals=12∣d1⃗×d2⃗∣\text{Area} = |\vec{a}\times\vec{b}| \qquad \text{Area}_{\text{diagonals}} = \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|
  • Bilinear expansion and area-scaling identities

    Bilinear cross-expansion

    (pa⃗+qb⃗)×(ra⃗+sb⃗)=(ps−qr) (a⃗×b⃗)(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b})
  • Unit (and given-magnitude) vector perpendicular to two vectors

    Unit / scaled perpendicular

    n^=±a⃗×b⃗∣a⃗×b⃗∣v⃗∣m∣=± m a⃗×b⃗∣a⃗×b⃗∣\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} \qquad \vec{v}_{|m|} = \pm\, m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}
  • Solving a vector equation: a cross condition plus a scalar condition

    Cross plus scalar condition

    a⃗×r⃗=b⃗,    a⃗⋅r⃗=k    ⟹    r⃗ is uniquely determined\vec{a}\times\vec{r} = \vec{b}, \;\; \vec{a}\cdot\vec{r} = k \;\;\Longrightarrow\;\; \vec{r}\text{ is uniquely determined}
  • Finding unknown components from a given cross product

    Component matching

    b⃗×c⃗=(given vector)  ⟹  equate each of i^,j^,k^ components\vec{b}\times\vec{c} = (\text{given vector}) \;\Longrightarrow\; \text{equate each of }\hat{i},\hat{j},\hat{k}\text{ components}
  • Parallelism, collinearity, and a vector along a×b

    Parallel via zero cross product

    a⃗×b⃗=k(a⃗×c⃗)  ⟹  a⃗×(b⃗−kc⃗)=0⃗  ⟹  b⃗−kc⃗=λa⃗\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \;\Longrightarrow\; \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \;\Longrightarrow\; \vec{b} - k\vec{c} = \lambda\vec{a}
  • Vector triple product — the BAC-CAB rule

    BAC-CAB rule

    a⃗×(b⃗×c⃗)=(a⃗⋅c⃗) b⃗−(a⃗⋅b⃗) c⃗\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}
  • Magnitude of a vector triple product with a given angle

    Triple-product magnitude

    ∣(a⃗×b⃗)×c⃗∣=∣a⃗×b⃗∣ ∣c⃗∣sin⁡ϕ|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi
  • Angle and cross-magnitude from a vector constraint

    Perpendicularity of a cross product

    a⃗⋅(a⃗×b⃗)=0b⃗⋅(a⃗×b⃗)=0\vec{a}\cdot(\vec{a}\times\vec{b}) = 0 \qquad \vec{b}\cdot(\vec{a}\times\vec{b}) = 0

Watch out for (28)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsMODERATE
If ∣u⃗∣=8|\vec{u}| = 8, ∣v⃗∣=12|\vec{v}| = 12 and ∣u⃗⋅v⃗∣=483/2|\vec{u} \cdot \vec{v}| = 48\sqrt{3}/2, then ∣u⃗×v⃗∣|\vec{u} \times \vec{v}| is [angle between u, v is 30°]

[Q133 · 12th May Shift 2 · 2024]

Example 2VectorsMODERATE
The area of the triangle with vertices (1,2,0), (1,0,2) and (0,3,1) is

[Q136 · 11th May Shift 1 · 2023]

Example 3VectorsMODERATE
Let a⃗=3i^−αj^+k^\vec{a} = 3\hat{i} - \alpha\hat{j} + \hat{k} and b⃗=i^+αj^+3k^\vec{b} = \hat{i} + \alpha\hat{j} + 3\hat{k}. If the area of the parallelogram whose adjacent sides are represented by the vectors a⃗\vec{a} and b⃗\vec{b}, is 838\sqrt{3} sq. units, then a⃗⋅b⃗\vec{a} \cdot \vec{b} is equal to

[Q127 · 9th May Shift 2 · 2023]

Example 4VectorsMODERATE
If the area of the parallelogram with a⃗\vec{a} and b⃗\vec{b} as two adjacent sides is 15 square units, then the area (in square units) of the parallelogram, having 3a⃗+2b⃗3\vec{a}+2\vec{b} and a⃗+3b⃗\vec{a}+3\vec{b} as two adjacent sides, is

[Q142 · 9th May Shift 1 · 2023]

Example 5VectorsEASY
The number of unit vectors perpendicular to a⃗=(1,1,0)\vec{a}=(1,1,0) and b⃗=(0,1,1)\vec{b}=(0,1,1) is

[Q134 · 11th May Shift 2 · 2023]

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