MHT-CET Maths · Vectors

Linear Combinations, Collinearity, and Coplanarity

Building one vector out of others as m·a + n·b, and the structural conditions hiding inside that idea — collinear vectors need ONE scalar, coplanar vectors need TWO, and a dependent set is exactly a coplanar one.

All Vectors notes MHT-CET Maths strategy

Why this matters

This is the most-tested cluster in MHT-CET Vectors — 13 PYQs sit here, spread evenly across EASY, MODERATE and HARD. Two recurring shapes dominate: the collinearity / coplanarity conditions (one-scalar vs two-scalar, and the 'no two collinear' chain systems), and the linear-system shapes — express a vector as m·b + n·c, find components against a transformed basis, or test linear dependence by a determinant. Get the counting right (collinear ⇒ 1 scalar, coplanar / dependent ⇒ 2 scalars) and almost every question here reduces to a small system you can solve by equating components.

Concept 1 of 10

Linear combination of vectors

Intuition

A linear combination is just \"scale each vector by a number, then add them up.\" If you have vectors

\vec{a}

and

\vec{b}

, then

m\vec{a} + n\vec{b}

— for any real numbers

m, n

— is a linear combination of them. Every expression you'll meet in this subtopic (

2\vec{a} - \vec{b} + 3\vec{c}

3\vec{a} + \vec{b} - 2\vec{c}

, …) is one of these. The whole topic is about what happens when ONE vector turns out to BE a linear combination of others.

Definition

A vector $\vec{r}$ is a linear combination of vectors $\vec{a}_1, \vec{a}_2, \dots, \vec{a}_k$ if there exist scalars $c_1, c_2, \dots, c_k$ such that $\vec{r} = c_1\vec{a}_1 + c_2\vec{a}_2 + \dots + c_k\vec{a}_k$ . In the standard basis you build a combination componentwise: $m\vec{a} + n\vec{b}$ has $\hat{i}$ -component $ma_1 + nb_1$ , $\hat{j}$ -component $ma_2 + nb_2$ , and so on. The scalars are called the coefficients of the combination.

Linear combination

\vec{r} = m\vec{a} + n\vec{b} + p\vec{c} \qquad r_i = ma_i + nb_i + pc_i

$m, n, p$ real scalar coefficients (any sign, including zero)
$\vec{r}$ the combined vector — built componentwise

Diagram · component form (drag to rotate)

Step along x (2.4 î), then y ( 1.6 ĵ), then z ( 2.0 k̂) to reach the tip: v = 2.4 î + 1.6 ĵ + 2.0 k̂. Any vector is the sum of its axis components, and |v| = √(x² + y² + z²).

Worked example

For

\vec{a} = \hat{i} + \hat{k}

\vec{b} = 2\hat{i} - \hat{j}

and

\vec{c} = \hat{j} + 3\hat{k}

, compute the linear combination

2\vec{a} - \vec{b} + \vec{c}

in component form.

Scale each: $2\vec{a} = 2\hat{i} + 2\hat{k}$ ; $-\vec{b} = -2\hat{i} + \hat{j}$ ; $\vec{c} = \hat{j} + 3\hat{k}$ .
Add the $\hat{i}$ -parts: $2 - 2 + 0 = 0$ . The $\hat{j}$ -parts: $0 + 1 + 1 = 2$ . The $\hat{k}$ -parts: $2 + 0 + 3 = 5$ .
So the combination is $0\hat{i} + 2\hat{j} + 5\hat{k} = 2\hat{j} + 5\hat{k}$ .

Answer:

2\vec{a} - \vec{b} + \vec{c} = 2\hat{j} + 5\hat{k}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\vec{a} = \hat{i} + \hat{j}$ , $\vec{b} = \hat{i} - \hat{j}$ . Find $3\vec{a} + \vec{b}$ .
2.
$\vec{a} = 2\hat{i}$ , $\vec{b} = \hat{j}$ , $\vec{c} = \hat{k}$ . Find $\vec{a} - 2\vec{b} + 3\vec{c}$ .
3.
What is the $\hat{i}$ -component of $m\vec{a} + n\vec{b}$ ?
4.
Is $\vec{0}$ a linear combination of any $\vec{a}, \vec{b}$ ?

A linear combination is built componentwise — three sums, not one

When you form

m\vec{a} + n\vec{b} + p\vec{c}

in 3-D you are doing three independent scalar sums: one for

\hat{i}

, one for

\hat{j}

, one for

\hat{k}

. Combine the wrong components and the answer is silently wrong — keep the three columns separated.

Concept 2 of 10

Collinear vectors and collinear points (one scalar)

Intuition

Two vectors are collinear (parallel) when they point along the SAME line of direction — same way or exactly opposite. That happens exactly when one is a scalar multiple of the other:

\vec{a} = k\vec{b}

. ONE scalar settles it. For three POINTS, collinearity is the point-version:

A, B, C

lie on one line when the displacement

\vec{AC}

is a scalar multiple of

\vec{AB}

Definition

Collinear vectors: $\vec{a}$ and $\vec{b}$ ( $\vec{b} \neq \vec{0}$ ) are collinear iff $\vec{a} = k\vec{b}$ for some scalar $k$ . In components, this means the components are proportional: $\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}$ .
Collinear points: $A, B, C$ (position vectors $\vec{a}, \vec{b}, \vec{c}$ ) are collinear iff $\vec{AB} \parallel \vec{AC}$ , i.e. $\vec{b} - \vec{a} = k(\vec{c} - \vec{a})$ for some scalar $k$ .
\"is collinear with\": the phrase $\vec{u}$ is collinear with $\vec{v}$ means $\vec{u} = t\vec{v}$ — introduce ONE unknown scalar and solve.

Collinearity (one scalar)

\vec{a} = k\vec{b} \qquad \dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}

$k$ the single scalar; $k > 0$ same direction, $k < 0$ opposite

Worked example

Are

\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}

and

\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}

collinear? If so, find

k

with

\vec{b} = k\vec{a}

Look for one scalar $k$ with $\vec{b} = k\vec{a}$ . From $\hat{i}$ : $-4 = 2k \Rightarrow k = -2$ .
Check the other components with $k = -2$ : $\hat{j}$ : $-2(-3) = 6$ $\checkmark$ ; $\hat{k}$ : $-2(4) = -8$ $\checkmark$ .
All three components agree on the SAME $k$ , so the vectors are collinear (anti-parallel, $k < 0$ ).

Answer:Collinear;

\vec{b} = -2\vec{a}

, so

k = -2

Practice this conceptself-check · 4 quick reps

Try it yourself

For what value of

\lambda

\vec{a} = \lambda\hat{i} + 2\hat{j}

collinear with

\vec{b} = 3\hat{i} + 6\hat{j}

Practice — Level 1 (4 reps)

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

1.
Are $\hat{i} + \hat{j}$ and $3\hat{i} + 3\hat{j}$ collinear?
2.
How many scalars settle whether two vectors are collinear?
3.
Find $k$ if $6\hat{i} - 9\hat{j} = k(2\hat{i} - 3\hat{j})$ .
4.
Points $A, B, C$ collinear means $\vec{AB}$ is a ___ of $\vec{AC}$ .

Collinear needs ONE scalar; coplanar needs TWO — keep the count straight

Whenever a question says \"collinear / parallel,\" introduce ONE unknown scalar (

\vec{u} = t\vec{v}

). If it says \"coplanar / lies in the plane of,\" introduce TWO (

\vec{u} = m\vec{v} + n\vec{w}

). Mixing up the count is the single biggest source of wrong setups in this subtopic.

Parallel VECTORS vs collinear POINTS

Vectors are parallel when they share a direction; points are collinear when they share a LINE. Test points by displacements:

A, B, C

collinear iff

\vec{AB} \parallel \vec{AC}

(they share point

A

), not merely \"some pair of vectors is parallel.\"

Concept 3 of 10

Coplanar vectors (two scalars)

Intuition

Three vectors are coplanar when you can draw all three flat in one plane. If

\vec{b}

and

\vec{c}

are not collinear, they span a plane; a third vector

\vec{a}

lies in that plane exactly when it's a linear combination of them:

\vec{a} = m\vec{b} + n\vec{c}

. TWO scalars, because the plane is 2-dimensional. A vector that lies in / bisects / is perpendicular within the plane of

\vec{b}, \vec{c}

is always of this form.

Definition

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar iff one is a linear combination of the other two: $\vec{a} = m\vec{b} + n\vec{c}$ for some scalars $m, n$ (assuming $\vec{b}, \vec{c}$ non-collinear). Equivalently, their scalar triple product vanishes: $[\vec{a}\ \vec{b}\ \vec{c}] = 0$ , i.e. the determinant of their components is zero. \"Lies in the plane of $\vec{b}$ and $\vec{c}$ \" is the phrase that signals this: write the unknown vector as $m\vec{b} + n\vec{c}$ and solve for $m, n$ .

Coplanarity (two scalars)

\vec{a} = m\vec{b} + n\vec{c} \qquad [\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} = 0

$m, n$ the two scalars — one per spanning vector
$[\vec{a}\ \vec{b}\ \vec{c}]$ scalar triple product; zero ⟺ coplanar

Diagram · orthonormal triple î, ĵ, k̂ (drag to rotate)

î, ĵ, k̂ are mutually perpendicular unit vectors: each pair has dot product 0 (î·ĵ = ĵ·k̂ = k̂·î = 0) and each has length 1. They form the standard basis — any vector is a unique combination x î + y ĵ + z k̂.

Worked example

Show that

\vec{a} = 3\hat{i} + \hat{j}

is coplanar with

\vec{b} = \hat{i} + \hat{j}

and

\vec{c} = \hat{i} - \hat{j}

by writing

\vec{a} = m\vec{b} + n\vec{c}

Set $3\hat{i} + \hat{j} = m(\hat{i} + \hat{j}) + n(\hat{i} - \hat{j})$ and equate components.
$\hat{i}$ : $m + n = 3$ . $\hat{j}$ : $m - n = 1$ .
Add: $2m = 4 \Rightarrow m = 2$ ; then $n = 1$ . A solution exists, so $\vec{a}$ is coplanar with $\vec{b}, \vec{c}$ .

Answer:

\vec{a} = 2\vec{b} + \vec{c}

(so

m = 2,\ n = 1

); the three are coplanar.

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
How many scalars express a vector in the plane of two others?
2.
Coplanar ⟺ scalar triple product equals ___.
3.
Write $2\hat{i}$ as $m(\hat{i}+\hat{j}) + n(\hat{i}-\hat{j})$ . Find $m, n$ .
4.
If $[\vec{a}\ \vec{b}\ \vec{c}] \neq 0$ , are they coplanar?

Coplanar / dependent ⇒ TWO scalars and a vanishing determinant

For coplanarity you have two equivalent tests: express the vector as

m\vec{b} + n\vec{c}

(good when you need

m, n

), or set the

3\times 3

determinant of the components to zero (good for a yes/no or for finding an unknown component). Pick the test that matches what the question asks for.

Concept 4 of 10

Forming a combination, then normalising / measuring it

Intuition

The most common EASY-to-MODERATE shape: you're handed

\vec{a}, \vec{b}, \vec{c}

and asked about a combination like

2\vec{a} - \vec{b} + 3\vec{c}

— find a unit vector along it, a parallel vector of given magnitude, or the angle it makes with another combination. The skill is the same every time: BUILD the combination componentwise first, then apply magnitude / unit-vector / angle as a second step.

Definition

Once a linear combination $\vec{r} = m\vec{a} + n\vec{b} + p\vec{c}$ is built componentwise, the follow-up is routine:

**Unit vector along $\vec{r}$ :** $\hat{r} = \vec{r}/|\vec{r}|$ .
**Vector of magnitude $M$ parallel to $\vec{r}$ :** $M\,\hat{r} = \dfrac{M}{|\vec{r}|}\vec{r}$ .
**Angle between two combinations $\vec{u}, \vec{v}$ :** $\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}|\,|\vec{v}|}$ .

Scale a combination to a target magnitude

M\,\hat{r} = \dfrac{M}{|\vec{r}|}\,\vec{r} \qquad \hat{r} = \dfrac{\vec{r}}{|\vec{r}|}

$\vec{r}$ the linear combination, built first
$M$ the required magnitude of the parallel vector

Worked example

Given

\vec{a} = \hat{i} + 2\hat{j}

\vec{b} = 3\hat{i} - \hat{j}

, find a vector of magnitude

10

parallel to

\vec{a} + \vec{b}

Build the combination: $\vec{r} = \vec{a} + \vec{b} = (1+3)\hat{i} + (2-1)\hat{j} = 4\hat{i} + 3\hat{j}$ .
Magnitude: $|\vec{r}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ .
Scale to magnitude $10$ : $\dfrac{10}{5}\vec{r} = 2(4\hat{i} + 3\hat{j}) = 8\hat{i} + 6\hat{j}$ .

Answer:

8\hat{i} + 6\hat{j}

(the negative

-8\hat{i} - 6\hat{j}

is also parallel).

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\vec{r} = 3\hat{i} + 4\hat{j}$ . Find the unit vector along $\vec{r}$ .
2.
A vector of magnitude $6$ parallel to $\vec{r} = \hat{i} - 2\hat{j} + 2\hat{k}$ ?
3.
First step before normalising $2\vec{a} - \vec{b}$ ?
4.
$\cos\theta$ between $\vec{u}, \vec{v}$ ?

From the bank · past-year question

Example 4VectorsEASY

Given

\vec{a} = \hat{i}+\hat{j}+\hat{k}

\vec{b} = 4\hat{i}-2\hat{j}+3\hat{k}

\vec{c} = \hat{i}-2\hat{j}+\hat{k}

, then a vector of magnitude 6 units, which is parallel to the vector

2\vec{a}-\vec{b}+3\vec{c}

, is

[Q124 · 10th May Shift 2 · 2023]

Build the combination BEFORE you normalise — don't normalise the parts

To find a unit vector along

3\vec{a} + \vec{b} - 2\vec{c}

you must combine first, get one vector, THEN divide by its magnitude. Normalising

\vec{a}, \vec{b}, \vec{c}

separately and combining the unit vectors gives a different (wrong) direction.

\"Parallel of magnitude $M$ \" has TWO answers — $\pm M\hat{r}$

A vector parallel to

\vec{r}

can point the same way OR the opposite way, so both

+M\hat{r}

and

-M\hat{r}

qualify. MCQs usually list only one; if your computed direction isn't an option, check its negative before assuming an error.

Drill 2 more on forming a combination, then normalising / measuring it

Concept 5 of 10

Collinearity of three points (and who lies between)

Intuition

Three points

P, Q, R

are collinear when the segments between them lie on one straight line. Test it by displacements: compute

\vec{PQ}

and

\vec{QR}

(or

\vec{PR}

); if one is a scalar multiple of the other AND they share a common point, the points are collinear. The sign and size of the multiplier even tells you the ORDER — who sits between whom.

Definition

Points $P, Q, R$ with position vectors $\vec{p}, \vec{q}, \vec{r}$ are collinear iff $\vec{QR} = k\,\vec{PQ}$ (or any two displacements are proportional), where $\vec{PQ} = \vec{q} - \vec{p}$ , $\vec{QR} = \vec{r} - \vec{q}$ . Reading the multiplier: if $\vec{QR} = k\,\vec{PQ}$ with $k > 0$ , the line goes $P \to Q \to R$ so ** $Q$ lies between $P$ and $R$ **; a negative $k$ means $R$ doubles back, putting a different point in the middle.

Three-point collinearity

\vec{QR} = k\,\vec{PQ} \quad\text{where}\quad \vec{PQ} = \vec{q} - \vec{p},\ \ \vec{QR} = \vec{r} - \vec{q}

$k$ the scalar; its sign/size fixes the order of the points

Worked example

Are

P(\hat{i} + \hat{j})

Q(3\hat{i} + 3\hat{j})

R(6\hat{i} + 6\hat{j})

collinear? Who lies between?

Displacements: $\vec{PQ} = \vec{q} - \vec{p} = 2\hat{i} + 2\hat{j}$ ; $\vec{QR} = \vec{r} - \vec{q} = 3\hat{i} + 3\hat{j}$ .
Test proportionality: $\vec{QR} = \tfrac{3}{2}\vec{PQ}$ — yes, a scalar multiple, and they share point $Q$ . So collinear.
The multiplier $\tfrac{3}{2} > 0$ , so the order along the line is $P \to Q \to R$ : $Q$ lies between $P$ and $R$ .

Answer:Collinear;

Q

lies between

P

and

R

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\vec{PQ} = \hat{i} + \hat{j}$ , $\vec{QR} = 2\hat{i} + 2\hat{j}$ . Collinear?
2.
Displacement $\vec{PQ}$ for $P(2\hat{i})$ , $Q(5\hat{i})$ ?
3.
If $\vec{QR} = 3\vec{PQ}$ , who is between $P$ and $R$ ?
4.
Are $\vec{PQ}$ and $\vec{QR}$ non-proportional ⇒ points are?

From the bank · past-year question

Example 5VectorsMODERATE

If the points

P, Q

and

R

are with the position vectors

\hat{i}-2\hat{j}+3\hat{k},\,-2\hat{i}+3\hat{j}+2\hat{k}

and

-8\hat{i}+13\hat{j}

respectively, then these points are

[Q125 · 15th May Shift 2 · 2023]

Use displacements that SHARE a point

To conclude collinearity from

\vec{QR} = k\vec{PQ}

the two displacements must share a common point (here

Q

). Two parallel displacements that don't share a point only say the segments are parallel — not that all the points lie on ONE line.

Concept 6 of 10

Express a vector as a combination of two others

Intuition

Given

\vec{a} = m\vec{b} + n\vec{c}

, the question hands you

\vec{a}, \vec{b}, \vec{c}

and asks for the scalars

m, n

. Equate components and you get a small linear system — usually two equations from two of the three components are enough; the third is a free consistency check. Solve, then read off whatever the question wants (often

m + n

mn

Definition

To write $\vec{a} = m\vec{b} + n\vec{c}$ : equate each component to get the system $a_1 = mb_1 + nc_1$ , $a_2 = mb_2 + nc_2$ , $a_3 = mb_3 + nc_3$ . Pick the two simplest equations, solve for $m$ and $n$ , then verify with the remaining one. If the third equation also holds, the representation is valid (the three vectors are coplanar); if it fails, no such $m, n$ exist.

Two-equation linear system

\begin{cases} a_1 = mb_1 + nc_1 \\ a_2 = mb_2 + nc_2 \end{cases}\ \Rightarrow\ m,\ n

$m, n$ coefficients to solve for; the 3rd component is the check

Worked example

Write

\vec{a} = 5\hat{i} + \hat{j}

m\vec{b} + n\vec{c}

with

\vec{b} = \hat{i} + \hat{j}

\vec{c} = \hat{i} - \hat{j}

. Find

m + n

Equate components: $\hat{i}$ : $m + n = 5$ ; $\hat{j}$ : $m - n = 1$ .
Add: $2m = 6 \Rightarrow m = 3$ ; subtract: $2n = 4 \Rightarrow n = 2$ .
So $m + n = 3 + 2 = 5$ . (Check: $3\vec{b} + 2\vec{c} = 3\hat{i}+3\hat{j} + 2\hat{i}-2\hat{j} = 5\hat{i} + \hat{j}$ $\checkmark$ .)

Answer:

m = 3,\ n = 2

, so

m + n = 5

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$2\hat{i} = m(\hat{i}+\hat{j}) + n(\hat{i}-\hat{j})$ . Find $m, n$ .
2.
How many equations does equating components in 3-D give?
3.
What does the third component equation provide?
4.
If the check fails, the representation ___.

From the bank · past-year question

Example 6VectorsMODERATE

\vec{a}=m\vec{b}+n\vec{c}

, where

\vec{a}=4\hat{i}+13\hat{j}-18\hat{k}

\vec{b}=\hat{i}-2\hat{j}+3\hat{k}

\vec{c}=2\hat{i}+3\hat{j}-4\hat{k}

, then

m+n=

[Q136 · 10th May Shift 2 · 2024]

Solve from two equations, but ALWAYS verify with the third

Two component-equations pin down

m, n

— but in 3-D there's a third equation. If it doesn't hold, no valid

m, n

exist (the vectors aren't coplanar). Skipping the check can hand you scalars that don't actually reproduce

\vec{a}

Concept 7 of 10

Linear dependence vs independence (the determinant test)

Intuition

A set of vectors is dependent when one of them is \"redundant\" — expressible from the others, so they don't truly fill the space. In 3-D, three vectors are dependent exactly when they're coplanar (one lies in the plane of the other two), and that shows up as a ZERO determinant. Three vectors are independent when only the all-zero combination gives the zero vector — nothing is redundant.

Definition

Vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent if there exist scalars $x, y, z$ , not all zero, with $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ . They are independent if $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ forces $x = y = z = 0$ . Test in 3-D: dependent ⟺ coplanar ⟺ the determinant of their components is zero:

$[\vec{a}\ \vec{b}\ \vec{c}] = 0$ ⇒ dependent / coplanar.
$[\vec{a}\ \vec{b}\ \vec{c}] \neq 0$ ⇒ independent / non-coplanar (they form a basis).

Dependence ⟺ vanishing determinant

x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}\ (\text{not all }0) \iff \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0

$x, y, z$ scalars; a non-trivial solution means dependent

Worked example

For what value of

\lambda

are

\vec{a} = \hat{i} + \hat{j} + \hat{k}

\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}

\vec{c} = \hat{i} + \hat{j} + \lambda\hat{k}

linearly dependent?

Dependent ⟺ determinant zero. Write the components as rows: $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 1 & \lambda \end{vmatrix} = 0$ .
Rows 1 and 3 differ only in the last entry, so expand: $1(3\lambda - 4) - 1(2\lambda - 4) + 1(2 - 3) = 0$ .
Simplify: $(3\lambda - 4) - (2\lambda - 4) - 1 = \lambda - 1 = 0 \Rightarrow \lambda = 1$ .

Answer:

\lambda = 1

(then

\vec{c} = \vec{a}

, trivially dependent).

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
Three coplanar vectors are linearly ___.
2.
$[\vec{a}\ \vec{b}\ \vec{c}] = 0$ means the vectors are?
3.
If $x\vec{a}+y\vec{b}+z\vec{c}=\vec{0}$ only when $x=y=z=0$ , they are?
4.
Three independent 3-D vectors form a ___.

From the bank · past-year question

Example 7VectorsMODERATE

\vec{a} = \hat{i}+\hat{j}+\hat{k}

\vec{b} = 4\hat{i}+3\hat{j}+4\hat{k}

and

\vec{c} = \hat{i}+\alpha\hat{j}+\beta\hat{k}

are linearly dependent vectors and

|\vec{c}|=\sqrt{3}

, then the value of

\alpha

and

\beta

are respectively.

[Q114 · 11th May Shift 2 · 2024]

Linearly dependent = coplanar = zero determinant — three names, one idea

In 3-D these are the SAME condition. A question may phrase it as \"linearly dependent,\" \"coplanar,\" or \"scalar triple product is zero\" — all three send you to the same

3\times 3

determinant set to zero.

A second condition (like $|\vec{c}| = \sqrt{3}$ ) is part of the same problem

Dependence often fixes only ONE unknown (e.g.

\beta = 1

). A magnitude condition supplies the OTHER (e.g.

1 + \alpha^2 + 1 = 3 \Rightarrow \alpha = \pm 1

). Use both givens — don't stop after the determinant.

Concept 8 of 10

Chained-collinearity systems (\"no two collinear\")

Intuition

The signature HARD shape:

\vec{a}, \vec{b}, \vec{c}

are three vectors, no two collinear, and you're told two collinearity facts — \"

\vec{a} + 2\vec{b}

is collinear with

\vec{c}

\" and \"

\vec{b} + 3\vec{c}

is collinear with

\vec{a}

.\" Each fact gives ONE scalar equation; equate coefficients of the (independent)

\vec{a}, \vec{b}, \vec{c}

on both sides, solve the little system, and the result is forced.

Definition

Turn each \"collinear with\" into a scalar multiple, then use independence to match coefficients:

$\vec{a} + 2\vec{b} = t\vec{c}$ (one scalar $t$ ).
$\vec{b} + 3\vec{c} = \lambda\vec{a}$ (one scalar $\lambda$ ).

Substitute one into the other and compare coefficients of the linearly-independent $\vec{a}, \vec{b}, \vec{c}$ (each side's coefficient of a given basis vector must match). This pins down $t, \lambda$ — typically $t = -6,\ \lambda = -\tfrac{1}{2}$ — and yields the target, e.g. $\vec{a} + 2\vec{b} = -6\vec{c}$ , so $\vec{a} + 2\vec{b} + 6\vec{c} = \vec{0}$ .

Chained collinearity

\vec{a} + 2\vec{b} = t\vec{c}, \quad \vec{b} + 3\vec{c} = \lambda\vec{a} \;\Rightarrow\; t = -6,\ \lambda = -\tfrac{1}{2}

$t, \lambda$ one scalar per collinearity fact; matched via coefficient comparison

Worked example

\vec{a}, \vec{b}, \vec{c}

: no two collinear.

\vec{a} + 2\vec{b}

is collinear with

\vec{c}

, and

\vec{b} + 3\vec{c}

is collinear with

\vec{a}

. Find

\vec{a} + 2\vec{b}

in terms of

\vec{c}

Write the two facts: $\vec{a} + 2\vec{b} = t\vec{c}$ and $\vec{b} + 3\vec{c} = \lambda\vec{a}$ .
From the first, $\vec{a} = t\vec{c} - 2\vec{b}$ . Substitute into the second: $\vec{b} + 3\vec{c} = \lambda(t\vec{c} - 2\vec{b}) = -2\lambda\vec{b} + \lambda t\,\vec{c}$ .
Match coefficients ( $\vec{b}, \vec{c}$ independent): $\vec{b}$ : $1 = -2\lambda \Rightarrow \lambda = -\tfrac{1}{2}$ . $\vec{c}$ : $3 = \lambda t = -\tfrac{1}{2}t \Rightarrow t = -6$ .
So $\vec{a} + 2\vec{b} = t\vec{c} = -6\vec{c}$ .

Answer:

\vec{a} + 2\vec{b} = -6\vec{c}

(equivalently

\vec{a} + 2\vec{b} + 6\vec{c} = \vec{0}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
\" $\vec{u}$ is collinear with $\vec{v}$ \" becomes which equation?
2.
From $\vec{a}+2\vec{b} = -6\vec{c}$ , find $\vec{a}+2\vec{b}+6\vec{c}$ .
3.
How many unknown scalars do two collinearity facts give?
4.
Why can you match coefficients of $\vec{a}, \vec{b}, \vec{c}$ ?

From the bank · past-year question

Example 8VectorsMODERATE

\vec{a}, \vec{b}, \vec{c}

are three non-zero vectors, no two of them are collinear,

\vec{a}+2\vec{b}

is collinear with

\vec{c}

\vec{b}+3\vec{c}

is collinear with

\vec{a}

, then

\vec{a}+2\vec{b}

[Q144 · 12th May Shift 2 · 2024]

Match coefficients only because the vectors are independent

\"No two collinear\" (in fact, all three independent) is the hypothesis that LETS you equate coefficients of

\vec{a}, \vec{b}, \vec{c}

on both sides. Without independence you couldn't conclude that equal vector-expressions force equal coefficients.

Read the target carefully: $\vec{a}+2\vec{b}$ vs $\vec{a}+2\vec{b}+6\vec{c}$

The same setup answers two different MCQ targets. If

\vec{a}+2\vec{b} = -6\vec{c}

, then the value of

\vec{a}+2\vec{b}

-6\vec{c}

, but the value of

\vec{a}+2\vec{b}+6\vec{c}

\vec{0}

. Answer the one actually asked.

Drill 2 more on chained-collinearity systems (\"no two collinear\")

Concept 9 of 10

A vector lying in the plane of two others

Intuition

When a question says a vector \"lies in the plane of

\vec{b}

and

\vec{c}

\" (or is coplanar with them), write it as

\vec{v} = \vec{b} + \lambda\vec{c}

(or

m\vec{b} + n\vec{c}

) — TWO scalars. Then a SECOND condition (it bisects the angle, is perpendicular to one of them, has a given projection or magnitude) pins down the scalar(s). Coplanarity supplies the FORM; the extra condition supplies the NUMBERS.

Definition

A vector $\vec{v}$ coplanar with non-collinear $\vec{b}, \vec{c}$ has the form $\vec{v} = m\vec{b} + n\vec{c}$ . Common second conditions and how to use them:

**Bisects the angle between $\vec{b}, \vec{c}$ :** $\vec{v}$ is parallel to $\hat{b} + \hat{c}$ (sum of the UNIT vectors).
**Perpendicular to $\vec{q}$ :** impose $\vec{v}\cdot\vec{q} = 0$ .
**Projection on $\vec{c}$ equals $k$ :** impose $\dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = k$ .
**Magnitude $M$ :** scale the resulting direction to length $M$ .

Coplanar form + a second condition

\vec{v} = m\vec{b} + n\vec{c} \quad\text{with}\quad \vec{v}\cdot\vec{q} = 0 \ \text{ or }\ \dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = k

$m, n$ two scalars from coplanarity
second conditionbisector / perpendicular / projection / magnitude — fixes the scalars

Diagram · orthonormal triple î, ĵ, k̂ (drag to rotate)

Worked example

Find a vector

\vec{v}

in the plane of

\vec{b} = \hat{i} + \hat{j}

and

\vec{c} = \hat{j} + \hat{k}

that is perpendicular to

\vec{c}

Coplanar form: $\vec{v} = \vec{b} + \lambda\vec{c} = \hat{i} + (1+\lambda)\hat{j} + \lambda\hat{k}$ .
Perpendicular to $\vec{c}$ : $\vec{v}\cdot\vec{c} = 0$ . Here $\vec{v}\cdot\vec{c} = (1+\lambda)\cdot 1 + \lambda\cdot 1 = 1 + 2\lambda$ .
Set $1 + 2\lambda = 0 \Rightarrow \lambda = -\tfrac{1}{2}$ . Then $\vec{v} = \hat{i} + \tfrac{1}{2}\hat{j} - \tfrac{1}{2}\hat{k}$ , or scaling by 2, $2\hat{i} + \hat{j} - \hat{k}$ .

Answer:

\vec{v} = 2\hat{i} + \hat{j} - \hat{k}

(any non-zero multiple works).

Practice this conceptself-check · 4 quick reps

Try it yourself

Find a vector in the plane of

\vec{b} = \hat{i} + \hat{j}

and

\vec{c} = \hat{i} - \hat{j}

perpendicular to

\vec{b}

Practice — Level 1 (4 reps)

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

1.
Form of a vector coplanar with $\vec{b}, \vec{c}$ ?
2.
Angle-bisector of $\vec{b}, \vec{c}$ is parallel to?
3.
\"Perpendicular to $\vec{q}$ \" gives which equation?
4.
Coplanarity fixes the ___; the second condition fixes the ___.

From the bank · past-year question

Example 9VectorsHARD

The vector

\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}

lies in the plane of the vectors

\vec{b} = \hat{i}+\hat{j}

and

\vec{c} = \hat{j}+\hat{k}

and bisects the angle between

\vec{b}

and

\vec{c}

. Then which one of the following gives possible values of

\alpha

and

\beta

[Q111 · 9th May Shift 2 · 2023]

Angle bisector uses UNIT vectors, not the raw vectors

The internal bisector of

\vec{b}

and

\vec{c}

is along

\hat{b} + \hat{c} = \dfrac{\vec{b}}{|\vec{b}|} + \dfrac{\vec{c}}{|\vec{c}|}

. Adding

\vec{b} + \vec{c}

directly only bisects when

|\vec{b}| = |\vec{c}|

— otherwise you get the wrong direction.

Coplanar form first, condition second

Always WRITE the coplanar form

m\vec{b} + n\vec{c}

before imposing the perpendicular / projection / magnitude condition. Trying to satisfy the condition without restricting to the plane gives a vector that doesn't actually lie where the question requires.

Drill 2 more on a vector lying in the plane of two others

Concept 10 of 10

Components of a vector against a transformed basis

Intuition

Three non-coplanar vectors form a basis, so any vector

\vec{s}

has UNIQUE components along them. The HARD twist: you're given

\vec{s}

's components along

\vec{p}, \vec{q}, \vec{r}

, and asked for its components along COMBINATIONS of them (like

-\vec{p}+\vec{q}+\vec{r}

). Write

\vec{s}

the new way, substitute, and match coefficients of

\vec{p}, \vec{q}, \vec{r}

— a clean linear system.

Definition

If $\vec{p}, \vec{q}, \vec{r}$ are non-coplanar and $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$ (the given components), then writing $\vec{s} = x(-\vec{p}+\vec{q}+\vec{r}) + y(\vec{p}-\vec{q}+\vec{r}) + z(-\vec{p}-\vec{q}+\vec{r})$ and collecting like terms gives a system: the coefficient of $\vec{p}$ must equal $4$ , of $\vec{q}$ equal $3$ , of $\vec{r}$ equal $5$ . Solve the $3\times 3$ system for $x, y, z$ ; independence of $\vec{p}, \vec{q}, \vec{r}$ is what makes the coefficient-matching valid.

Match coefficients of a basis

\begin{cases} -x + y - z = 4 \\ \ \ x - y - z = 3 \\ \ \ x + y + z = 5 \end{cases}

rowscoefficients of $\vec{p}, \vec{q}, \vec{r}$ equated to the given components

Worked example

\vec{p}, \vec{q}, \vec{r}

non-coplanar;

\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}

. Find

x, y, z

\vec{s} = x(-\vec{p}+\vec{q}+\vec{r}) + y(\vec{p}-\vec{q}+\vec{r}) + z(-\vec{p}-\vec{q}+\vec{r})

, and evaluate

2x + y + z

Collect coefficients. $\vec{p}$ : $-x + y - z = 4$ . $\vec{q}$ : $x - y - z = 3$ . $\vec{r}$ : $x + y + z = 5$ .
Add the $\vec{q}$ - and $\vec{r}$ -equations: $(x-y-z) + (x+y+z) = 3 + 5 \Rightarrow 2x = 8 \Rightarrow x = 4$ .
Add the $\vec{p}$ - and $\vec{r}$ -equations: $(-x+y-z) + (x+y+z) = 4 + 5 \Rightarrow 2y = 9 \Rightarrow y = \tfrac{9}{2}$ .
Add the $\vec{p}$ - and $\vec{q}$ -equations: $(-x+y-z) + (x-y-z) = 4 + 3 \Rightarrow -2z = 7 \Rightarrow z = -\tfrac{7}{2}$ .
Now $2x + y + z = 8 + \tfrac{9}{2} - \tfrac{7}{2} = 8 + 1 = 9$ .

Answer:

x = 4,\ y = \tfrac{9}{2},\ z = -\tfrac{7}{2}

;

2x + y + z = 9

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
Three non-coplanar vectors form a ___.
2.
Are the components of $\vec{s}$ along a basis unique?
3.
Why may we equate coefficients of $\vec{p}, \vec{q}, \vec{r}$ ?
4.
Solve $x - y - z = 3,\ x + y + z = 5$ for $x$ .

From the bank · past-year question

Example 10VectorsHARD

Suppose that

\vec{p}

\vec{q}

and

\vec{r}

are three non-coplanar vectors in

\mathbb{R}^3

. Let the components of a vector

\vec{s}

along

\vec{p}

\vec{q}

and

\vec{r}

be 4, 3 and 5 respectively. If the components of this vector

\vec{s}

along

(-\vec{p}+\vec{q}+\vec{r})

(\vec{p}-\vec{q}+\vec{r})

and

(-\vec{p}-\vec{q}+\vec{r})

are

x

y

and

z

respectively, then the value of

2x+y+z

[Q130 · 9th May Shift 1 · 2023]

Equating components needs an INDEPENDENT basis

Matching coefficients of

\vec{p}, \vec{q}, \vec{r}

on both sides is valid ONLY because they are non-coplanar (independent), giving a unique representation. If they were coplanar the components wouldn't be unique and the method collapses.

Answer the asked combination, not the raw $x, y, z$

Many of these end by asking for a combination like

2x + y + z

, not the individual scalars. Solve the system fully, then plug into the requested expression — a fraction in

y

and

z

often cancels cleanly there.

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 (10)

Linear combination of vectors
Linear combination
$\vec{r} = m\vec{a} + n\vec{b} + p\vec{c} \qquad r_i = ma_i + nb_i + pc_i$
Collinear vectors and collinear points (one scalar)
Collinearity (one scalar)
$\vec{a} = k\vec{b} \qquad \dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}$
Coplanar vectors (two scalars)
Coplanarity (two scalars)
$\vec{a} = m\vec{b} + n\vec{c} \qquad [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$
Forming a combination, then normalising / measuring it
Scale a combination to a target magnitude
$M\,\hat{r} = \dfrac{M}{|\vec{r}|}\,\vec{r} \qquad \hat{r} = \dfrac{\vec{r}}{|\vec{r}|}$
Collinearity of three points (and who lies between)
Three-point collinearity
$\vec{QR} = k\,\vec{PQ} \quad\text{where}\quad \vec{PQ} = \vec{q} - \vec{p},\ \ \vec{QR} = \vec{r} - \vec{q}$
Express a vector as a combination of two others
Two-equation linear system
$\begin{cases} a_1 = mb_1 + nc_1 \\ a_2 = mb_2 + nc_2 \end{cases}\ \Rightarrow\ m,\ n$
Linear dependence vs independence (the determinant test)
Dependence ⟺ vanishing determinant
$x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}\ (\text{not all }0) \iff \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$
Chained-collinearity systems (\"no two collinear\")
Chained collinearity
$\vec{a} + 2\vec{b} = t\vec{c}, \quad \vec{b} + 3\vec{c} = \lambda\vec{a} \;\Rightarrow\; t = -6,\ \lambda = -\tfrac{1}{2}$
A vector lying in the plane of two others
Coplanar form + a second condition
$\vec{v} = m\vec{b} + n\vec{c} \quad\text{with}\quad \vec{v}\cdot\vec{q} = 0 \ \text{ or }\ \dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = k$
Components of a vector against a transformed basis
Match coefficients of a basis
$\begin{cases} -x + y - z = 4 \\ \ \ x - y - z = 3 \\ \ \ x + y + z = 5 \end{cases}$

Watch out for (16)

A linear combination is built componentwise — three sums, not one
→ Linear combination of vectors
Collinear needs ONE scalar; coplanar needs TWO — keep the count straight
→ Collinear vectors and collinear points (one scalar)
Parallel VECTORS vs collinear POINTS
→ Collinear vectors and collinear points (one scalar)
Coplanar / dependent ⇒ TWO scalars and a vanishing determinant
→ Coplanar vectors (two scalars)
Build the combination BEFORE you normalise — don't normalise the parts
→ Forming a combination, then normalising / measuring it
\"Parallel of magnitude $M$ \" has TWO answers — $\pm M\hat{r}$
→ Forming a combination, then normalising / measuring it
Use displacements that SHARE a point
→ Collinearity of three points (and who lies between)
Solve from two equations, but ALWAYS verify with the third
→ Express a vector as a combination of two others
Linearly dependent = coplanar = zero determinant — three names, one idea
→ Linear dependence vs independence (the determinant test)
A second condition (like $|\vec{c}| = \sqrt{3}$ ) is part of the same problem
→ Linear dependence vs independence (the determinant test)
Match coefficients only because the vectors are independent
→ Chained-collinearity systems (\"no two collinear\")
Read the target carefully: $\vec{a}+2\vec{b}$ vs $\vec{a}+2\vec{b}+6\vec{c}$
→ Chained-collinearity systems (\"no two collinear\")
Angle bisector uses UNIT vectors, not the raw vectors
→ A vector lying in the plane of two others
Coplanar form first, condition second
→ A vector lying in the plane of two others
Equating components needs an INDEPENDENT basis
→ Components of a vector against a transformed basis
Answer the asked combination, not the raw $x, y, z$
→ Components of a vector against a transformed basis

Mastery check — 5 interleaved questions

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

Example 1VectorsMODERATE

\vec{a} = \hat{i}-2\hat{j}+3\hat{k}

and

\vec{b} = 2\hat{i}+3\hat{j}-\hat{k}

, then the angle between the vectors

(2\vec{a}+\vec{b})

and

(\vec{a}+2\vec{b})

[Q103 · 4th May Shift 2 · 2023]

Example 2VectorsHARD

Let

\vec{a},\vec{b}

, and

\vec{c}

be three non-zero vectors such that no two of these are collinear. If the vector

\vec{a}+2\vec{b}

is collinear with

\vec{c}

and

\vec{b}+3\vec{c}

is collinear with

\vec{a}

, then

\vec{a}+2\vec{b}+6\vec{c}

equals

[Q120 · 11th May Shift 2 · 2023]

Example 3VectorsHARD

\vec{a}=\hat{i}+2\hat{j}+\hat{k}

\vec{b}=\hat{i}-\hat{j}+\hat{k}

\vec{c}=\hat{i}+\hat{j}-\hat{k}

. Vector in plane of

\vec{a}

and

\vec{b}

, projection on

\vec{c}

\frac{1}{\sqrt{3}}

[Q135 · 10th May Shift 1 · 2024]

Example 4VectorsMODERATE

\vec{a} = 2\hat{i}-\hat{j}+\hat{k}

\vec{b} = \hat{i}+\hat{j}-2\hat{k}

and

\vec{c} = 4\hat{i}-2\hat{j}+\hat{k}

, then the unit vector in the direction of

3\vec{a}+\vec{b}-2\vec{c}

[Q122 · 16th May Shift 2 · 2023]

Example 5VectorsHARD

\vec{a},\vec{b},\vec{c}

are three non-zero vectors, no two of them are collinear,

\vec{a}+2\vec{b}

is collinear with

\vec{c}

\vec{b}+3\vec{c}

is collinear with

\vec{a}

, then

\vec{a}+2\vec{b}+6\vec{c}

[Q144 · 13th May Shift 1 · 2024]

Drill every past-year question on this subtopic

13 questions from the bank — paginated, with cart and Word-export support.

Drill the 13 questions

Drill every past-year question on this subtopic

Related notes