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MHT-CET Maths · Vectors

Scalar Triple Product, Coplanarity, and Volume

The single number [a b c] = a·(b×c) — the signed volume of the box on three vectors. It is zero exactly when they are coplanar, its modulus is the parallelepiped volume, and it powers the chapter's hardest pool: volumes, coplanarity, and the vector triple product.

All Vectors notesMHT-CET Maths strategy

Why this matters

At 56 PYQs this is the chapter's BIGGEST subtopic and its hardest — about 77% are rated HARD. The scalar triple product [a b c] is the workhorse: it is the signed volume of the parallelepiped, the modulus is the actual volume, one-sixth of it is the tetrahedron volume, and it vanishes precisely when the three vectors are coplanar. The same determinant drives parameter-finding (coplanarity), min/max volume problems, and the linearity identities like [a+b b+c c+a] = 2[a b c]; the vector triple product (BAC-CAB rule) finishes the set with angle and orthogonal-coplanar problems. Master the determinant, the cyclic/sign rules, and BAC-CAB and the chapter's HARD tail collapses.

Concept 1 of 11

The scalar triple product — dot-cross and determinant form

Intuition

The scalar triple product takes THREE vectors and returns a single NUMBER: dot one vector into the cross product of the other two. You almost never compute it from the geometry — you write the three vectors as the rows of a 3x3 determinant and evaluate it. Geometrically that number is the signed volume of the box (parallelepiped) built on the three vectors.

Definition

For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}, c⃗=c1i^+c2j^+c3k^\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}:

  • Notation: [a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) — a single scalar
  • Determinant form (the workhorse): [a⃗ b⃗ c⃗]=∣a1a2a3b1b2b3c1c2c3∣[\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}
  • Dot-cross interchange: a⃗⋅(b⃗×c⃗)=(a⃗×b⃗)⋅c⃗\vec{a}\cdot(\vec{b}\times\vec{c}) = (\vec{a}\times\vec{b})\cdot\vec{c} — the dot and cross can swap places without changing the value
  • Geometric meaning: [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] is the SIGNED volume of the parallelepiped with edge vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}

Scalar triple product as a determinant

[a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)=∣a1a2a3b1b2b3c1c2c3∣[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}
  • Rowsthe components of a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} in order
  • (a⃗×b⃗)⋅c⃗(\vec{a}\times\vec{b})\cdot\vec{c}equal value — dot and cross interchange
  • [i^ j^ k^][\hat{i}\ \hat{j}\ \hat{k}]=1= 1, the unit box

Worked example

Evaluate [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] for a⃗=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k}, b⃗=2i^−j^+3k^\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} and c⃗=i^+k^\vec{c} = \hat{i} + \hat{k}.
  1. Stack the components as rows: ∣1212−13101∣\begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ 1 & 0 & 1 \end{vmatrix}.
  2. Expand along the first row: 1⋅[(−1)(1)−(3)(0)]−2⋅[(2)(1)−(3)(1)]+1⋅[(2)(0)−(−1)(1)]1\cdot[(-1)(1) - (3)(0)] - 2\cdot[(2)(1) - (3)(1)] + 1\cdot[(2)(0) - (-1)(1)].
  3. Simplify: 1⋅(−1)−2⋅(−1)+1⋅(1)=−1+2+1=21\cdot(-1) - 2\cdot(-1) + 1\cdot(1) = -1 + 2 + 1 = 2.
Answer:[a⃗ b⃗ c⃗]=2[\vec{a}\ \vec{b}\ \vec{c}] = 2
Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

  1. 1.
    Write [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] as a dot-cross.
  2. 2.
    [i^ j^ k^]=?[\hat{i}\ \hat{j}\ \hat{k}] = ?
  3. 3.
    a⃗⋅(b⃗×c⃗)\vec{a}\cdot(\vec{b}\times\vec{c}) equals which other dot-cross?
  4. 4.
    The scalar triple product is a number or a vector?

The scalar triple product is a SCALAR

[a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) is a single number. The vector triple product a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) is a VECTOR. Identify which one the question asks for before choosing an identity.

Dot and cross can swap, but keep the order of the three vectors

a⃗⋅(b⃗×c⃗)=(a⃗×b⃗)⋅c⃗\vec{a}\cdot(\vec{b}\times\vec{c}) = (\vec{a}\times\vec{b})\cdot\vec{c} — moving the dot/cross is free. But swapping two of the three vectors themselves flips the sign (see the cyclic-and-sign concept next).

Concept 2 of 11

Cyclic and sign properties of the scalar triple product

Intuition

Rotating the three vectors round in a cycle leaves the scalar triple product unchanged; swapping any two of them flips its sign. And if any vector repeats, the box is flat — the product is zero. These three rules let you simplify almost any [a b c] expression before touching a determinant.

Definition

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

  • Cyclic (rotation is free): [a⃗ b⃗ c⃗]=[b⃗ c⃗ a⃗]=[c⃗ a⃗ b⃗][\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}]
  • Swap flips the sign: [a⃗ b⃗ c⃗]=−[b⃗ a⃗ c⃗]=−[a⃗ c⃗ b⃗][\vec{a}\ \vec{b}\ \vec{c}] = -[\vec{b}\ \vec{a}\ \vec{c}] = -[\vec{a}\ \vec{c}\ \vec{b}]
  • Repeated vector ⇒ zero: [a⃗ a⃗ b⃗]=0[\vec{a}\ \vec{a}\ \vec{b}] = 0 (two equal rows make the determinant vanish)
  • Linearity in each slot: [αu⃗+βv⃗  b⃗  c⃗]=α[u⃗ b⃗ c⃗]+β[v⃗ b⃗ c⃗][\alpha\vec{u}+\beta\vec{v}\ \ \vec{b}\ \ \vec{c}] = \alpha[\vec{u}\ \vec{b}\ \vec{c}] + \beta[\vec{v}\ \vec{b}\ \vec{c}]

Cyclic and swap rules

[a⃗ b⃗ c⃗]=[b⃗ c⃗ a⃗]=[c⃗ a⃗ b⃗]=−[b⃗ a⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}] = -[\vec{b}\ \vec{a}\ \vec{c}]
  • Cyclic rotationa⃗→b⃗→c⃗→a⃗\vec{a}\to\vec{b}\to\vec{c}\to\vec{a}: value unchanged
  • One swapvalue negated
  • Repeated rowvalue =0= 0

Worked example

Simplify [a⃗ b⃗ c⃗]+[b⃗ c⃗ a⃗]−[b⃗ a⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] + [\vec{b}\ \vec{c}\ \vec{a}] - [\vec{b}\ \vec{a}\ \vec{c}].
  1. Cyclic: [b⃗ c⃗ a⃗]=[a⃗ b⃗ c⃗][\vec{b}\ \vec{c}\ \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}].
  2. One swap: [b⃗ a⃗ c⃗]=−[a⃗ b⃗ c⃗][\vec{b}\ \vec{a}\ \vec{c}] = -[\vec{a}\ \vec{b}\ \vec{c}], so −[b⃗ a⃗ c⃗]=+[a⃗ b⃗ c⃗]-[\vec{b}\ \vec{a}\ \vec{c}] = +[\vec{a}\ \vec{b}\ \vec{c}].
  3. Total: [a⃗ b⃗ c⃗]+[a⃗ b⃗ c⃗]+[a⃗ b⃗ c⃗]=3[a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] + [\vec{a}\ \vec{b}\ \vec{c}] + [\vec{a}\ \vec{b}\ \vec{c}] = 3[\vec{a}\ \vec{b}\ \vec{c}].
Answer:3[a⃗ b⃗ c⃗]3[\vec{a}\ \vec{b}\ \vec{c}]
Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

  1. 1.
    [b⃗ c⃗ a⃗]=?[\vec{b}\ \vec{c}\ \vec{a}] = ? in terms of [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}].
  2. 2.
    [a⃗ c⃗ b⃗]=?[\vec{a}\ \vec{c}\ \vec{b}] = ?
  3. 3.
    [a⃗ a⃗ c⃗]=?[\vec{a}\ \vec{a}\ \vec{c}] = ?
  4. 4.
    [2a⃗ b⃗ c⃗]=?[2\vec{a}\ \vec{b}\ \vec{c}] = ?

Cyclic keeps the value; ANY single swap negates it

[a⃗ b⃗ c⃗]=[b⃗ c⃗ a⃗][\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] (cyclic, same value), but [a⃗ c⃗ b⃗]=−[a⃗ b⃗ c⃗][\vec{a}\ \vec{c}\ \vec{b}] = -[\vec{a}\ \vec{b}\ \vec{c}] (one swap, flipped). Lose track of a swap and your final λ\lambda comes out with the wrong sign.

A repeated vector kills the product — spot it early

Whenever expanding by linearity, any term that ends up with two identical vectors (like [a⃗ a⃗ c⃗][\vec{a}\ \vec{a}\ \vec{c}]) is zero. Most coplanarity-combo problems are solved entirely by deleting these zero terms.

Concept 3 of 11

Computing the value of a scalar triple product

Intuition

Sometimes you compute [a b c] not from raw components but from magnitudes and a perpendicularity setup, or you must decide which variables it depends on. When one vector is perpendicular to the plane of the other two, the box is a clean prism and [a⃗ b⃗ c⃗]=∣a⃗∣ ∣b⃗×c⃗∣[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}|. When the determinant simplifies to a constant, the answer is independent of the parameters in the vectors.

Definition

  • Perpendicular case: if a⃗\vec{a} is perpendicular to both b⃗\vec{b} and c⃗\vec{c}, then a⃗\vec{a} is parallel to b⃗×c⃗\vec{b}\times\vec{c}, so [a⃗ b⃗ c⃗]=±∣a⃗∣ ∣b⃗×c⃗∣=±∣a⃗∣ ∣b⃗∣∣c⃗∣sin⁡θ[\vec{a}\ \vec{b}\ \vec{c}] = \pm|\vec{a}|\,|\vec{b}\times\vec{c}| = \pm|\vec{a}|\,|\vec{b}||\vec{c}|\sin\theta where θ\theta is the angle between b⃗\vec{b} and c⃗\vec{c}.
  • Independence: if the determinant of the component matrix reduces to a constant, [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] does not depend on the parameters inside the vectors.
  • Magnitude relations: a datum like ∣b⃗×c⃗∣|\vec{b}\times\vec{c}| fixes sin⁡θ\sin\theta, and a relation b⃗=2c⃗+λa⃗\vec{b} = 2\vec{c} + \lambda\vec{a} is solved by taking magnitudes: ∣b⃗−2c⃗∣2=λ2∣a⃗∣2|\vec{b} - 2\vec{c}|^2 = \lambda^2|\vec{a}|^2.

STP when one vector is perpendicular to the other two

[a⃗ b⃗ c⃗]=∣a⃗∣ ∣b⃗×c⃗∣=∣a⃗∣∣b⃗∣∣c⃗∣sin⁡θ[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}| = |\vec{a}||\vec{b}||\vec{c}|\sin\theta
  • a⃗⊥b⃗,a⃗⊥c⃗\vec{a}\perp\vec{b}, \vec{a}\perp\vec{c}so a⃗ ∥ b⃗×c⃗\vec{a}\,\|\,\vec{b}\times\vec{c}
  • θ\thetaangle between b⃗\vec{b} and c⃗\vec{c}
  • sin⁡θ\sin\thetafrom ∣b⃗×c⃗∣=∣b⃗∣∣c⃗∣sin⁡θ|\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\sin\theta

Worked example

a⃗\vec{a} is perpendicular to both b⃗\vec{b} and c⃗\vec{c}. If ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=2|\vec{b}| = 2, ∣c⃗∣=5|\vec{c}| = 5 and the angle between b⃗\vec{b} and c⃗\vec{c} is π6\frac{\pi}{6}, find [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}].
  1. a⃗⊥b⃗\vec{a}\perp\vec{b} and a⃗⊥c⃗\vec{a}\perp\vec{c} means a⃗\vec{a} is parallel to b⃗×c⃗\vec{b}\times\vec{c}.
  2. So [a⃗ b⃗ c⃗]=∣a⃗∣ ∣b⃗×c⃗∣=∣a⃗∣∣b⃗∣∣c⃗∣sin⁡θ[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}| = |\vec{a}||\vec{b}||\vec{c}|\sin\theta.
  3. =3⋅2⋅5⋅sin⁡π6=30⋅12=15= 3\cdot 2\cdot 5\cdot\sin\frac{\pi}{6} = 30\cdot\tfrac{1}{2} = 15.
Answer:[a⃗ b⃗ c⃗]=15[\vec{a}\ \vec{b}\ \vec{c}] = 15
Practice this conceptself-check · 4 quick reps

Try it yourself

a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are mutually perpendicular with magnitudes 2,3,42, 3, 4. Find [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] (take it positive).

Practice — Level 1 (4 reps)

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

  1. 1.
    If a⃗⊥b⃗\vec{a}\perp\vec{b} and a⃗⊥c⃗\vec{a}\perp\vec{c}, then [a⃗ b⃗ c⃗]=?[\vec{a}\ \vec{b}\ \vec{c}] = ?
  2. 2.
    For mutually perpendicular a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, [a⃗ b⃗ c⃗]=?[\vec{a}\ \vec{b}\ \vec{c}] = ?
  3. 3.
    If a determinant reduces to a constant, the STP depends on its parameters?
  4. 4.
    ∣b⃗×c⃗∣=∣b⃗∣∣c⃗∣ ?|\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\,?

From the bank · past-year question

Example 3VectorsMODERATE
If a⃗\vec{a} is perpendicular to b⃗\vec{b} and c⃗\vec{c}, ∣a⃗∣=2|\vec{a}|=2, ∣b⃗∣=3|\vec{b}|=3, ∣c⃗∣=4|\vec{c}|=4 and the angle between b⃗\vec{b} and c⃗\vec{c} is π3\frac{\pi}{3}, then [a⃗ b⃗ c⃗]=[\vec{a}\,\vec{b}\,\vec{c}] =

[Q117 · 16th May Shift 2 · 2023]

Perpendicular to BOTH means parallel to the cross product

If a⃗⊥b⃗\vec{a}\perp\vec{b} and a⃗⊥c⃗\vec{a}\perp\vec{c}, then a⃗\vec{a} lies along b⃗×c⃗\vec{b}\times\vec{c}, so the box is a right prism and [a⃗ b⃗ c⃗]=∣a⃗∣ ∣b⃗×c⃗∣[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}|. Don't try to plug a single sin⁡\sin of the angle between a⃗\vec{a} and b⃗\vec{b} — the relevant angle is between b⃗\vec{b} and c⃗\vec{c}.

"Depends on x and y" — expand the determinant first

When the vectors carry parameters, compute [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] as a determinant and simplify. It frequently collapses to a constant, meaning the answer is independent of every parameter — a deliberately surprising option.
Drill 7 more on computing the value of a scalar triple product

Concept 4 of 11

Coplanarity of three vectors (and solving for a parameter)

Intuition

Three vectors lie in a common plane through the origin exactly when the box on them is flat — zero volume. So the test is simply [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0: set the determinant to zero and solve for the unknown. It also detects how many values of a parameter make them coplanar, and a vector lying in the plane of two others satisfies the same equation.

Definition

a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are coplanar   ⟺  [a⃗ b⃗ c⃗]=0\iff [\vec{a}\ \vec{b}\ \vec{c}] = 0, i.e. the determinant of their components is zero. If c⃗\vec{c} lies in the plane of a⃗\vec{a} and b⃗\vec{b}, then [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0 too. To find a parameter λ\lambda (or xx) that makes them coplanar, set the determinant equal to zero and solve the resulting equation; a parameter that appears squared can give two distinct real values.

Coplanarity criterion

a⃗,b⃗,c⃗ coplanar  ⟺  ∣a1a2a3b1b2b3c1c2c3∣=0\vec{a},\vec{b},\vec{c}\text{ coplanar} \iff \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0
  • [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0zero box volume ⇒ all three lie in one plane
  • Parametersolve the determinant-equals-zero equation for it

Worked example

Find λ\lambda so that a⃗=i^+j^−k^\vec{a} = \hat{i} + \hat{j} - \hat{k}, b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k} and c⃗=i^+λj^+3k^\vec{c} = \hat{i} + \lambda\hat{j} + 3\hat{k} are coplanar.
  1. Coplanar ⇒ [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0: ∣11−12−111λ3∣=0\begin{vmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 1 & \lambda & 3 \end{vmatrix} = 0.
  2. Expand along the first row: 1⋅(−3−λ)−1⋅(6−1)+(−1)⋅(2λ+1)1\cdot(-3 - \lambda) - 1\cdot(6 - 1) + (-1)\cdot(2\lambda + 1).
  3. Simplify: (−3−λ)−5−(2λ+1)=−9−3λ=0(-3 - \lambda) - 5 - (2\lambda + 1) = -9 - 3\lambda = 0.
  4. Solve: λ=−3\lambda = -3.
Answer:λ=−3\lambda = -3
Practice this conceptself-check · 4 quick reps

Try it yourself

For how many real values of λ\lambda are λi^+j^\lambda\hat{i} + \hat{j}, i^+λj^\hat{i} + \lambda\hat{j} and i^+j^+λk^\hat{i} + \hat{j} + \lambda\hat{k} coplanar?

Practice — Level 1 (4 reps)

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

  1. 1.
    Three vectors are coplanar iff [a⃗ b⃗ c⃗]=?[\vec{a}\ \vec{b}\ \vec{c}] = ?
  2. 2.
    If c⃗\vec{c} lies in the plane of a⃗,b⃗\vec{a}, \vec{b}, then [a⃗ b⃗ c⃗]=?[\vec{a}\ \vec{b}\ \vec{c}] = ?
  3. 3.
    To find a coplanarity parameter, set the determinant equal to?
  4. 4.
    [i^ j^ (i^+j^)]=?[\hat{i}\ \hat{j}\ (\hat{i}+\hat{j})] = ? — coplanar?

From the bank · past-year question

Example 4VectorsMODERATE
If the vectors pi^+j^+k^,i^+qj^+k^p\hat{i}+\hat{j}+\hat{k},\hat{i}+q\hat{j}+\hat{k} and i^+j^+rk^ (p≠q≠r≠1)\hat{i}+\hat{j}+r\hat{k}\ (p\neq q\neq r\neq1) are coplanar, then the value of pqr−(p+q+r)pqr-(p+q+r) is

[Q119 · 13th May Shift 2 · 2024]

Coplanar ⇒ STP = 0, NOT "two of them are parallel"

Three vectors coplanar means they fit in one plane; they need not be parallel to each other. A parallel pair also makes the STP zero, but it is a stronger, separate condition — don't confuse the two.

A squared parameter can give TWO coplanarity values

When the unknown appears as λ2\lambda^2 (e.g. −λ2i^+…-\lambda^2\hat{i}+\dots), the determinant-equals-zero equation can have two distinct real roots. Count carefully — "number of values of λ\lambda" questions hinge exactly on this.
Drill 3 more on coplanarity of three vectors (and solving for a parameter)

Concept 5 of 11

Scalar triple product of linear combinations

Intuition

When the three slots of a scalar triple product are themselves sums of a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, expand by linearity — every term with a repeated vector vanishes, leaving a clean multiple of [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}]. The pattern [a⃗+b⃗  b⃗+c⃗  c⃗+a⃗]=2[a⃗ b⃗ c⃗][\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] is the signature case. More generally, the coefficient is a 3x3 determinant of the combination coefficients.

Definition

Expanding by linearity and deleting repeated-vector terms:

  • [a⃗+b⃗  b⃗+c⃗  c⃗+a⃗]=2[a⃗ b⃗ c⃗][\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}]
  • [ma⃗+b⃗  mb⃗+c⃗  mc⃗+a⃗]=(m3+1)[a⃗ b⃗ c⃗][m\vec{a}+\vec{b}\ \ m\vec{b}+\vec{c}\ \ m\vec{c}+\vec{a}] = (m^3 + 1)[\vec{a}\ \vec{b}\ \vec{c}]
  • In general the new STP equals det⁡(M) [a⃗ b⃗ c⃗]\det(M)\,[\vec{a}\ \vec{b}\ \vec{c}], where MM is the 3x3 matrix of the combination coefficients.

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are coplanar, [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0 so every such combination is also zero.

Linear-combination identities

[a⃗+b⃗  b⃗+c⃗  c⃗+a⃗]=2[a⃗ b⃗ c⃗][ma⃗+b⃗  mb⃗+c⃗  mc⃗+a⃗]=(m3+1)[a⃗ b⃗ c⃗][\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \qquad [m\vec{a}+\vec{b}\ \ m\vec{b}+\vec{c}\ \ m\vec{c}+\vec{a}] = (m^3+1)[\vec{a}\ \vec{b}\ \vec{c}]
  • Repeated-vector termsall vanish on expansion
  • Coefficient=det⁡= \det of the combination-coefficient matrix

Worked example

Simplify [a⃗+b⃗  b⃗+c⃗  c⃗+a⃗][\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] in terms of [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}].
  1. Expand the first slot by linearity: [a⃗  b⃗+c⃗  c⃗+a⃗]+[b⃗  b⃗+c⃗  c⃗+a⃗][\vec{a}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] + [\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}].
  2. Keep expanding and delete every term with a repeated vector (e.g. [a⃗ c⃗ c⃗]=0[\vec{a}\ \vec{c}\ \vec{c}] = 0, [b⃗ b⃗ … ]=0[\vec{b}\ \vec{b}\ \dots] = 0).
  3. The surviving terms are [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] and [b⃗ c⃗ a⃗]=[a⃗ b⃗ c⃗][\vec{b}\ \vec{c}\ \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}], giving 2[a⃗ b⃗ c⃗]2[\vec{a}\ \vec{b}\ \vec{c}].
Answer:2[a⃗ b⃗ c⃗]2[\vec{a}\ \vec{b}\ \vec{c}]
Practice this conceptself-check · 4 quick reps

Try it yourself

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are coplanar unit vectors, find [2a⃗−b⃗  2b⃗−c⃗  2c⃗−a⃗][2\vec{a}-\vec{b}\ \ 2\vec{b}-\vec{c}\ \ 2\vec{c}-\vec{a}].

Practice — Level 1 (4 reps)

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

  1. 1.
    [a⃗+b⃗  b⃗+c⃗  c⃗+a⃗]=?[\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = ?
  2. 2.
    [3a⃗+b⃗  3b⃗+c⃗  3c⃗+a⃗]=?[3\vec{a}+\vec{b}\ \ 3\vec{b}+\vec{c}\ \ 3\vec{c}+\vec{a}] = ?
  3. 3.
    If a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} coplanar, [2a⃗−b⃗ … ]=?[2\vec{a}-\vec{b}\ \dots] = ?
  4. 4.
    A term like [a⃗ a⃗ c⃗][\vec{a}\ \vec{a}\ \vec{c}] contributes what to the expansion?

From the bank · past-year question

Example 5VectorsHARD
If a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}, c⃗=c1i^+c2j^+c3k^\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k} and [3a⃗+b⃗,3b⃗+c⃗,3c⃗+a⃗]=λ[a⃗ b⃗ c⃗][3\vec{a}+\vec{b}, 3\vec{b}+\vec{c}, 3\vec{c}+\vec{a}] = \lambda[\vec{a}\,\vec{b}\,\vec{c}], then the value of λ\lambda is

[Q129 · 16th May Shift 2 · 2023]

[ma⃗+b⃗ … ]=(m3+1)[a⃗ b⃗ c⃗][m\vec{a}+\vec{b}\ \dots] = (m^3+1)[\vec{a}\ \vec{b}\ \vec{c}], not m3m^3

The cyclic m ⋅m\,\cdot pattern gives m3+1m^3 + 1 — the extra +1+1 comes from the [b⃗ c⃗ a⃗][\vec{b}\ \vec{c}\ \vec{a}] cross-term. So m=3m = 3 gives 2828, not 2727. Dropping the +1+1 is a deliberate distractor.

Track every sign through the swaps

Expansions like [a⃗+2b⃗+3c⃗ … ][\vec{a}+2\vec{b}+3\vec{c}\ \dots] generate many cross-terms; a single mis-signed swap throws off the coefficient. Use cyclic to standardise every surviving term to [a⃗ b⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] before summing.
Drill 8 more on scalar triple product of linear combinations

Concept 6 of 11

Volume of a parallelepiped (and min/max problems)

Intuition

The volume of the box (parallelepiped) on three edge vectors is the MODULUS of their scalar triple product — the sign just tells you the orientation. Set the determinant equal to a given volume to find an unknown component, or differentiate the determinant in a parameter to find where the volume is smallest or largest.

Definition

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

  • Volume =∣[a⃗ b⃗ c⃗]∣=∣det⁡(component matrix)∣= |[\vec{a}\ \vec{b}\ \vec{c}]| = \left|\det\text{(component matrix)}\right|.
  • Find a component: set ∣[a⃗ b⃗ c⃗]∣=V|[\vec{a}\ \vec{b}\ \vec{c}]| = V (given) and solve.
  • Min/max in a parameter: write the determinant as a function V(m)V(m), set V′(m)=0V'(m) = 0, and use the sign of V′′(m)V''(m) to classify minimum vs maximum.
  • The dihedral angle between two faces of a tetrahedron is the angle between the two face normals, each found as a cross product of edge vectors.

Parallelepiped volume

V=∣[a⃗ b⃗ c⃗]∣=∣∣a1a2a3b1b2b3c1c2c3∣∣V = |[\vec{a}\ \vec{b}\ \vec{c}]| = \left|\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\right|
  • ∣⋅∣|\cdot|modulus — volume is always non-negative
  • V′(m)=0V'(m)=0stationary volume; V′′V'' sign decides min/max

Diagram · triple product = box volume (SVG, drag to rotate)

abc
[a b c] = a · (b × c) = 13.22 (signed volume of the box)

The box spanned by a, b, c has volume |[a b c]|. Painter's-ordered faces fake the solidity — edges don't truly hide behind nearer faces, which is the SVG limit this comparison is testing.

Worked example

Find the value of mm for which the volume of the parallelepiped on i^+mj^+k^\hat{i}+m\hat{j}+\hat{k}, j^+mk^\hat{j}+m\hat{k} and mi^+k^m\hat{i}+\hat{k} is minimum.
  1. Volume function: ∣1m101mm01∣=1(1)−m(0−m2)+1(0−m)=m3−m+1\begin{vmatrix} 1 & m & 1 \\ 0 & 1 & m \\ m & 0 & 1 \end{vmatrix} = 1(1) - m(0 - m^2) + 1(0 - m) = m^3 - m + 1.
  2. Take V(m)=m3−m+1V(m) = m^3 - m + 1; set V′(m)=3m2−1=0⇒m=±13V'(m) = 3m^2 - 1 = 0 \Rightarrow m = \pm\frac{1}{\sqrt{3}}.
  3. V′′(m)=6mV''(m) = 6m: positive at m=13m = \frac{1}{\sqrt{3}}, so that gives the minimum.
Answer:m=13m = \dfrac{1}{\sqrt{3}}
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the volume of the parallelepiped with edges a⃗=i^\vec{a} = \hat{i}, b⃗=j^\vec{b} = \hat{j} and c⃗=i^+j^+2k^\vec{c} = \hat{i} + \hat{j} + 2\hat{k}.

Practice — Level 1 (4 reps)

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

  1. 1.
    Volume of the parallelepiped on a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}?
  2. 2.
    If [a⃗ b⃗ c⃗]=−7[\vec{a}\ \vec{b}\ \vec{c}] = -7, the volume is?
  3. 3.
    To find a min/max volume in mm, set what to zero?
  4. 4.
    Dihedral angle between two faces = angle between their?

From the bank · past-year question

Example 6VectorsHARD
The volume of parallelopiped formed by vectors i^+mj^+k^\hat{i} + m\hat{j} + \hat{k}, j^+mk^\hat{j} + m\hat{k} and mi^+k^m\hat{i} + \hat{k} becomes minimum when mm is

[Q113 · 2nd May Shift 1 · 2023]

Volume is the MODULUS — never a negative number

The scalar triple product can be negative (it is a SIGNED volume), but a physical volume is ∣[a⃗ b⃗ c⃗]∣|[\vec{a}\ \vec{b}\ \vec{c}]|. When a volume is given (e.g. 158 cu units), set the modulus equal to it, which may give two parameter values ±\pm.

Min vs max: check the second derivative

V′(m)=0V'(m) = 0 at m=±13m = \pm\frac{1}{\sqrt{3}}, but only one is a minimum. V′′(m)=6m>0V''(m) = 6m > 0 at m=+13m = +\frac{1}{\sqrt{3}} (minimum) and <0< 0 at m=−13m = -\frac{1}{\sqrt{3}} (maximum). The two questions share roots but want opposite answers.
Drill 6 more on volume of a parallelepiped (and min/max problems)

Concept 7 of 11

Volume of a tetrahedron

Intuition

A tetrahedron is one-sixth of the parallelepiped on the same three edge vectors. So its volume is one-sixth the modulus of the scalar triple product of the three edges meeting at one vertex. Build the edges by subtracting the chosen vertex from the other three, then apply the formula.

Definition

For a tetrahedron with vertices A,B,C,DA, B, C, D, form the edge vectors from one vertex: AB→,AC→,AD→\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}. Then Volume =16 ∣[AB→ AC→ AD→]∣= \dfrac{1}{6}\,|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]|. Setting this equal to a given volume yields an equation for an unknown coordinate.

Tetrahedron volume

V=16 ∣[AB→ AC→ AD→]∣V = \tfrac{1}{6}\,\bigl|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\bigr|
  • 16\tfrac{1}{6}a tetrahedron is one-sixth of the parallelepiped
  • AB→,AC→,AD→\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}three edges from the SAME vertex AA

Worked example

Find the volume of the tetrahedron with vertices A(0,0,0)A(0,0,0), B(2,0,0)B(2,0,0), C(0,3,0)C(0,3,0) and D(0,0,4)D(0,0,4).
  1. Edges from AA: AB→=2i^\overrightarrow{AB} = 2\hat{i}, AC→=3j^\overrightarrow{AC} = 3\hat{j}, AD→=4k^\overrightarrow{AD} = 4\hat{k}.
  2. [AB→ AC→ AD→]=∣200030004∣=24[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{vmatrix} = 24.
  3. Volume =16⋅24=4= \tfrac{1}{6}\cdot 24 = 4.
Answer:Volume =4= 4 cubic units
Practice this conceptself-check · 4 quick reps

Try it yourself

The volume of the tetrahedron with vertices A(1,1,1)A(1,1,1), B(2,1,1)B(2,1,1), C(1,3,1)C(1,3,1), D(1,1,5)D(1,1,5) is?

Practice — Level 1 (4 reps)

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

  1. 1.
    Tetrahedron volume from edges a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}?
  2. 2.
    A tetrahedron is what fraction of the parallelepiped on the same edges?
  3. 3.
    If [AB→ AC→ AD→]=30[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = 30, the volume is?
  4. 4.
    From vertices A,B,C,DA,B,C,D, which three edges do you use?

From the bank · past-year question

Example 7VectorsHARD
If the volume of tetrahedron whose vertices are A≡(1,−6,10)A\equiv(1,-6,10), B≡(−1,−3,7)B\equiv(-1,-3,7), C≡(5,−1,k)C\equiv(5,-1,k) and D≡(7,−4,7)D\equiv(7,-4,7) is 11 cu. units, then the value of k is

[Q141 · 11th May Shift 1 · 2023]

The one-sixth is on the tetrahedron, not the parallelepiped

Tetrahedron volume =16∣[a⃗ b⃗ c⃗]∣= \tfrac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]|; the parallelepiped is the full ∣[a⃗ b⃗ c⃗]∣|[\vec{a}\ \vec{b}\ \vec{c}]|. Forgetting the 16\tfrac{1}{6} over-counts the volume six-fold — a classic distractor when a volume is given.

Build all three edges from the SAME vertex

Use AB→,AC→,AD→\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD} — all starting at AA. Mixing base points (e.g. AB→,BC→,…\overrightarrow{AB}, \overrightarrow{BC}, \dots) gives the wrong determinant and a wrong unknown.
Drill 1 more on volume of a tetrahedron

Concept 8 of 11

Reciprocal-basis identities and the STP-squared rule

Intuition

From a non-coplanar triple a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, the vectors b⃗×c⃗[a⃗ b⃗ c⃗]\dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} and its cyclic partners form a "reciprocal" set that dots cleanly with the originals: each matched pair gives 11, each mismatched pair gives 00. That single fact collapses long-looking sums to small integers. A close relative is the identity [a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=[a⃗ b⃗ c⃗]2[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2.

Definition

For non-coplanar a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, set p⃗=b⃗×c⃗[a⃗ b⃗ c⃗]\vec{p} = \dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]}, q⃗=c⃗×a⃗[a⃗ b⃗ c⃗]\vec{q} = \dfrac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]}, r⃗=a⃗×b⃗[a⃗ b⃗ c⃗]\vec{r} = \dfrac{\vec{a}\times\vec{b}}{[\vec{a}\ \vec{b}\ \vec{c}]}:

  • Matched pairs: a⃗⋅p⃗=b⃗⋅q⃗=c⃗⋅r⃗=1\vec{a}\cdot\vec{p} = \vec{b}\cdot\vec{q} = \vec{c}\cdot\vec{r} = 1
  • Mismatched pairs: a⃗⋅q⃗=a⃗⋅r⃗=b⃗⋅p⃗=⋯=0\vec{a}\cdot\vec{q} = \vec{a}\cdot\vec{r} = \vec{b}\cdot\vec{p} = \dots = 0 (the cross product is perpendicular to its own factors)
  • STP-squared rule: [a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=[a⃗ b⃗ c⃗]2[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2

Reciprocal pairings and STP-squared

a⃗⋅p⃗=b⃗⋅q⃗=c⃗⋅r⃗=1,a⃗⋅q⃗=b⃗⋅p⃗=⋯=0[a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=[a⃗ b⃗ c⃗]2\vec{a}\cdot\vec{p} = \vec{b}\cdot\vec{q} = \vec{c}\cdot\vec{r} = 1, \quad \vec{a}\cdot\vec{q} = \vec{b}\cdot\vec{p} = \dots = 0 \qquad [\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2
  • Matcheda⃗⋅p⃗=1\vec{a}\cdot\vec{p} = 1 etc.
  • Mismatcheda⃗⋅q⃗=0\vec{a}\cdot\vec{q} = 0 (perpendicularity)
  • STP-squaredcross-of-pairs box =[a⃗ b⃗ c⃗]2= [\vec{a}\ \vec{b}\ \vec{c}]^2

Worked example

With p⃗=b⃗×c⃗[a⃗ b⃗ c⃗]\vec{p} = \dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]}, q⃗=c⃗×a⃗[a⃗ b⃗ c⃗]\vec{q} = \dfrac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]}, r⃗=a⃗×b⃗[a⃗ b⃗ c⃗]\vec{r} = \dfrac{\vec{a}\times\vec{b}}{[\vec{a}\ \vec{b}\ \vec{c}]}, find a⃗⋅p⃗+b⃗⋅q⃗+c⃗⋅r⃗\vec{a}\cdot\vec{p} + \vec{b}\cdot\vec{q} + \vec{c}\cdot\vec{r}.
  1. a⃗⋅p⃗=a⃗⋅(b⃗×c⃗)[a⃗ b⃗ c⃗]=[a⃗ b⃗ c⃗][a⃗ b⃗ c⃗]=1\vec{a}\cdot\vec{p} = \dfrac{\vec{a}\cdot(\vec{b}\times\vec{c})}{[\vec{a}\ \vec{b}\ \vec{c}]} = \dfrac{[\vec{a}\ \vec{b}\ \vec{c}]}{[\vec{a}\ \vec{b}\ \vec{c}]} = 1.
  2. By the same cyclic argument, b⃗⋅q⃗=1\vec{b}\cdot\vec{q} = 1 and c⃗⋅r⃗=1\vec{c}\cdot\vec{r} = 1.
  3. Sum =1+1+1=3= 1 + 1 + 1 = 3.
Answer:33
Practice this conceptself-check · 4 quick reps

Try it yourself

With the same p⃗,q⃗,r⃗\vec{p}, \vec{q}, \vec{r}, evaluate a⃗⋅q⃗+b⃗⋅r⃗+c⃗⋅p⃗\vec{a}\cdot\vec{q} + \vec{b}\cdot\vec{r} + \vec{c}\cdot\vec{p}.

Practice — Level 1 (4 reps)

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

  1. 1.
    a⃗⋅b⃗×c⃗[a⃗ b⃗ c⃗]=?\vec{a}\cdot\dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} = ?
  2. 2.
    b⃗⋅b⃗×c⃗[a⃗ b⃗ c⃗]=?\vec{b}\cdot\dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} = ?
  3. 3.
    [a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=?[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = ?
  4. 4.
    2a⃗⋅p⃗+b⃗⋅q⃗+c⃗⋅r⃗=?2\vec{a}\cdot\vec{p} + \vec{b}\cdot\vec{q} + \vec{c}\cdot\vec{r} = ?

From the bank · past-year question

Example 8VectorsMODERATE
Let a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} be three non-coplanar vectors and p⃗,q⃗,r⃗\vec{p},\vec{q},\vec{r} defined by p⃗=b⃗×c⃗[a⃗ b⃗ c⃗], q⃗=c⃗×a⃗[a⃗ b⃗ c⃗], r⃗=a⃗×b⃗[a⃗ b⃗ c⃗]\vec{p}=\frac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]},\ \vec{q}=\frac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]},\ \vec{r}=\frac{\vec{a}\times\vec{b}}{[\vec{a}\ \vec{b}\ \vec{c}]}, then the value of (a⃗+b⃗)⋅p⃗+(b⃗+c⃗)⋅q⃗+(c⃗+a⃗)⋅r⃗(\vec{a}+\vec{b})\cdot\vec{p}+(\vec{b}+\vec{c})\cdot\vec{q}+(\vec{c}+\vec{a})\cdot\vec{r} is equal to

[Q148 · 3rd May 2nd Shift · 2023]

Matched pairs are 1, mismatched pairs are 0

a⃗⋅b⃗×c⃗[a⃗ b⃗ c⃗]=1\vec{a}\cdot\dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} = 1, but a⃗⋅c⃗×a⃗[a⃗ b⃗ c⃗]=0\vec{a}\cdot\dfrac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]} = 0 because c⃗×a⃗⊥a⃗\vec{c}\times\vec{a}\perp\vec{a}. Read each dot pair to decide whether it survives — most of the terms in these sums vanish.

Cross-of-pairs box is the SQUARE, not the cube

[a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=[a⃗ b⃗ c⃗]2[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2 — so the coefficient λ\lambda in λ[a⃗ b⃗ c⃗]2\lambda[\vec{a}\ \vec{b}\ \vec{c}]^2 is 11, not 33 or 22.
Drill 3 more on reciprocal-basis identities and the stp-squared rule

Concept 9 of 11

Vector triple product (BAC-CAB rule)

Intuition

A triple product with TWO crosses returns a VECTOR, expanded by the 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}. Read it as "outer-dot-far times near, minus outer-dot-near times far". The result lies in the plane of the two innermost vectors — a structural fact that powers most of the chapter's angle problems.

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}
  • The vector triple product is NOT associative: a⃗×(b⃗×c⃗)≠(a⃗×b⃗)×c⃗\vec{a}\times(\vec{b}\times\vec{c}) \neq (\vec{a}\times\vec{b})\times\vec{c} in general — the two lie in different planes.

Equating a given expansion to a stated multiple of b⃗\vec{b} and c⃗\vec{c} lets you read off a⃗⋅b⃗\vec{a}\cdot\vec{b} (hence the angle), since b⃗,c⃗\vec{b}, \vec{c} are independent.

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
  • b⃗,c⃗\vec{b}, \vec{c}the plane the result lies in
  • Resultperpendicular to a⃗\vec{a}, inside the b⃗\vec{b}-c⃗\vec{c} plane

Worked example

Three unit vectors satisfy a⃗×(b⃗×c⃗)=12(b⃗+c⃗)\vec{a}\times(\vec{b}\times\vec{c}) = \dfrac{1}{2}(\vec{b} + \vec{c}) with b⃗\vec{b} not parallel to c⃗\vec{c}. Find the angle between a⃗\vec{a} and b⃗\vec{b}.
  1. BAC-CAB: (a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗=12b⃗+12c⃗(\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = \tfrac{1}{2}\vec{b} + \tfrac{1}{2}\vec{c}.
  2. Since b⃗,c⃗\vec{b}, \vec{c} are independent, match coefficients: a⃗⋅c⃗=12\vec{a}\cdot\vec{c} = \tfrac{1}{2} and −(a⃗⋅b⃗)=12-(\vec{a}\cdot\vec{b}) = \tfrac{1}{2}, so a⃗⋅b⃗=−12\vec{a}\cdot\vec{b} = -\tfrac{1}{2}.
  3. Unit vectors: cos⁡θ=a⃗⋅b⃗=−12\cos\theta = \vec{a}\cdot\vec{b} = -\tfrac{1}{2}, so θ=2π3\theta = \dfrac{2\pi}{3}.
Answer:θ=2π3\theta = \dfrac{2\pi}{3}
Practice this conceptself-check · 4 quick reps

Try it yourself

Using BAC-CAB, find a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) for a⃗=i^\vec{a} = \hat{i}, b⃗=j^\vec{b} = \hat{j}, c⃗=i^+k^\vec{c} = \hat{i} + \hat{k}.

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}) = ? (BAC-CAB)
  2. 2.
    a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) lies in the plane of?
  3. 3.
    Is the vector triple product associative?
  4. 4.
    If unit a⃗\vec{a} gives a⃗⋅b⃗=−12\vec{a}\cdot\vec{b} = -\tfrac{1}{2}, the angle between them is?

From the bank · past-year question

Example 9VectorsHARD
a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} are three unit vectors such that a⃗×(b⃗×c⃗)=32(b⃗+c⃗)\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{\sqrt{3}}{2}(\vec{b} + \vec{c}). If b⃗\vec{b} is not parallel to c⃗\vec{c}, then the angle between a⃗\vec{a} and b⃗\vec{b} is

[Q127 · 2nd May Shift 1 · 2023]

Inner pair sets the plane: a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) is in the b⃗\vec{b}-c⃗\vec{c} plane

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} lies in the plane of b⃗,c⃗\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} lies in the plane of a⃗,b⃗\vec{a}, \vec{b}. They are DIFFERENT vectors.

Match coefficients only when the basis vectors are independent

Reading off a⃗⋅b⃗\vec{a}\cdot\vec{b} by comparing coefficients of b⃗\vec{b} and c⃗\vec{c} is valid because "b⃗\vec{b} not parallel to c⃗\vec{c}" makes them independent. Watch the sign: the coefficient of c⃗\vec{c} is −(a⃗⋅b⃗)-(\vec{a}\cdot\vec{b}), so a positive RHS coefficient gives a NEGATIVE dot product (obtuse angle).

Two crosses ⇒ BAC-CAB; one cross + one dot ⇒ scalar triple product

If the shape is a⃗⋅(b⃗×c⃗)\vec{a}\cdot(\vec{b}\times\vec{c}) (a number), it is NOT a BAC-CAB case. Count the crosses and dots first to pick the right identity.
Drill 10 more on vector triple product (bac-cab rule)

Concept 10 of 11

Vector orthogonal to one vector and coplanar with two others

Intuition

A standard PYQ shape: find a unit vector that is perpendicular to a given vector AND lies in the plane of two others. The vector c⃗×(a⃗×b⃗)\vec{c}\times(\vec{a}\times\vec{b}) is exactly that — it is coplanar with a⃗,b⃗\vec{a}, \vec{b} (a BAC-CAB combination of them) and perpendicular to c⃗\vec{c}. Equivalently, write the unknown as v⃗=λa⃗+μb⃗\vec{v} = \lambda\vec{a} + \mu\vec{b} (coplanar) and impose the perpendicularity / projection condition.

Definition

To find a vector **coplanar with a⃗,b⃗\vec{a}, \vec{b} and perpendicular to c⃗\vec{c}**:

  • Shortcut: c⃗×(a⃗×b⃗)=(c⃗⋅b⃗)a⃗−(c⃗⋅a⃗)b⃗\vec{c}\times(\vec{a}\times\vec{b}) = (\vec{c}\cdot\vec{b})\vec{a} - (\vec{c}\cdot\vec{a})\vec{b} is coplanar with a⃗,b⃗\vec{a}, \vec{b} and perpendicular to c⃗\vec{c}; normalise it for a unit vector.
  • Linear-combination method: set v⃗=λa⃗+μb⃗\vec{v} = \lambda\vec{a} + \mu\vec{b} (this makes it coplanar with a⃗,b⃗\vec{a}, \vec{b}), then impose v⃗⋅c⃗=0\vec{v}\cdot\vec{c} = 0 (perpendicular) or a given projection v⃗⋅c⃗∣c⃗∣\dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} to fix the ratio λ:μ\lambda : \mu.
  • For three given POINTS, coplanarity of position-difference vectors is the same [⋯ ]=0[\cdots] = 0 test.

Orthogonal-and-coplanar vector

c⃗×(a⃗×b⃗)=(c⃗⋅b⃗) a⃗−(c⃗⋅a⃗) b⃗\vec{c}\times(\vec{a}\times\vec{b}) = (\vec{c}\cdot\vec{b})\,\vec{a} - (\vec{c}\cdot\vec{a})\,\vec{b}
  • Coplanar with a⃗,b⃗\vec{a}, \vec{b}it is a combination λa⃗+μb⃗\lambda\vec{a} + \mu\vec{b}
  • Perpendicular to c⃗\vec{c}by construction of the outer cross
  • Normalisedivide by its magnitude for a UNIT answer

Worked example

Find a unit vector coplanar with a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k} and b⃗=2i^+j^+k^\vec{b} = 2\hat{i} + \hat{j} + \hat{k}, and perpendicular to c⃗=i^+j^−k^\vec{c} = \hat{i} + \hat{j} - \hat{k}.
  1. Write v⃗=λa⃗+μb⃗=(λ+2μ)i^+(λ+μ)j^+(λ+μ)k^\vec{v} = \lambda\vec{a} + \mu\vec{b} = (\lambda + 2\mu)\hat{i} + (\lambda + \mu)\hat{j} + (\lambda + \mu)\hat{k}.
  2. Perpendicular to c⃗\vec{c}: v⃗⋅c⃗=(λ+2μ)+(λ+μ)−(λ+μ)=λ+2μ=0\vec{v}\cdot\vec{c} = (\lambda + 2\mu) + (\lambda + \mu) - (\lambda + \mu) = \lambda + 2\mu = 0, so λ=−2μ\lambda = -2\mu.
  3. Take μ=1,λ=−2\mu = 1, \lambda = -2: v⃗=0i^−j^−k^=−(j^+k^)\vec{v} = 0\hat{i} - \hat{j} - \hat{k} = -(\hat{j} + \hat{k}).
  4. Normalise: ∣v⃗∣=2|\vec{v}| = \sqrt{2}, so v^=±12(j^+k^)\hat{v} = \pm\dfrac{1}{\sqrt{2}}(\hat{j} + \hat{k}).
Answer:v^=±12(−j^−k^)\hat{v} = \pm\dfrac{1}{\sqrt{2}}(-\hat{j} - \hat{k})
Practice this conceptself-check · 4 quick reps

Try it yourself

Find a unit vector coplanar with i^+j^\hat{i} + \hat{j} and j^+k^\hat{j} + \hat{k}, perpendicular to i^+k^\hat{i} + \hat{k}.

Practice — Level 1 (4 reps)

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

  1. 1.
    A vector coplanar with a⃗,b⃗\vec{a}, \vec{b} is written as?
  2. 2.
    c⃗×(a⃗×b⃗)\vec{c}\times(\vec{a}\times\vec{b}) is perpendicular to?
  3. 3.
    c⃗×(a⃗×b⃗)\vec{c}\times(\vec{a}\times\vec{b}) is coplanar with?
  4. 4.
    To impose perpendicular to c⃗\vec{c}, set what to zero?

From the bank · past-year question

Example 10VectorsHARD
The unit vector which is orthogonal to the vector 5i^+2j^+6k^5\hat{i}+2\hat{j}+6\hat{k} and is coplanar with the vectors 2i^+j^+k^2\hat{i}+\hat{j}+\hat{k} and i^−j^+k^\hat{i}-\hat{j}+\hat{k} is

[Q104 · 2nd May Shift 2 · 2023]

Coplanar means a COMBINATION, not just "in the same plane"

Encode coplanarity with a⃗,b⃗\vec{a}, \vec{b} by writing the unknown as λa⃗+μb⃗\lambda\vec{a} + \mu\vec{b} from the start. Then the only freedom left is the ratio λ:μ\lambda:\mu, which the perpendicularity or projection condition pins down.

Both signs of the UNIT answer are valid

The normalised result is ±v^\pm\hat{v} — both directions satisfy "perpendicular and coplanar" unless the question fixes orientation. Pick whichever sign appears in the options.

Confirm orthogonality AND coplanarity at the end

A tempting distractor satisfies one condition but not the other. Verify v⃗⋅c⃗=0\vec{v}\cdot\vec{c} = 0 (perpendicular) and that v⃗\vec{v} is a combination of a⃗,b⃗\vec{a}, \vec{b} (coplanar) before selecting.
Drill 5 more on vector orthogonal to one vector and coplanar with two others

Concept 11 of 11

Solving a vector equation: a cross condition plus a magnitude/dot condition

Intuition

A condition like r⃗×b⃗=c⃗×b⃗\vec{r}\times\vec{b} = \vec{c}\times\vec{b} does not pin down r⃗\vec{r} — it only forces r⃗−c⃗\vec{r} - \vec{c} to be parallel to b⃗\vec{b}. Pairing it with a scalar (dot) condition such as r⃗⋅a⃗=0\vec{r}\cdot\vec{a} = 0 closes the system and gives a unique r⃗\vec{r}. Problems mixing a⃗×b⃗\vec{a}\times\vec{b}, a magnitude, and an angle are solved the same way: peel off the vector identity, then use the scalar data.

Definition

To solve r⃗×b⃗=c⃗×b⃗\vec{r}\times\vec{b} = \vec{c}\times\vec{b} with a scalar condition:

  • Rewrite as (r⃗−c⃗)×b⃗=0⃗(\vec{r} - \vec{c})\times\vec{b} = \vec{0}, so r⃗−c⃗=tb⃗\vec{r} - \vec{c} = t\vec{b}, i.e. r⃗=c⃗+tb⃗\vec{r} = \vec{c} + t\vec{b}.
  • Substitute into the scalar condition (e.g. r⃗⋅a⃗=0\vec{r}\cdot\vec{a} = 0) to find tt, then r⃗\vec{r}.
  • For ∣(a⃗×b⃗)×c⃗∣|(\vec{a}\times\vec{b})\times\vec{c}| given ∣c⃗∣|\vec{c}| and an angle: ∣(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}, and the auxiliary conditions fix ∣c⃗∣|\vec{c}|.

Cross condition reduces to a parallel offset

r⃗×b⃗=c⃗×b⃗  ⟹  r⃗=c⃗+t b⃗,then use the scalar condition for t\vec{r}\times\vec{b} = \vec{c}\times\vec{b} \;\Longrightarrow\; \vec{r} = \vec{c} + t\,\vec{b}, \quad\text{then use the scalar condition for } t
  • (r⃗−c⃗)×b⃗=0⃗(\vec{r} - \vec{c})\times\vec{b} = \vec{0}so r⃗−c⃗ ∥ b⃗\vec{r} - \vec{c}\,\|\,\vec{b}
  • ttthe one free scalar — fixed by the dot/magnitude condition

Worked example

Let A⃗=2i^+k^\vec{A} = 2\hat{i} + \hat{k}, B⃗=i^+j^+k^\vec{B} = \hat{i} + \hat{j} + \hat{k}, C⃗=4i^−3j^+7k^\vec{C} = 4\hat{i} - 3\hat{j} + 7\hat{k}. Find R⃗\vec{R} with R⃗×B⃗=C⃗×B⃗\vec{R}\times\vec{B} = \vec{C}\times\vec{B} and R⃗⋅A⃗=0\vec{R}\cdot\vec{A} = 0.
  1. R⃗×B⃗=C⃗×B⃗\vec{R}\times\vec{B} = \vec{C}\times\vec{B} ⇒ (R⃗−C⃗)×B⃗=0⃗(\vec{R} - \vec{C})\times\vec{B} = \vec{0} ⇒ R⃗=C⃗+tB⃗\vec{R} = \vec{C} + t\vec{B}.
  2. So R⃗=(4+t)i^+(−3+t)j^+(7+t)k^\vec{R} = (4 + t)\hat{i} + (-3 + t)\hat{j} + (7 + t)\hat{k}.
  3. Impose R⃗⋅A⃗=0\vec{R}\cdot\vec{A} = 0: 2(4+t)+0(−3+t)+1(7+t)=0⇒8+2t+7+t=0⇒3t=−15⇒t=−52(4 + t) + 0(-3 + t) + 1(7 + t) = 0 \Rightarrow 8 + 2t + 7 + t = 0 \Rightarrow 3t = -15 \Rightarrow t = -5.
  4. Substitute: R⃗=−i^−8j^+2k^\vec{R} = -\hat{i} - 8\hat{j} + 2\hat{k}.
Answer:R⃗=−i^−8j^+2k^\vec{R} = -\hat{i} - 8\hat{j} + 2\hat{k}
Practice this conceptself-check · 4 quick reps

Try it yourself

Find r⃗\vec{r} with r⃗×b⃗=c⃗×b⃗\vec{r}\times\vec{b} = \vec{c}\times\vec{b} and r⃗⋅a⃗=4\vec{r}\cdot\vec{a} = 4, where a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}, b⃗=k^\vec{b} = \hat{k} and c⃗=2i^+j^−k^\vec{c} = 2\hat{i} + \hat{j} - \hat{k}.

Practice — Level 1 (4 reps)

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

  1. 1.
    r⃗×b⃗=c⃗×b⃗\vec{r}\times\vec{b} = \vec{c}\times\vec{b} implies r⃗−c⃗\vec{r} - \vec{c} is?
  2. 2.
    So r⃗=?\vec{r} = ?
  3. 3.
    What fixes the scalar tt?
  4. 4.
    ∣(a⃗×b⃗)×c⃗∣=?|(\vec{a}\times\vec{b})\times\vec{c}| = ?

From the bank · past-year question

Example 11VectorsHARD
Let A⃗=2i^+k^\vec{A} = 2\hat{i}+\hat{k}, B⃗=i^+j^+k^\vec{B} = \hat{i}+\hat{j}+\hat{k} and C⃗=4i^−3j^+7k^\vec{C} = 4\hat{i}-3\hat{j}+7\hat{k}. If a vector R⃗\vec{R} satisfies R⃗×B⃗=C⃗×B⃗\vec{R}\times\vec{B} = \vec{C}\times\vec{B} and R⃗⋅A⃗=0\vec{R}\cdot\vec{A} = 0, then R⃗\vec{R} is given by

[Q108 · 2nd May Shift 2 · 2023]

You cannot cancel the cross product

r⃗×b⃗=c⃗×b⃗\vec{r}\times\vec{b} = \vec{c}\times\vec{b} does NOT give r⃗=c⃗\vec{r} = \vec{c}. The correct deduction is (r⃗−c⃗)×b⃗=0⃗(\vec{r} - \vec{c})\times\vec{b} = \vec{0}, i.e. r⃗=c⃗+tb⃗\vec{r} = \vec{c} + t\vec{b}; the scalar condition then fixes tt.

The cross condition alone leaves one free scalar

A single cross equation can never determine r⃗\vec{r} uniquely — any multiple of b⃗\vec{b} can be added. Always use the accompanying scalar (dot or magnitude) condition to close the system.
Drill 4 more on solving a vector equation: a cross condition plus a magnitude/dot condition

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 (11)

  • The scalar triple product — dot-cross and determinant form

    Scalar triple product as a determinant

    [a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)=∣a1a2a3b1b2b3c1c2c3∣[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}
  • Cyclic and sign properties of the scalar triple product

    Cyclic and swap rules

    [a⃗ b⃗ c⃗]=[b⃗ c⃗ a⃗]=[c⃗ a⃗ b⃗]=−[b⃗ a⃗ c⃗][\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}] = -[\vec{b}\ \vec{a}\ \vec{c}]
  • Computing the value of a scalar triple product

    STP when one vector is perpendicular to the other two

    [a⃗ b⃗ c⃗]=∣a⃗∣ ∣b⃗×c⃗∣=∣a⃗∣∣b⃗∣∣c⃗∣sin⁡θ[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}| = |\vec{a}||\vec{b}||\vec{c}|\sin\theta
  • Coplanarity of three vectors (and solving for a parameter)

    Coplanarity criterion

    a⃗,b⃗,c⃗ coplanar  ⟺  ∣a1a2a3b1b2b3c1c2c3∣=0\vec{a},\vec{b},\vec{c}\text{ coplanar} \iff \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0
  • Scalar triple product of linear combinations

    Linear-combination identities

    [a⃗+b⃗  b⃗+c⃗  c⃗+a⃗]=2[a⃗ b⃗ c⃗][ma⃗+b⃗  mb⃗+c⃗  mc⃗+a⃗]=(m3+1)[a⃗ b⃗ c⃗][\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \qquad [m\vec{a}+\vec{b}\ \ m\vec{b}+\vec{c}\ \ m\vec{c}+\vec{a}] = (m^3+1)[\vec{a}\ \vec{b}\ \vec{c}]
  • Volume of a parallelepiped (and min/max problems)

    Parallelepiped volume

    V=∣[a⃗ b⃗ c⃗]∣=∣∣a1a2a3b1b2b3c1c2c3∣∣V = |[\vec{a}\ \vec{b}\ \vec{c}]| = \left|\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\right|
  • Volume of a tetrahedron

    Tetrahedron volume

    V=16 ∣[AB→ AC→ AD→]∣V = \tfrac{1}{6}\,\bigl|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\bigr|
  • Reciprocal-basis identities and the STP-squared rule

    Reciprocal pairings and STP-squared

    a⃗⋅p⃗=b⃗⋅q⃗=c⃗⋅r⃗=1,a⃗⋅q⃗=b⃗⋅p⃗=⋯=0[a⃗×b⃗  b⃗×c⃗  c⃗×a⃗]=[a⃗ b⃗ c⃗]2\vec{a}\cdot\vec{p} = \vec{b}\cdot\vec{q} = \vec{c}\cdot\vec{r} = 1, \quad \vec{a}\cdot\vec{q} = \vec{b}\cdot\vec{p} = \dots = 0 \qquad [\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2
  • Vector triple product (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}
  • Vector orthogonal to one vector and coplanar with two others

    Orthogonal-and-coplanar vector

    c⃗×(a⃗×b⃗)=(c⃗⋅b⃗) a⃗−(c⃗⋅a⃗) b⃗\vec{c}\times(\vec{a}\times\vec{b}) = (\vec{c}\cdot\vec{b})\,\vec{a} - (\vec{c}\cdot\vec{a})\,\vec{b}
  • Solving a vector equation: a cross condition plus a magnitude/dot condition

    Cross condition reduces to a parallel offset

    r⃗×b⃗=c⃗×b⃗  ⟹  r⃗=c⃗+t b⃗,then use the scalar condition for t\vec{r}\times\vec{b} = \vec{c}\times\vec{b} \;\Longrightarrow\; \vec{r} = \vec{c} + t\,\vec{b}, \quad\text{then use the scalar condition for } t

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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 a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three vectors, ∣a⃗∣=2|\vec{a}|=2, ∣b⃗∣=4|\vec{b}|=4, ∣c⃗∣=1|\vec{c}|=1, ∣b⃗×c⃗∣=15|\vec{b}\times\vec{c}|=\sqrt{15} and b⃗=2c⃗+λa⃗\vec{b}=2\vec{c}+\lambda\vec{a}, then the value of λ\lambda is

[Q101 · 12th May Shift 1 · 2024]

Example 2VectorsMODERATE
If the vector c⃗\vec{c} lies in the plane of a⃗\vec{a} and b⃗\vec{b}, where a⃗=i^−j^+2k^, b⃗=i^+j^+k^\vec{a}=\hat{i}-\hat{j}+2\hat{k},\, \vec{b}=\hat{i}+\hat{j}+\hat{k} and c⃗=xi^−(2−x)j^−k^\vec{c}=x\hat{i}-(2-x)\hat{j}-\hat{k}, then the value of xx is

[Q139 · 15th May Shift 2 · 2023]

Example 3VectorsHARD
If a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} are mutually perpendicular vectors having magnitudes 1, 2, 3 respectively, then the value of [a⃗+b⃗+c⃗    b⃗−a⃗    c⃗][\vec{a}+\vec{b}+\vec{c} \;\; \vec{b}-\vec{a} \;\; \vec{c}] is

[Q105 · 10th May Shift 2 · 2023]

Example 4VectorsHARD
If the volume of the parallelepiped is 158 cu. units whose coterminous edges are given by the vectors a⃗=(i^+j^+nk^)\vec{a} = (\hat{i}+\hat{j}+n\hat{k}), b⃗=2i^+4j^−nk^\vec{b} = 2\hat{i}+4\hat{j}-n\hat{k} and c⃗=i^+nj^+3k^\vec{c} = \hat{i}+n\hat{j}+3\hat{k}, where n≥0n \geq 0, then the value of n is

[Q114 · 11th May Shift 1 · 2024]

Example 5VectorsHARD
If the volume of tetrahedron, whose vertices are with position vectors i^−6j^+10k^, −i^−3j^+7k^\hat{i}-6\hat{j}+10\hat{k},\,-\hat{i}-3\hat{j}+7\hat{k}, 5i^−j^+λk^5\hat{i}-\hat{j}+\lambda\hat{k} and 7i^−4j^+7k^7\hat{i}-4\hat{j}+7\hat{k} is 11 cubic units, then value of λ\lambda is

[Q101 · 10th May Shift 2 · 2024]

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