MHT-CET Maths · Line and Plane

Intersection, Coplanarity, and Skew Lines

When do two 3-D lines meet, lie in one plane, or fly past each other? The single tool is the scalar-triple-product determinant of the joining vector and the two direction vectors — zero means coplanar (they intersect or are parallel), non-zero means skew, and dividing by the cross-product magnitude gives the shortest distance.

All Line and Plane notes MHT-CET Maths strategy

Why this matters

This is the HARDEST subtopic in the chapter — about two-thirds of its PYQs are HARD — and the most repeated single stem across the whole Line and Plane chapter is 'these two lines intersect (are coplanar), find k'. That stem hides a signature trap: the coplanarity determinant is QUADRATIC in the unknown, so it has TWO answers (the bank's correct option lists both, e.g. k = 1, 2 or k = 0, -3) and a single-value distractor is the planted wrong answer. Master one determinant — the scalar triple product [joining vector, direction-1, direction-2] = 0 — and you own coplanarity, the intersect-find-k template, four-point coplanarity, and (divided by the cross-product magnitude) the shortest distance between skew lines. The shortest-distance questions reverse the same machinery: 'SD given, find the parameter' is again quadratic.

Concept 1 of 9

A general point on a line

Intuition

Every point on a line is its fixed point plus some multiple of its direction. Set the symmetric ratios equal to a single parameter

t

(or

\lambda

) and you can slide along the whole line by changing

t

. This one move — turning a line into a moving point — unlocks every intersection and coplanarity question.

Definition

A line through $A(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ is written symmetrically as $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=t$ . Setting that common ratio to a parameter $t$ gives the general point:

$x = x_1 + at,\quad y = y_1 + bt,\quad z = z_1 + ct$ .

In vector form, $\vec{r} = \vec{a} + t\,\vec{d}$ : a fixed position vector $\vec{a}$ plus a scalar $t$ times the direction $\vec{d}$ . Every later technique starts here — substitute this moving point into whatever condition the question gives.

General point on a line

P = (x_1 + at,\; y_1 + bt,\; z_1 + ct) \qquad \vec{r} = \vec{a} + t\,\vec{d}

$(x_1,y_1,z_1)$ a fixed point on the line
$(a,b,c)$ direction ratios of the line
$t$ parameter — sweeps out every point on the line

Worked example

Write the general point on the line

\dfrac{x-2}{3}=\dfrac{y+1}{-1}=\dfrac{z-4}{2}

, then find the point at which it leaves the fixed point (

t = 0

) and the point at

t = 2

Set each ratio equal to $t$ : $x = 2 + 3t,\; y = -1 - t,\; z = 4 + 2t$ .
At $t = 0$ : the fixed point $(2, -1, 4)$ .
At $t = 2$ : $(2 + 6,\; -1 - 2,\; 4 + 4) = (8, -3, 8)$ .

Answer:General point

(2+3t,\,-1-t,\,4+2t)

; at

t=2

it is

(8,-3,8)

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
General point on $\dfrac{x-1}{2}=\dfrac{y}{3}=\dfrac{z+1}{4}$ ?
2.
On the line through $(0,0,0)$ with direction $(1,2,3)$ , the point at $t=2$ ?
3.
If a line's general point is $(3+t,\,1-2t,\,5t)$ , its direction ratios are?
4.
A line in vector form $\vec{r}=\vec{a}+t\vec{d}$ : which part is the direction?

A NEGATIVE denominator is a negative direction ratio

\dfrac{y+1}{-1}

the direction component is

-1

, so

y = -1 - t

, NOT

-1 + t

. Carry the sign from the denominator straight into the general point — flipping it is the most common silent error.

Use DIFFERENT parameters for two different lines

When you parametrise two lines and look for an intersection, call them

t

and

s

(not both

t

). A single parameter forces them to move in lock-step and gives a wrong system — the lines meet at a point each reaches at its OWN parameter value.

Concept 2 of 9

Point where a line meets a plane

Intuition

A line pierces a plane at exactly one point (unless it is parallel). Write the line's general point, plug it into the plane equation, and you get ONE linear equation in the parameter

t

. Solve for

t

, put it back, and you have the piercing point. The coordinate planes are just special planes: the XZ-plane is

y = 0

, the XY-plane is

z = 0

, the YZ-plane is

x = 0

Definition

To find where the line $\vec{r} = \vec{a} + t\,\vec{d}$ meets the plane:

Substitute the general point $(x_1+at,\,y_1+bt,\,z_1+ct)$ into the plane equation $Ax + By + Cz = D$ .
Solve the resulting linear equation for $t$ .
Back-substitute $t$ into the general point to get the coordinates.

For a coordinate plane, set the relevant coordinate to 0: XZ-plane $\Rightarrow y = 0$ , XY-plane $\Rightarrow z = 0$ , YZ-plane $\Rightarrow x = 0$ . Solve for $t$ from that single equation. A variable plane in intercept form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ meets the axes at $(a,0,0),(0,b,0),(0,0,c)$ .

Line meets plane

A(x_1+at)+B(y_1+bt)+C(z_1+ct)=D \;\Rightarrow\; t \;\Rightarrow\; P

$(x_1{+}at,\,\ldots)$ general point on the line
$Ax+By+Cz=D$ the plane (set a coordinate $=0$ for a coordinate plane)
$t$ the single parameter value at the piercing point

Diagram · line piercing a plane (drag to rotate)

Substitute the line's point (x₀+at, y₀+bt, z₀+ct) into the plane equation → one equation in t → solve → back-substitute to get the pierce point P.

Worked example

Find where the line

\dfrac{x-1}{2}=\dfrac{y-2}{-3}=\dfrac{z+5}{4}

meets the plane

2x + 4y - z = 3

General point: $(1+2t,\;2-3t,\;-5+4t)$ .
Substitute into the plane: $2(1+2t)+4(2-3t)-(-5+4t)=3$ .
Simplify: $2+4t+8-12t+5-4t = 3 \Rightarrow 15 - 12t = 3 \Rightarrow t = 1$ .
Back-substitute $t = 1$ : $(3,\,-1,\,-1)$ .

Answer:The line meets the plane at

(3, -1, -1)

Practice this conceptself-check · 4 quick reps

Try it yourself

The line through

A(3,4,1)

and

B(5,1,6)

crosses the XZ-plane at which point?

Practice — Level 1 (4 reps)

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

1.
The XY-plane is which equation?
2.
Line $(t,\,1+t,\,2t)$ meets $z = 0$ at?
3.
An intercept-form plane meets the X-axis at?
4.
After solving for $t$ , what is the last step?

From the bank · past-year question

Example 2Line and PlaneMODERATE

If the line

\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}

meets the plane

x+2y+3z=15

at the point P, then the distance of P from the origin is:

[Q106 · 11th May Shift 2 · 2024]

XZ-plane is $y = 0$ , not $z = 0$

The plane is named by the two axes it CONTAINS, so the MISSING axis is the one set to zero. XZ-plane omits

y

\Rightarrow y = 0

; YZ-plane omits

x

\Rightarrow x = 0

. Mixing these up sends you to the wrong coordinate every time.

The question may want a derived quantity, not the point itself

Many stems ask for the DISTANCE of the piercing point from the origin, or its reflection in a plane. Find the point first, then do the extra step —

\sqrt{x^2+y^2+z^2}

for distance, or negate one coordinate for a reflection. Stopping at the point is a half-finished answer.

Drill 4 more on point where a line meets a plane

Concept 3 of 9

Point of intersection of two lines

Intuition

Two lines meet where their general points coincide. Parametrise each with its OWN parameter (

t

and

s

), set the three coordinates equal, and you get three equations in two unknowns. Solve any two for

t, s

, then the THIRD equation must hold automatically — if it does, the lines truly intersect; substitute back for the point.

Definition

To find the intersection of $\vec{r}_1 = \vec{a}_1 + t\,\vec{d}_1$ and $\vec{r}_2 = \vec{a}_2 + s\,\vec{d}_2$ :

Write both general points and equate coordinate-by-coordinate $\Rightarrow$ three equations in $t, s$ .
Solve two of them for $t$ and $s$ .
Check the third equation is satisfied (consistency = they really intersect).
Substitute $t$ (or $s$ ) into its line to get the point.

Intersection by equating

x_1+a_1t = x_2+a_2s,\quad y_1+b_1t = y_2+b_2s,\quad z_1+c_1t = z_2+c_2s

$t,\,s$ the two SEPARATE parameters (one per line)
3 equations, 2 unknownssolve 2, the 3rd must check out for a real intersection

Worked example

Find the distance of

(2,4,0)

from the intersection of

\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}

and

\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}

Line 1: $(3t-6,\;2t,\;t-1)$ ; Line 2: $(4s+7,\;3s+9,\;2s+4)$ .
Equate: $3t-4s = 13,\;\; 2t-3s = 9,\;\; t-2s = 5$ . Solving the first two gives $t = 3,\; s = -1$ .
Check the third: $3 - 2(-1) = 5$ ✓. Intersection point $(3,6,2)$ .
Distance from $(2,4,0)$ : $\sqrt{1^2+2^2+2^2} = \sqrt{9} = 3$ .

Answer:Distance

= 3

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the point of intersection of

\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{1}

and

\dfrac{x-4}{1}=\dfrac{y-5}{2}=\dfrac{z-3}{3}

Practice — Level 1 (4 reps)

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

1.
Two lines giving 3 equations in 2 unknowns — how many must you solve, how many must check?
2.
If the third equation FAILS, the lines are?
3.
Reflection of $(2,-4,7)$ in the XY-plane?
4.
Why use $t$ and $s$ , not $t$ for both lines?

From the bank · past-year question

Example 3Line and PlaneMODERATE

Two lines

\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}

and

\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}

intersect at the point R. Then reflection of R in the xy-plane has co-ordinates

[Q103 · 10th May Shift 2 · 2024]

Always verify the THIRD equation

Solving two of the three coordinate equations ALWAYS gives some

t, s

— even for skew lines. The lines only truly intersect if those values also satisfy the third equation. Skipping the check can hand you a phantom 'intersection point' for lines that never meet.

Read the FINAL ask

Most of these stems don't want the intersection point — they want the distance from it to another point, or its reflection. Find the point, then finish the requested operation. The intersection is the midpoint of the work, not the answer.

Drill 1 more on point of intersection of two lines

Concept 4 of 9

Coplanarity and intersect-find-k by the scalar triple product

Intuition

Two lines lie in one plane (are coplanar) exactly when the vector joining a point of one to a point of the other lies in the plane spanned by their two directions — i.e. the three vectors are flat (zero box volume). That flatness is the scalar triple product = 0. This is THE most-tested template in the chapter: 'these lines intersect / are coplanar, find the unknown'. Crucially, when the unknown sits in BOTH direction vectors, the determinant comes out QUADRATIC — so there are usually TWO values.

Definition

Two lines with points $A_1, A_2$ and directions $\vec{d}_1=(a_1,b_1,c_1)$ , $\vec{d}_2=(a_2,b_2,c_2)$ are coplanar (they intersect or are parallel) iff:

\big[\,\vec{A_1A_2},\;\vec{d_1},\;\vec{d_2}\,\big]=\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0

For NON-parallel lines, coplanar ** $\Leftrightarrow$ they intersect**. So 'find $k$ for which the lines intersect' uses the SAME determinant $=0$ .
If the unknown appears in both direction rows, expanding gives a quadratic $\Rightarrow$ TWO answers.
A non-zero value $\Rightarrow$ the lines are skew (do not intersect).

Coplanarity / intersection determinant = 0

\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0

Row 1joining vector $\vec{A_1A_2}=A_2-A_1$
Row 2direction ratios of line 1
Row 3direction ratios of line 2

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

[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

The lines

\dfrac{x-3}{1}=\dfrac{y-2}{1}=\dfrac{z-5}{-k}

and

\dfrac{x-4}{k}=\dfrac{y-3}{1}=\dfrac{z-3}{2}

are coplanar. Find

k

Points $(3,2,5)$ , $(4,3,3)$ $\Rightarrow$ joining vector $(1,1,-2)$ . Directions $(1,1,-k)$ and $(k,1,2)$ .
Set the determinant to zero: $\begin{vmatrix} 1 & 1 & -2 \\ 1 & 1 & -k \\ k & 1 & 2 \end{vmatrix}=0$ .
Expand along row 1: $1(2+k) - 1(2+k^2) - 2(1-k) = 0$ .
Simplify: $2+k-2-k^2-2+2k = -k^2+3k-2 = 0 \Rightarrow k^2-3k+2 = 0$ .
Factor: $(k-1)(k-2)=0 \Rightarrow k = 1$ or $k = 2$ .

Answer:

k = 1

k = 2

— BOTH values (the determinant is quadratic)

Practice this conceptself-check · 5 quick reps

Try it yourself

For what

k

are

\dfrac{x-2}{1}=\dfrac{y-3}{1}=\dfrac{z-4}{-k}

and

\dfrac{x-1}{k}=\dfrac{y-4}{2}=\dfrac{z-5}{1}

coplanar?

Practice — Level 1 (5 reps)

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

1.
Coplanarity condition for two lines, in one phrase?
2.
For non-parallel lines, coplanar is equivalent to?
3.
If the determinant is NON-zero, the lines are?
4.
$k^2-3k+2=0$ factors to?
5.
Why does intersect-find-k often give two answers?

From the bank · past-year question

Example 4Line and PlaneHARD

If the lines

\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}

and

\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}

intersect, then the value of k is:

[Q104 · 11th May Shift 2 · 2024]

The QUADRATIC trap — there are usually TWO values of k

When the unknown sits in both direction rows, expanding the determinant gives

k^2 + \ldots = 0

, which has TWO roots. The bank's correct option lists both (e.g. '1, 2' or '0, -3'); the planted distractor gives only ONE root. If your working produces a single

k

, you almost certainly dropped a term — re-expand and look for the squared term.

Joining vector is $A_2 - A_1$ , and it is ROW 1

The determinant's first row is the vector between the two fixed points (head minus tail), NOT a direction. Putting a direction vector in row 1, or subtracting the points the wrong way, flips signs and corrupts the whole expansion. Layout: joining vector on top, then the two directions.

'Intersect' uses the SAME determinant as 'coplanar'

For non-parallel lines, intersecting and being coplanar are the same condition. So whether the stem says 'intersect' or 'coplanar', set the same scalar-triple-product determinant to zero — don't hunt for a different formula.

Drill 9 more on coplanarity and intersect-find-k by the scalar triple product

Concept 5 of 9

Four points coplanar

Intuition

Four points lie in one plane when the three edge-vectors drawn from one of them are flat — zero box volume again. Pick a base point

A

, build

\vec{AB}, \vec{AC}, \vec{AD}

, and set their scalar triple product to zero. Same engine as line coplanarity, just with point-to-point vectors instead of directions.

Definition

Points $A, B, C, D$ are coplanar iff the three vectors from $A$ have zero scalar triple product:

[\,\vec{AB},\;\vec{AC},\;\vec{AD}\,]=\begin{vmatrix} \vec{AB} \\ \vec{AC} \\ \vec{AD} \end{vmatrix}=0

where each row holds the components of that edge vector (

\vec{AB}=B-A

, etc.). If an unknown sits in one coordinate, this gives a LINEAR equation for it (one answer), unlike the line-coplanarity case which is usually quadratic.

Four points coplanar

[\,\vec{AB},\;\vec{AC},\;\vec{AD}\,]=0

$\vec{AB}=B-A$ edge vector from base point $A$ to $B$
scalar triple productdeterminant of the three edge vectors as rows

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

[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

\lambda

so that the points

A(1,0,1),\,B(2,1,3),\,C(0,2,1),\,D(\lambda,1,2)

are coplanar.

$\vec{AB} = (1,1,2)$ , $\vec{AC} = (-1,2,0)$ , $\vec{AD} = (\lambda-1,1,1)$ .
Set $[\vec{AB},\vec{AC},\vec{AD}] = \begin{vmatrix} 1 & 1 & 2 \\ -1 & 2 & 0 \\ \lambda-1 & 1 & 1 \end{vmatrix} = 0$ .
Expand along row 1: $1(2-0) - 1(-1-0) + 2(-1-2(\lambda-1)) = 2 + 1 + 2(1-2\lambda)$ .
$= 3 + 2 - 4\lambda = 5 - 4\lambda = 0 \Rightarrow \lambda = \tfrac{5}{4}$ .

Answer:

\lambda = \dfrac{5}{4}

Practice this conceptself-check · 4 quick reps

Try it yourself

Are the points

A(0,0,0),\,B(1,2,3),\,C(2,4,6),\,D(1,1,1)

coplanar?

Practice — Level 1 (4 reps)

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

1.
Condition for 4 points to be coplanar?
2.
Edge vector $\vec{AB}$ equals?
3.
If two of the edge-vector rows are proportional, the determinant is?
4.
An unknown in ONE coordinate gives a ___ equation here?

From the bank · past-year question

Example 5Line and PlaneMODERATE

If the points

A(1,1,2),B(2,1,p),C(1,0,3)

and

D(2,2,0)

are coplanar then the value of

p

[Q109 · 20 April Shift II · 2025]

All edge vectors must start from the SAME base point

Use

\vec{AB}, \vec{AC}, \vec{AD}

— all from

A

. Mixing in

\vec{BC}

\vec{CD}

breaks the 'three edges of a box from one corner' picture and the determinant no longer tests coplanarity.

Four-point coplanarity is usually LINEAR in the unknown

Here the unknown sits in only one edge vector, so the expansion is linear — ONE value. Don't expect the two-root quadratic of the line-coplanarity template; if you get a quadratic, you probably put the unknown in two rows by mistake.

Concept 6 of 9

Shortest distance between skew lines

Intuition

Skew lines never meet, but there is a unique shortest segment between them — and it is perpendicular to BOTH. Its direction is

\vec{d_1}\times\vec{d_2}

. Project the joining vector onto that common perpendicular: the shortest distance is the scalar triple product of (joining vector,

\vec{d_1}

\vec{d_2}

) divided by

|\vec{d_1}\times\vec{d_2}|

. The numerator is the SAME coplanarity determinant — so SD

= 0

exactly when the lines are coplanar.

Definition

For lines $\vec{r}_1=\vec{a_1}+\lambda\vec{d_1}$ and $\vec{r}_2=\vec{a_2}+\mu\vec{d_2}$ , the shortest distance is

d = \frac{\big|\,(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})\,\big|}{|\vec{d_1}\times\vec{d_2}|}

The numerator is $\big|[\,\vec{a_2}-\vec{a_1},\,\vec{d_1},\,\vec{d_2}\,]\big|$ — the coplanarity determinant in absolute value.
** $d = 0 \Leftrightarrow$ coplanar** (intersecting or parallel).
'SD given, find a parameter' reverses it: set $d$ equal to the given value and solve — often a quadratic, giving two values whose SUM the question may ask for.

Shortest distance (skew lines)

d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}

$\vec{a_2}-\vec{a_1}$ vector joining the two fixed points
$\vec{d_1}\times\vec{d_2}$ common perpendicular direction
numeratorabsolute scalar triple product = coplanarity determinant

Worked example

Find the shortest distance between

\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}

and

\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}

$\vec{a_2}-\vec{a_1} = (2-1,\,4-2,\,5-3) = (1,2,2)$ . Directions $\vec{d_1}=(2,3,4)$ , $\vec{d_2}=(3,4,5)$ .
$\vec{d_1}\times\vec{d_2} = (3\cdot5-4\cdot4,\;4\cdot3-2\cdot5,\;2\cdot4-3\cdot3) = (-1,2,-1)$ ; magnitude $\sqrt{1+4+1} = \sqrt{6}$ .
Numerator: $|(1)(-1)+(2)(2)+(2)(-1)| = |-1+4-2| = 1$ .
$d = \dfrac{1}{\sqrt{6}}$ .

Answer:

d = \dfrac{1}{\sqrt{6}}

units

Practice this conceptself-check · 5 quick reps

Try it yourself

If the shortest distance between

\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{\lambda}

and

\dfrac{x-2}{1}=\dfrac{y-4}{4}=\dfrac{z-5}{5}

\dfrac{1}{\sqrt{3}}

, find the SUM of possible

\lambda

Practice — Level 1 (5 reps)

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

1.
Common perpendicular direction of two skew lines?
2.
Shortest distance $= 0$ means the lines are?
3.
Numerator of the SD formula equals which determinant?
4.
If SD given and $\lambda$ is in a direction, the equation in $\lambda$ is usually?
5.
$\vec{d_1}\times\vec{d_2}=(8,8,4)$ : its magnitude?

From the bank · past-year question

Example 6Line and PlaneHARD

The shortest distance (in units) between the lines

\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}

and

\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})

[Q103 · 12th May Shift 2 · 2024]

Divide by $|\vec{d_1}\times\vec{d_2}|$ , and take the ABSOLUTE value on top

The shortest distance is a length, so the numerator is in modulus and you divide by the cross-product's magnitude — NOT by

|\vec{a_2}-\vec{a_1}|

. Forgetting the absolute value can give a negative 'distance'; dividing by the wrong magnitude gives a plausible-looking but wrong fraction.

'SD given, find the parameter' is a QUADRATIC — expect two values

Reversing the formula (set

d

to a number, solve for the unknown) almost always squares into a quadratic. The question often asks for the SUM of the two values — read it off as

-b/a

without even finding the roots separately.

Drill 5 more on shortest distance between skew lines

Concept 7 of 9

Direction of the line of intersection of two planes

Intuition

Two non-parallel planes meet along a line. That line lies inside BOTH planes, so it is perpendicular to BOTH normals — which means its direction is the cross product of the two normals

\vec{n_1}\times\vec{n_2}

. You don't need a point on the line to answer 'a vector parallel to the line of intersection'.

Definition

Planes $\vec{r}\cdot\vec{n_1}=d_1$ and $\vec{r}\cdot\vec{n_2}=d_2$ intersect in a line whose direction is

\vec{d}=\vec{n_1}\times\vec{n_2}

because the line lies in both planes and is therefore perpendicular to each normal. Any non-zero scalar multiple of

\vec{n_1}\times\vec{n_2}

is an equally valid direction.

Direction of line of intersection of two planes

\vec{d}=\vec{n_1}\times\vec{n_2}

$\vec{n_1},\,\vec{n_2}$ normals of the two planes
$\vec{n_1}\times\vec{n_2}$ direction of their line of intersection

Worked example

Find a vector parallel to the line of intersection of the planes

\vec{r}\cdot(2\hat{i}+\hat{j}-\hat{k})=3

and

\vec{r}\cdot(\hat{i}-\hat{j}+2\hat{k})=1

Normals: $\vec{n_1}=(2,1,-1)$ , $\vec{n_2}=(1,-1,2)$ .
$\vec{n_1}\times\vec{n_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\1&-1&2\end{vmatrix}$ .
$= \hat{i}(2-1) - \hat{j}(4+1) + \hat{k}(-2-1) = \hat{i}-5\hat{j}-3\hat{k}$ .

Answer:

\hat{i}-5\hat{j}-3\hat{k}

Practice this conceptself-check · 4 quick reps

Try it yourself

A vector parallel to the line of intersection of the planes

x+y+z=1

and

x-y+z=2

Practice — Level 1 (4 reps)

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

1.
Direction of the line where two planes meet?
2.
Why is the line perpendicular to each normal?
3.
Cross product of $(1,0,0)$ and $(0,1,0)$ ?
4.
Is $2(\vec{n_1}\times\vec{n_2})$ also a valid direction?

From the bank · past-year question

Example 7Line and PlaneMODERATE

A vector parallel to the line of intersection of the planes

\vec{r}\cdot(3\hat{i}-\hat{j}+\hat{k})=1

and

\vec{r}\cdot(\hat{i}+4\hat{j}-2\hat{k})=2

[Q118 · 14th May Shift 2 · 2024]

Use the NORMALS, not the planes' constants

The direction depends only on

\vec{n_1}\times\vec{n_2}

— the right-hand-side constants

d_1, d_2

play no part. They would only matter if you wanted an actual POINT on the line.

Cross-product sign — keep the middle term's minus

Expanding

\vec{n_1}\times\vec{n_2}

, the

\hat{j}

component carries a leading minus sign in the cofactor expansion. Dropping it flips that component and you'll match the wrong option (the distractors are often the sign-flipped vector).

Concept 8 of 9

Transversal intersecting two given lines

Intuition

A transversal is a third line that crosses two given lines, with its OWN fixed direction. Take a general point

A

on line 1 and

B

on line 2 (each with its own parameter); the segment

\vec{AB}

must be PARALLEL to the transversal's given direction ratios. That parallel condition pins down both parameters, hence both points A and B.

Definition

To find where a line of given direction $(l,m,n)$ meets two lines at $A$ and $B$ :

Write $A$ as a general point of line 1 (parameter $\lambda$ ) and $B$ of line 2 (parameter $\mu$ ).
**Force $\vec{AB}\parallel (l,m,n)$ **: the components of $\vec{AB}$ are proportional to $(l,m,n)$ , giving $\dfrac{(\vec{AB})_x}{l}=\dfrac{(\vec{AB})_y}{m}=\dfrac{(\vec{AB})_z}{n}$ .
Solve the two independent proportion equations for $\lambda, \mu$ ; substitute back for $A$ and $B$ .

Transversal condition

\vec{AB}\parallel (l,m,n) \;\Rightarrow\; \frac{(\vec{AB})_x}{l}=\frac{(\vec{AB})_y}{m}=\frac{(\vec{AB})_z}{n}

$A,\,B$ general points on line 1 (param $\lambda$ ) and line 2 (param $\mu$ )
$(l,m,n)$ direction ratios of the transversal

Worked example

A line with direction ratios

(1,1,1)

meets

\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}

at A and

\dfrac{x-1}{2}=\dfrac{y-1}{1}=\dfrac{z-1}{1}

at B. Find A and B.

$A = (\lambda,\;2\lambda,\;3\lambda)$ ; $B = (1+2\mu,\;1+\mu,\;1+\mu)$ .
$\vec{AB} = (1+2\mu-\lambda,\;1+\mu-2\lambda,\;1+\mu-3\lambda)$ must be proportional to $(1,1,1)$ — so all three components are EQUAL.
Equate $x$ - and $y$ -components: $1+2\mu-\lambda = 1+\mu-2\lambda \Rightarrow \mu+\lambda = 0$ . Equate $y$ - and $z$ -components: $1+\mu-2\lambda = 1+\mu-3\lambda \Rightarrow \lambda = 0$ , hence $\mu = 0$ .
Substitute: $A = (0,0,0)$ , $B = (1,1,1)$ ; indeed $\vec{AB} = (1,1,1)$ ✓.

Answer:

A(0,0,0),\; B(1,1,1)

Practice this conceptself-check · 4 quick reps

Try it yourself

After finding the parameters in a transversal problem, what MUST you verify before trusting your A and B?

Practice — Level 1 (4 reps)

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

1.
The segment joining the two intersection points must be ___ to the transversal?
2.
How many parameters do you introduce for a transversal meeting two lines?
3.
$\vec{AB}\parallel(l,m,n)$ gives how many independent equations?
4.
$\vec{AB}=(2,-8,4)$ — is it parallel to $(1,-4,2)$ ?

From the bank · past-year question

Example 8Line and PlaneHARD

A line having direction ratios

1, -4, 2

intersects the lines

\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}

and

\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}

at the points A and B respectively. Then co-ordinates of points A and B are

[Q112 · Shift 1 · 2023]

VERIFY the parallel condition after solving

It is easy to solve two of the proportion equations and stop, but the found

\lambda, \mu

must make

\vec{AB}

genuinely proportional to

(l,m,n)

. Skip this and you can land on the sign-flipped distractor (e.g.

(8,6,7)

instead of

(-8,6,-7)

Two SEPARATE parameters, one per line

Point A uses line 1's parameter

\lambda

; point B uses line 2's parameter

\mu

. Re-using one symbol collapses the system and gives no valid transversal.

Drill 1 more on transversal intersecting two given lines

Concept 9 of 9

Condition for a line to lie in a plane

Intuition

A line lies entirely in a plane when TWO things hold at once: the line's direction is perpendicular to the plane's normal (so the line is parallel to the plane), AND one point of the line actually sits on the plane (so it is not just parallel-but-off). Both conditions together force every point of the line onto the plane.

Definition

The line through $P_0$ with direction $\vec{d}$ lies in the plane $Ax+By+Cz+D=0$ (normal $\vec{n}=(A,B,C)$ ) iff BOTH:

Direction perpendicular to normal: $\vec{d}\cdot\vec{n}=0$ (line is parallel to the plane), AND
Point on plane: $P_0$ satisfies the plane equation.

Either condition alone is not enough — $\vec{d}\cdot\vec{n}=0$ without the point gives a line parallel to but OUTSIDE the plane.

Line lies in plane (both conditions)

\vec{d}\cdot\vec{n}=0 \quad\text{and}\quad P_0 \in \text{plane}

$\vec{d}\cdot\vec{n}=0$ direction ⟂ normal ⇒ line parallel to plane
$P_0 \in$ planea point of the line satisfies the plane equation

Worked example

Find

a

and

b

so that the line

\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{a}

lies in the plane

x - 2y + z = b

Direction $(2,1,a)$ ⟂ normal $(1,-2,1)$ : $2(1)+1(-2)+a(1)=0 \Rightarrow a = 0$ .
Point $(1,-1,2)$ on the plane: $1 - 2(-1) + 2 = b \Rightarrow b = 5$ .
So the line $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{0}$ lies in $x - 2y + z = 5$ .

Answer:

a = 0,\; b = 5

Practice this conceptself-check · 4 quick reps

Try it yourself

Does the line through

(1,0,0)

with direction

(1,1,1)

lie in the plane

x+y-2z=1

Practice — Level 1 (4 reps)

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

1.
Two conditions for a line to lie in a plane?
2.
$\vec{d}\cdot\vec{n}=0$ alone means the line is?
3.
Direction $(2,1,1)$ , normal $(1,-1,-1)$ : is $\vec{d}\cdot\vec{n}=0$ ?
4.
Which condition do you use to find the constant term $\beta$ ?

From the bank · past-year question

Example 9Line and PlaneHARD

Line

\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}

lies in plane

x+3y-\alpha z+\beta=0

. Value of

\alpha^2+\alpha\beta+\beta^2

[Q140 · 10th May Shift 1 · 2024]

BOTH conditions are required

Solving only

\vec{d}\cdot\vec{n}=0

gives

\alpha

, but you still need the point-on-plane condition to get

\beta

(and to confirm the line is actually IN, not merely parallel to, the plane). One condition gives one unknown — you need both for problems with two unknowns.

Perpendicular DIRECTIONS, parallel LINE

It's the line's direction that is perpendicular to the normal, which makes the LINE parallel to the plane. Don't confuse 'direction ⟂ normal' with 'line ⟂ plane' — those are opposite situations.

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 (9)

A general point on a line
General point on a line
$P = (x_1 + at,\; y_1 + bt,\; z_1 + ct) \qquad \vec{r} = \vec{a} + t\,\vec{d}$
Point where a line meets a plane
Line meets plane
$A(x_1+at)+B(y_1+bt)+C(z_1+ct)=D \;\Rightarrow\; t \;\Rightarrow\; P$
Point of intersection of two lines
Intersection by equating
$x_1+a_1t = x_2+a_2s,\quad y_1+b_1t = y_2+b_2s,\quad z_1+c_1t = z_2+c_2s$
Coplanarity and intersect-find-k by the scalar triple product
Coplanarity / intersection determinant = 0
$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0$
Four points coplanar
Four points coplanar
$[\,\vec{AB},\;\vec{AC},\;\vec{AD}\,]=0$
Shortest distance between skew lines
Shortest distance (skew lines)
$d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}$
Direction of the line of intersection of two planes
Direction of line of intersection of two planes
$\vec{d}=\vec{n_1}\times\vec{n_2}$
Transversal intersecting two given lines
Transversal condition
$\vec{AB}\parallel (l,m,n) \;\Rightarrow\; \frac{(\vec{AB})_x}{l}=\frac{(\vec{AB})_y}{m}=\frac{(\vec{AB})_z}{n}$
Condition for a line to lie in a plane
Line lies in plane (both conditions)
$\vec{d}\cdot\vec{n}=0 \quad\text{and}\quad P_0 \in \text{plane}$

Watch out for (19)

A NEGATIVE denominator is a negative direction ratio
→ A general point on a line
Use DIFFERENT parameters for two different lines
→ A general point on a line
XZ-plane is $y = 0$ , not $z = 0$
→ Point where a line meets a plane
The question may want a derived quantity, not the point itself
→ Point where a line meets a plane
Always verify the THIRD equation
→ Point of intersection of two lines
Read the FINAL ask
→ Point of intersection of two lines
The QUADRATIC trap — there are usually TWO values of k
→ Coplanarity and intersect-find-k by the scalar triple product
Joining vector is $A_2 - A_1$ , and it is ROW 1
→ Coplanarity and intersect-find-k by the scalar triple product
'Intersect' uses the SAME determinant as 'coplanar'
→ Coplanarity and intersect-find-k by the scalar triple product
All edge vectors must start from the SAME base point
→ Four points coplanar
Four-point coplanarity is usually LINEAR in the unknown
→ Four points coplanar
Divide by $|\vec{d_1}\times\vec{d_2}|$ , and take the ABSOLUTE value on top
→ Shortest distance between skew lines
'SD given, find the parameter' is a QUADRATIC — expect two values
→ Shortest distance between skew lines
Use the NORMALS, not the planes' constants
→ Direction of the line of intersection of two planes
Cross-product sign — keep the middle term's minus
→ Direction of the line of intersection of two planes
VERIFY the parallel condition after solving
→ Transversal intersecting two given lines
Two SEPARATE parameters, one per line
→ Transversal intersecting two given lines
BOTH conditions are required
→ Condition for a line to lie in a plane
Perpendicular DIRECTIONS, parallel LINE
→ Condition for a line to lie in a plane

Mastery check — 5 interleaved questions

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

Example 1Line and PlaneMODERATE

The co-ordinates of the point, where the line

\dfrac{x-1}{2}=\dfrac{y-2}{-3}=\dfrac{z+5}{4}

meets the plane

2x+4y-z=3

, are

[Q139 · 9th May Shift 2 · 2024]

Example 2Line and PlaneMODERATE

The distance of the point

(2,4,0)

from the point of intersection of the lines

\frac{x+ 6}{3}=\frac{y}{2}=\frac{z+ 1}{1}

and

\frac{x- 7}{4}=\frac{y- 9}{3}=\frac{z- 4}{2}

[Q129 · 20 April Shift II · 2025]

Example 3Line and PlaneHARD

If the lines

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}

and

\frac{x-3}{-1}=\frac{y-k}{2}=\frac{z}{1}

intersect, then k is equal to

[Q122 · 10th May Shift 1 · 2023]

Example 4Line and PlaneHARD

The shortest distance between lines

\vec{r} = (\hat{i}+2\hat{j}-\hat{k}) + \lambda(2\hat{i}+\hat{j}-3\hat{k})

and

\vec{r} = (2\hat{i}-\hat{j}+2\hat{k}) + \mu(\hat{i}-\hat{j}+\hat{k})

[Q113 · 10th May Shift 2 · 2023]

Example 5Line and PlaneHARD

A line having direction ratios

1, -4, 2

intersects the lines

\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}

and

\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}

at points A and B respectively. Then co-ordinates of A and B are

[Q112 · 16th May Shift 2 · 2023]

Drill every past-year question on this subtopic

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

Drill the 28 questions

Drill every past-year question on this subtopic

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