MHT-CET Maths · Formula sheet

Line and Plane formulas

55 formulas and 118 common traps for MHT-CET Maths Line and Plane, grouped by subtopic.

Line — Equation, Direction Cosines, and Vector Form

Direction ratios, direction cosines, and the l² + m² + n² = 1 identity

Direction cosines and their identity

(l, m, n) = \frac{(a, b, c)}{\sqrt{a^2 + b^2 + c^2}}, \qquad l^2 + m^2 + n^2 = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

$(a, b, c)$ direction ratios — any unscaled triple along the line
$(l, m, n)$ direction cosines — the normalised (unit) triple
$\alpha, \beta, \gamma$ angles the line makes with the X, Y, Z axes

Symmetric Cartesian form and vector form of a line

Symmetric and vector form

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \quad\Longleftrightarrow\quad \vec{r} = \vec{a} + \lambda\vec{b}

$(x_1, y_1, z_1)$ a fixed point on the line (the numerators)
$(a, b, c)$ direction ratios of the line (the denominators)
$\lambda$ scalar parameter sweeping along the line

Line through two points

Direction from two points

\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})

$\vec{a}, \vec{b}$ position vectors of the two points
$\vec{b} - \vec{a}$ the line's direction (displacement $A \to B$ )

Normalising a non-standard Cartesian equation

Reduce to symmetric form

px - p_0 = qy - q_0 = rz - r_0 \;\Rightarrow\; \frac{x - p_0/p}{1/p} = \frac{y - q_0/q}{1/q} = \frac{z - r_0/r}{1/r}

point $\left(\tfrac{p_0}{p}, \tfrac{q_0}{q}, \tfrac{r_0}{r}\right)$
direction $\left(\tfrac{1}{p}, \tfrac{1}{q}, \tfrac{1}{r}\right)$ , scaled to integers

Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes

Cross product (determinant expansion)

\vec{u} \times \vec{v} = (u_2 v_3 - u_3 v_2)\,\hat{i} - (u_1 v_3 - u_3 v_1)\,\hat{j} + (u_1 v_2 - u_2 v_1)\,\hat{k}

$\vec{u}, \vec{v}$ the two directions (line directions or plane normals)
$\hat{j}$ termcarries a MINUS sign — the cofactor expansion's alternating sign

Unit vector perpendicular to two lines

Unit normal to two lines

\hat{n} = \frac{\vec{d}_1 \times \vec{d}_2}{|\vec{d}_1 \times \vec{d}_2|}

$\vec{d}_1 \times \vec{d}_2$ vector perpendicular to both lines
$|\vec{d}_1 \times \vec{d}_2|$ its magnitude — divide to normalise

Angle a line makes with an axis; equal inclination

Angle with an axis

\cos\alpha = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \qquad \text{equally inclined} \Rightarrow |a| = |b| = |c|,\ \cos = \tfrac{1}{\sqrt 3}

$\alpha$ angle between the line and the X-axis
$a$ the X-direction ratio of the line

Direction cosines from a linear + quadratic constraint pair

Method (linear eliminate, quadratic factor, normalise)

l = 5m - 3n \;\xrightarrow{\text{sub}}\; \text{quadratic in } m, n \;\Rightarrow\; l : m : n \;\xrightarrow{\div\sqrt{l^2+m^2+n^2}}\; (l, m, n)

lineareliminate one of $l, m, n$
quadraticfactor for the surviving ratio (two roots → two lines)

Common traps

Direction ratios are NOT direction cosines until you normalise

(2, -1, 2)

are direction ratios; the direction cosines are

(2, -1, 2)/3

. The identity

l^2 + m^2 + n^2 = 1

holds only for the normalised triple — never for raw ratios.

The $\pm$ sign decides acute vs obtuse

Solving

\cos^2\gamma = \tfrac14

gives

\cos\gamma = \pm\tfrac12

, i.e.

\gamma = 60^\circ

120^\circ

. Read the question: if it asks for the OBTUSE angle, take the negative root and report

120^\circ

, not

60^\circ

Numerators give the point, denominators give the direction — don't swap them

\dfrac{x-3}{2} = \cdots

the point coordinate is

+3

(sign flipped from

x - 3

) and the direction is

2

. A frequent slip is reading the denominator as part of the point or carrying the wrong sign on the constant.

A fixed coordinate means a zero direction component

For

y = 2

, the line never moves in

y

, so the direction's

\hat{j}

-component is

0

— the direction is

(3, 0, 4)

, not

(3, 2, 4)

. The constant

2

belongs to the POINT, not the direction.

Head minus tail — keep the order consistent

Direction is

\vec{b} - \vec{a}

; reversing gives the opposite direction. For a line that is fine (a line has no preferred sense), but the option must MATCH a scalar multiple of your direction — a sign flip on only some components is a different vector.

"Parallel to the line joining P, Q" ≠ "through P or Q"

Borrow only the DIRECTION

Q - P

; the base point is the separately-given point. Plugging

P

Q

in as the base point gives the wrong line.

You must factor BEFORE reading the point

From

2x - 2 = \cdots

, the point coordinate is

x = 1

(set

2x - 2 = 0

), NOT

x = 2

and NOT

x = -2

. Factor out the coefficient first; the constant alone is misleading.

Direction ratios are the RECIPROCALS of the coefficients

For

2x = 3y = 6z

the direction is

\left(\tfrac12, \tfrac13, \tfrac16\right) \propto (3, 2, 1)

— the reciprocal pattern. The classic distractor leaves the direction as the coefficients

(2, 3, 6)

, which points the wrong way.

Parallel to two PLANES → cross the NORMALS, not the planes

A line parallel to two planes is perpendicular to both normals, so its direction is

\vec{n}_1 \times \vec{n}_2

. Likewise the line of intersection of two planes uses

\vec{n}_1 \times \vec{n}_2

. Don't confuse a plane's normal

(a,b,c)

with the plane itself.

The $\hat{j}$ component flips sign

The middle term of

\vec{u} \times \vec{v}

-(u_1 v_3 - u_3 v_1)

. Dropping that minus sign is the single most common cross-product error — and the wrong-sign option is always sitting right there as a distractor.

Direction ratios are only defined up to a scalar

(−4, −7, −13)

and

(4, 7, 13)

point along the same line, and

(2, -7, 4)

equals

(4, -14, 8)/2

. When matching options, scale your answer before declaring it absent — a common multiple may be the listed choice.

Factor the cross product before normalising

(7, -21, 7) = 7(1, -3, 1)

. Normalising the unfactored vector still works (

\sqrt{49 + 441 + 49} = 7\sqrt{11}

), but factoring first turns the magnitude into the clean

\sqrt{11}

the options use and avoids arithmetic slips.

Both signs are valid — read the option set

+\hat{n}

and

-\hat{n}

are both perpendicular to the two lines. The bank's key picks ONE; match the sign pattern of the cross product exactly (especially that flipped

\hat{j}

sign) rather than assuming all-positive.

Divide by the magnitude — $\cos\alpha = a/|\vec{d}|$ , not $a$ alone

For direction

(3, -3, 3)

\cos\alpha = 3/(3\sqrt 3) = 1/\sqrt 3

, NOT

3

. The cosine is the NORMALISED first component (the direction cosine), so always divide by

|\vec{d}|

"Equally inclined" means equal magnitudes, watch the signs

Equal inclination forces

|a| = |b| = |c|

, but signs can differ if the question specifies acute/obtuse angles with particular axes. For the basic case take all components equal and positive.

Solve for the RATIO first, then normalise — don't skip normalisation

The quadratic gives

l : m : n

(e.g.

1 : 2 : 3

), which are direction RATIOS. Direction cosines require dividing by

\sqrt{l^2 + m^2 + n^2} = \sqrt{14}

. Reporting

l + m + n = 6

instead of

6/\sqrt{14}

is the classic miss.

Two roots → two answers; report both if asked

The quadratic factors into two cases, giving two distinct lines (e.g.

\tfrac{6}{\sqrt{14}}

AND

\tfrac{2}{\sqrt 6}

). The correct option usually lists BOTH values — a single value is an incomplete distractor.

Plane — Equation, Normal, and Construction

Learn this subtopic in the notes

Equation of a plane and its normal

The three equivalent forms

ax + by + cz = d \;\Longleftrightarrow\; \vec{r}\cdot\vec{n} = d \;\Longleftrightarrow\; \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{n} = (a,b,c)$ normal — the coefficients of $x, y, z$
$\vec{a}$ position vector of a known point on the plane
$d$ constant, fixed by substituting the known point

Direction cosines of the normal

Direction-cosine identity

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

$\alpha, \beta, \gamma$ angles the normal makes with $X, Y, Z$ axes

Planes parallel to a coordinate plane or to a given plane

Same normal, new constant

ax + by + cz = d' \quad\text{where } d' = a x_0 + b y_0 + c z_0

$(a,b,c)$ normal copied from the given plane
$(x_0,y_0,z_0)$ point the new plane passes through

Plane from the foot of the perpendicular from the origin

Plane from foot of perpendicular

\vec{r}\cdot\overrightarrow{OM} = |\overrightarrow{OM}|^2 = x_0^2 + y_0^2 + z_0^2

$M(x_0,y_0,z_0)$ foot of perpendicular from the origin
$\overrightarrow{OM}$ the normal vector to the plane

Plane through a point with normal fixed by axis angles

Point-normal with angle-derived normal

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

$(a,b,c)$ normal recovered from the axis angles
$(x_0,y_0,z_0)$ the given point on the plane

Plane perpendicular to two given planes

Normal from two perpendicular planes

\vec{n} = \vec{n_1}\times\vec{n_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{n_1}, \vec{n_2}$ normals of the two given planes
$\vec{n}$ required normal = their cross product

Plane through a point parallel to two lines

Normal from two parallel lines

\vec{n} = \vec{d_1}\times\vec{d_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{d_1}, \vec{d_2}$ direction vectors of the two lines
$\vec{a}$ the point the plane passes through

Plane through three points

Three-point plane

\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\overrightarrow{AB}, \overrightarrow{AC}$ two edges from anchor $A$
$\vec{n}$ normal = their cross product

Perpendicular bisector plane of a segment

Perpendicular bisector plane

\vec{n} = \overrightarrow{PQ}, \quad M = \tfrac{1}{2}(\vec{p} + \vec{q}), \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0

$M$ midpoint of $PQ$ — the plane passes through it
$\overrightarrow{PQ}$ segment direction = the normal

Family of planes through a line of intersection (lambda engine)

Family of planes

P_1 + \lambda P_2 = 0, \qquad \vec{n}_\lambda = \big(a_1 + \lambda a_2,\; b_1 + \lambda b_2,\; c_1 + \lambda c_2\big)

$\lambda$ scalar fixed by the one extra condition
$\vec{n}_\lambda$ the family's normal, a function of $\lambda$

Intercept form, intercept triangle area and centroid

Intercept triangle: centroid and area

G = \left(\tfrac{a}{3}, \tfrac{b}{3}, \tfrac{c}{3}\right), \qquad \text{Area} = \tfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}

$a, b, c$ intercepts on the $X, Y, Z$ axes
$G$ centroid of the triangle of intercepts

Recovering a plane from a point and its mirror image

Plane from point and its image

M = \tfrac12(P + P'), \quad \vec{n} = P' - P, \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0

$P, P'$ the point and its mirror image
$M$ midpoint — lies on the plane

Common traps

The normal is the coefficient triple, not the point

ax+by+cz=d

the normal is

(a,b,c)

. Students sometimes grab the point's coordinates as the normal — those only fix

d

. Read direction from the coefficients, position from the given point.

$d$ is found by substituting, never left at the wrong sign

After

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

, expand fully before reading

d

. A sign slip on

-bx_0

etc. flips the constant — the most common reason a correct normal still lands on the wrong option.

"Acute angle" chooses the positive square root

\cos^2\gamma = \tfrac14

gives

\cos\gamma = \pm\tfrac12

. The word acute forces the

+

sign, so the Z-component of the normal is positive. Miss this and you may build the plane from the mirror-flipped normal.

Equally inclined means equal COSINES, not equal angles spread over 90 degrees

Equal inclination gives

(1,1,1)

, not

(1,1,0)

. All three cosines equal forces

\tfrac{1}{\sqrt3}

each — don't assume one axis drops out.

Parallel to XY-plane is $z = k$ , not $x + y = k$

The XY-plane is

z = 0

; any plane parallel to it freezes

z

. Match the right coordinate: parallel to YZ freezes

x

, parallel to ZX freezes

y

Re-use the WHOLE normal when copying a plane

A parallel plane shares all three coefficients, signs included. Substitute the point only into the constant — do not rescale or re-sign the normal, or it stops being parallel.

The constant is $|\overrightarrow{OM}|^2$ , not $|\overrightarrow{OM}|$

For foot

(2,1,-2)

the constant is

9

(the squared length), not

3

(the distance). The distance

\sqrt9 = 3

is the perpendicular length; the plane equation uses the squared value.

Don't move the foot to the wrong side of the equation

The form is

\vec{r}\cdot\overrightarrow{OM} = +|\overrightarrow{OM}|^2

, a POSITIVE constant. Distractors flip it to

\dots + 45 = 0

; that plane no longer passes through

M

Clear the irrational direction cosine into a clean ratio

Direction cosines

\left(\tfrac{1}{\sqrt2}, \tfrac12, \tfrac12\right)

become the ratio

(2,1,1)

— double everything until the

\sqrt2

is gone. Writing the normal as

(1,1,1)

here is wrong; only EQUAL angles give

(1,1,1)

Put the point into the expanded form, not the angle data

The angles fix only the normal's DIRECTION; the point alone fixes the constant. Don't try to use an angle to find

k

Cross product, not dot product, for the normal

Perpendicular-to-two-planes needs a direction perpendicular to BOTH normals — that is

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

. A dot product gives a number, not a direction; reaching for it here is the classic wrong start.

Keep the cross-product sign and middle-term flip straight

The

\hat{j}

component carries a minus sign in the determinant expansion. A sign error there sends you to a sibling option with the middle coefficient flipped (e.g.

6x+7y\dots

instead of

6x-7y\dots

Parallel to two LINES uses their directions, parallel to two PLANES uses their normals

Both reduce to a cross product, but read the right vectors: a line gives a direction

(p,q,s)

from its denominators; a plane gives a normal

(a,b,c)

from its coefficients. Mixing them up cross-products the wrong pair.

Read line directions from the denominators, signs included

\frac{z}{-4}

the Z-direction is

-4

, not

4

. A dropped minus on a single component changes the whole cross product.

Anchor BOTH edge vectors at the same point

Use

\overrightarrow{AB}

and

\overrightarrow{AC}

(both from

A

), not

\overrightarrow{AB}

and

\overrightarrow{BC}

mixed with the wrong anchor for the point-normal step. The normal is fine either way, but the substituted point must lie on the plane.

"Parallel to an axis" kills exactly one coefficient

Parallel to X-axis sets the

x

-coefficient to

0

(the axis direction must lie in the plane, so the normal has no

\hat{i}

part). Don't also zero

y

z

Pass through the MIDPOINT, not through $P$ or $Q$

The perpendicular bisector plane goes through

M

, the midpoint. Substituting

P

Q

gives a parallel plane with the wrong constant.

Simplify the normal before substituting

\overrightarrow{PQ} = (2,2,-2)

is fine, but

(1,1,-1)

is cleaner — just keep the constant consistent. Either way, the direction (signs) must match

\overrightarrow{PQ}

Perpendicular to the XY-plane means the Z-coefficient vanishes

A plane perpendicular to the XY-plane is 'vertical' — its normal lies IN the XY-plane, so it has no

z

-part: set

c_1 + \lambda c_2 = 0

. Students often confuse this with parallel (which would set

a, b

to make the normal point along Z). Vertical → kill

z

; horizontal → keep only

z

Parallel-to-axis kills the SAME-named coefficient

Parallel to the Y-axis means

\hat{j}

lies in the plane, so the normal has no

\hat{j}

b_1 + \lambda b_2 = 0

. Don't confuse 'parallel to Y-axis' (kill

b

) with 'perpendicular to ZX-plane' even though they coincide.

Clear the fractions before matching options

Solving for

\lambda

leaves fractional coefficients like

\tfrac25 x + \tfrac15 y

. Multiply through (here by 5) to reach

2x + y - 3 = 0

; the un-cleared version matches no option.

Read intercepts from the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

For

2x + 3y + 4z = 1

, the X-intercept is

\tfrac12

, NOT

2

. Divide through to make the RHS exactly

1

, then the denominators are the intercepts.

Use the squared intercepts in the area formula

Area

= \tfrac12\sqrt{a^2b^2 + b^2c^2 + c^2a^2}

— pairwise PRODUCTS of squares. A negative intercept like

c = -4

contributes

c^2 = 16

; the sign drops out, but don't forget to square it.

The normal is the segment, the plane is at the midpoint

Use

\overrightarrow{PP'}

as the normal and the MIDPOINT as the through-point — not

P

P'

. Substituting an endpoint gives a plane parallel to the true one.

Simplify the messy normal before testing option-points

\left(-\tfrac{10}{3}, -\tfrac{10}{3}, -\tfrac{10}{3}\right)

is just

(1,1,1)

. Reduce first; then substituting each option-point to find which lies on

x + y + z = 1

is trivial arithmetic.

Angles — Line, Plane, and Direction Conditions

Learn this subtopic in the notes

Direction ratios, direction cosines, and the dot/cross toolkit

The toolkit

l^2 + m^2 + n^2 = 1, \qquad \vec{u}\cdot\vec{v} = u_1v_1 + u_2v_2 + u_3v_3, \qquad |\vec{d}| = \sqrt{a^2 + b^2 + c^2}

$(a,b,c)$ direction ratios of a line
$(l,m,n)$ direction cosines (normalised d.r.s)
$(A,B,C)$ normal of the plane $Ax+By+Cz=d$

Angle between two lines

\cos\theta = \frac{|\vec{d_1}\cdot\vec{d_2}|}{|\vec{d_1}|\,|\vec{d_2}|}

$\vec{d_1}, \vec{d_2}$ direction vectors of the two lines
$|\cdots|$ (numerator)modulus — forces the acute angle

Angle between two planes (and solving for an unknown coefficient)

Angle between two planes

\cos\theta = \frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\,|\vec{n_2}|}, \qquad |\alpha_1 - \alpha_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

$\vec{n_1}, \vec{n_2}$ normals of the two planes
$|\alpha_1 - \alpha_2|$ difference of the two roots of the resulting quadratic

Angle between a line and a plane (the solve-for-lambda variant)

Angle between a line and a plane

\sin\theta = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}|\,|\vec{n}|}

$\vec{d}$ direction vector of the line
$\vec{n}$ normal of the plane
$\sin\theta$ sine (NOT cosine) — angle is with the plane, not the normal

Parallel and perpendicular conditions (lines, and line-parallel-to-plane)

Perpendicular / parallel conditions

\vec{d_1}\cdot\vec{d_2} = 0 \;\text{(}\perp\text{)}, \qquad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \;\text{(}\parallel\text{)}, \qquad \vec{d}\cdot\vec{n} = 0 \;\text{(line}\parallel\text{plane)}

$\vec{d_1}\cdot\vec{d_2} = 0$ two lines are perpendicular
$\vec{d}\cdot\vec{n} = 0$ line (or $\vec{PQ}$ ) parallel to the plane

Line lies in a plane (two conditions, solve the unknowns)

Line-lies-in-plane conditions

Ax_1 + By_1 + Cz_1 = d \quad \text{and} \quad aA + bB + cC = 0

$(x_1, y_1, z_1)$ a point on the line — must lie on the plane
$(a, b, c)$ line direction — must be $\perp$ the normal $(A,B,C)$

Direction-cosine systems and equal-angle lines

Direction-cosine identity

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1, \qquad \cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|

$\alpha, \beta, \gamma$ angles the line makes with the $x, y, z$ axes
$(l_i, m_i, n_i)$ direction cosines (unit) of the two lines

Line of intersection of two planes, and its angle with an axis

Line of intersection

\vec{d} = \vec{n_1}\times\vec{n_2}, \qquad \cos\alpha = \frac{|\vec{d}\cdot\hat{x}|}{|\vec{d}|}

$\vec{n_1}\times\vec{n_2}$ direction of the line where the planes meet
$\hat{x}$ unit vector along the axis, e.g. $(1,0,0)$

Common traps

Direction RATIOS are not direction COSINES

(2, -1, 2)

are direction ratios; you must divide by

|\vec{d}| = 3

to get the cosines. An angle formula written with cos uses the cosines (or, equivalently, ratios divided by the magnitudes inside the formula) — never raw ratios where unit vectors are required.

A zero denominator is a valid direction ratio

\dfrac{z+1}{0}

the

0

means the line is perpendicular to the

z

-axis, with d.r.s

(\cdot, \cdot, 0)

. Don't discard it — it contributes

0

to dot products and

0^2

to magnitudes, not an error.

Drop the modulus and you may report the obtuse angle

A negative dot product (like

-7

) gives a negative cosine and the obtuse angle. MHT-CET asks for the ACUTE angle — take

|\vec{d_1}\cdot\vec{d_2}|

\cos\theta

is positive.

Lines need DIRECTIONS, not points

When a line is given as a join of two points, first subtract to get

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

. Dotting the position vectors instead of the direction is a classic error.

Plane angle uses normals, not the plane's 'direction'

A plane has no single direction — it is fixed by its normal. Equating the angle to

\cos^{-1}

of a dot of in-plane vectors is wrong; always dot

\vec{n_1}\cdot\vec{n_2}

The question may want the DIFFERENCE of roots, not a root

Squaring the angle equation gives a quadratic with two valid

\alpha

. Use

\dfrac{\sqrt{b^2-4ac}}{|a|}

for the gap — solving each root separately wastes time and invites sign slips.

Use SINE for line–plane, COSINE for line–line and plane–plane

The single most common error here: writing

\cos\theta = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|}

. The line–plane angle is the complement of the line–normal angle, so the dot-product ratio equals

\sin\theta

, not

\cos\theta

Convert a $\cos^{-1}$ given-angle to $\sin\theta$ first

If the question states the angle as

\cos^{-1}(k)

, you need

\sin\theta = \sqrt{1-k^2}

before substituting, because the formula carries

\sin\theta

. Plugging

k

straight in gives the wrong

\lambda

Line PARALLEL to a plane means direction ⟂ NORMAL

Parallel-to-the-plane is a perpendicularity in disguise: the line's direction sits inside the plane, hence

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

. Students wrongly set

\vec{d}\parallel\vec{n}

(that would make the line perpendicular to the plane).

Normalise messy ratios before dotting

A term like

\dfrac{7y-14}{2p}

\dfrac{y-2}{2p/7}

: the

y

-direction ratio is

\tfrac{2p}{7}

, not

2p

. And

\dfrac{6-z}{5} = \dfrac{z-6}{-5}

flips the sign. Convert each fraction to

\dfrac{x-x_1}{a}

form first.

BOTH conditions are required — one is not enough

A point on the plane only puts ONE point of the line there; without

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

the line would pierce the plane. Conversely

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

alone only makes the line parallel to the plane (possibly floating above it). You need both.

Read the point off the numerators correctly

For

\dfrac{z-m}{2}

the point's

z

-coordinate is

+m

, but for

\dfrac{z+m}{2}

it is

-m

— the sign flips the answer (

m = 7

m = -7

). Similarly

\dfrac{2z-m}{3}

means

z = \tfrac{m}{2}

, not

z = m

Use the identity $\sum\cos^2 = 1$ , not $\sum\cos = 1$

Direction cosines square-sum to 1, they do not add to 1. For an equal-angle line, set

\cos^2\alpha + 2\cos^2\beta = 1

— squaring the cosines is essential.

A two-constraint system gives TWO directions — find the angle BETWEEN them

Solving

l + m + n = 0

with a quadratic yields two factor cases, i.e. two distinct lines. The question wants the angle between THOSE two, computed with the line–line formula — not a single direction.

The intersection direction is the CROSS product of the normals

The line lies in both planes, so it is perpendicular to both normals — that is exactly

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

. Using

\vec{n_1} + \vec{n_2}

or a dot product instead gives a meaningless direction.

Plane 'parallel to two vectors' ⟹ normal is THEIR cross product

When a plane is described by two vectors it contains, the normal is their cross product first; only then cross the two normals to get the common line. Skipping the inner cross product is the usual slip in the multi-plane variant.

Distances in 3-D

Learn this subtopic in the notes

Distance of a point from the axes and the origin

Distance from origin and axes

OP = \sqrt{x^2+y^2+z^2}, \qquad \sum(\text{axis distance})^2 = 2(x^2+y^2+z^2)

$x, y, z$ coordinates of the point $P$
$OP$ distance of $P$ from the origin

Distance of a point from a plane

Point-to-plane distance

d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

$(a, b, c)$ the plane's normal $\vec{n}$
$(x_1, y_1, z_1)$ the point
$d$ constant term, with the plane in $=0$ form

Equidistant points and the gap between parallel planes

Distance between parallel planes

d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}

$d_1, d_2$ constants of the two planes (identical normals)
$(a,b,c)$ the shared normal

Distance of a point from a line

Point-to-line distance

d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} = \sqrt{|\overrightarrow{AP}|^2 - \left(\frac{\overrightarrow{AP}\cdot\vec{b}}{|\vec{b}|}\right)^2}

$A$ a known point on the line
$\vec{b}$ direction ratios of the line
$\overrightarrow{AP}$ $P - A$ , point minus line-point

Distance between two parallel lines

Distance between parallel lines

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

$\vec{a}_1, \vec{a}_2$ base points of the two lines
$\vec{b}$ the common direction

Shortest distance between skew lines (and solving backwards for a parameter)

Shortest distance between skew lines

d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}

$\vec{a}_1, \vec{a}_2$ base points of the two lines
$\vec{b}_1, \vec{b}_2$ the two directions
$\vec{b}_1 \times \vec{b}_2$ common-perpendicular direction

Build a plane from conditions, then take a distance

Plane normal from a cross product

\vec{n} = \vec{p} \times \vec{q}, \qquad d = \frac{|\vec{n}\cdot(\vec{r}_1 - \vec{r}_0)|}{|\vec{n}|}

$\vec{p}, \vec{q}$ the two normals (or directions) the plane must respect
$\vec{r}_0$ a known point on the plane
$\vec{r}_1$ the point whose distance you want

Where a line meets a plane, and distance measured along a line

Line in parametric form

(x, y, z) = (x_0 + at,\ y_0 + bt,\ z_0 + ct)

$(x_0, y_0, z_0)$ a point on the line
$(a, b, c)$ the line's direction ratios
$t$ parameter solved from the plane/coordinate condition

Common traps

Axis distance DROPS one coordinate, origin distance keeps all three

Distance from the X-axis is

\sqrt{y^2+z^2}

— NOT

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

. Confusing the two is the most common slip; the axis you measure to is the coordinate you discard.

Sum-of-squares from axes is TWICE the origin-squared, not equal

Each of

x^2, y^2, z^2

appears in exactly two of the three axis-distance formulas, so the total is

2(x^2+y^2+z^2)

. Forgetting the factor of 2 gives an origin distance that is

\sqrt{2}

too large.

Move every term to one side first — the constant $d$ must be in $=0$ form

For

2x + y - 2z = 18

, rewrite as

2x + y - 2z - 18 = 0

d = -18

. Plugging into the un-rearranged equation, or forgetting to carry the constant, is the classic numerator error.

Absolute value on top — distance is never negative

The signed plug-in can come out negative; the distance takes

|\,\cdot\,|

. The SIGN matters only when you compare which side of a plane a point lies on (used in the equidistant-planes problems).

Equidistant gives TWO cases — keep both signs

Dropping the modulus on an equidistance condition yields

(\text{plug}_P) = +(\text{plug}_Q)

AND

=-(\text{plug}_Q)

. A 'sum of all values of

\lambda

' question is testing exactly whether you found both roots.

Parallel-plane gap needs MATCHING normals

2x - y + 2z = 1

and

4x - 2y + 4z = 10

look different but are parallel; halve the second to

2x - y + 2z = 5

before doing

|1 - 5|/3

. Subtracting raw constants from unscaled equations gives a wrong gap.

Divide by $|\vec{b}|$ , not by $|\vec{b}|^2$

The cross-product route is

|\overrightarrow{AP}\times\vec{b}| / |\vec{b}|

— ONE power of the direction length on the bottom. A frequent distractor squares the denominator, halving the order of magnitude of the answer.

$\overrightarrow{AP} = P - A$ — point minus the line's point

Read

A

off the numerators of the symmetric line (

\frac{x-6}{3}\Rightarrow A_x = 6

) and

\vec{b}

off the denominators. Mixing them up — or computing

A - P

— flips a sign that survives into the cross product.

Use the JOIN vector $\vec{a}_2 - \vec{a}_1$ , not a single point

The numerator crosses

(\vec{a}_2 - \vec{a}_1)

with

\vec{b}

. Plugging just one base point's position vector (instead of the difference) treats the wrong displacement and gives a meaningless length.

Confirm parallel FIRST

If the directions are NOT proportional the lines are skew, and this formula is wrong — you need the skew shortest-distance formula instead. Check

\vec{b}_1 \parallel \vec{b}_2

before reaching for

(\vec{a}_2-\vec{a}_1)\times\vec{b}

Numerator is a scalar (dot of difference with the cross), denominator is the cross's MAGNITUDE

Don't confuse the two cross products:

\vec{b}_1\times\vec{b}_2

appears in BOTH places — once dotted into

(\vec{a}_2-\vec{a}_1)

on top, once as a magnitude on the bottom. The top is a number; the bottom is a length.

Backwards problems often hide TWO roots — pick by the stated constraint

|60 + 8\alpha| = 108

gives

\alpha = 6

\alpha = -21

; the condition

\alpha > 0

selects

6

. Always read the constraint (

\alpha > 0

, 'positive value', etc.) before committing to a root.

The normal is the CROSS product, then the plane passes through the GIVEN point

Building the normal is only half the job — you still need a point on the plane to fix the constant

d

. For 'containing two lines', a point on either line works; for the ⊥-to-two-planes case, use the explicitly given point.

Simplify the normal before plugging in

A normal like

(-3, -3, 0)

is parallel to

(1, 1, 0)

; using the smaller proportional vector keeps the arithmetic clean and the

\sqrt{a^2+b^2+c^2}

honest — just be consistent in both numerator and denominator.

'Measured along the line' ≠ perpendicular distance

When a question says distance MEASURED ALONG

x = y = z

, you travel down that line to the plane and measure that slanted segment — it is LONGER than the perpendicular drop. Using the point-to-plane formula here is the headline trap.

Equal angles with the axes fixes the direction to $(1,1,1)$

A line with positive direction cosines making equal angles has

l = m = n = \tfrac{1}{\sqrt{3}}

, i.e. direction ratios

(1,1,1)

. Don't leave it as an unknown — that single fact unlocks the whole 'meet the plane, then PQ length' question.

Foot of Perpendicular, Image, and Projection

Learn this subtopic in the notes

Foot of the perpendicular from a point to a line

Foot on a line

F = \vec{a} + \lambda\vec{d}, \qquad \overrightarrow{PF}\cdot\vec{d} = 0 \;\Rightarrow\; \lambda

$\vec{a}$ a fixed point on the line
$\vec{d}$ direction vector of the line
$P$ the external point
$\overrightarrow{PF}$ $F - P$ , must be perpendicular to $\vec{d}$

Foot of the perpendicular from a point to a plane

Foot on a plane

F = P + t\,\vec{n}, \qquad t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}

$\vec{n} = (a,b,c)$ normal to the plane $ax+by+cz+d=0$
$t$ how far along the normal to reach the plane
$F$ foot of the perpendicular on the plane

Mirror image of a point in a plane

Image in a plane

P' = 2F - P = P + 2t\,\vec{n}, \qquad t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}

$F$ foot of perpendicular (midpoint of $PP'$ )
$P'$ mirror image of $P$ in the plane
$2t\,\vec{n}$ twice the foot's displacement along the normal

Mirror image of a point in a line

Image in a line (forward and backward)

P' = 2F - P; \qquad \text{midpoint } \tfrac{P+P'}{2} \in \text{line}, \;\; \overrightarrow{PP'}\cdot\vec{d} = 0

$F$ foot of perpendicular on the line
$P'$ mirror image of $P$ in the line
$\overrightarrow{PP'}$ $P' - P$ , perpendicular to the line direction

Image of a line in a plane (and planes through an image)

Image line — reflect point, preserve direction

\vec{d}\cdot\vec{n} = 0 \;\Rightarrow\; \text{line} \parallel \text{plane}, \quad \text{image line} = \{P',\; \vec{d}\}

$\vec{d} = (p,q,r)$ direction of the original line, preserved if $\vec{d}\cdot\vec{n}=0$
$P'$ image of a point on the line (via $2F - P$ )
$\vec{n}$ normal of the mirror plane

Projection of a segment onto a line

Projection onto a line

\text{proj} = \dfrac{|\overrightarrow{AB}\cdot\vec{d}|}{|\vec{d}|}

$\overrightarrow{AB}$ $B - A$ , the segment vector
$\vec{d}$ direction ratios of the line
$|\vec{d}|$ $\sqrt{a^2+b^2+c^2}$

Projection of a segment onto a plane

Projection onto a plane

\text{proj}_{\text{plane}} = \sqrt{|\overrightarrow{AB}|^2 - \left(\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2}

$\overrightarrow{AB}$ the segment vector $B - A$
$\vec{n}$ normal to the plane
$\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}$ the along-normal component (removed)

Common traps

Perpendicularity is $\overrightarrow{PF}\cdot\vec{d}=0$ , NOT $\overrightarrow{PF} = \vec{d}$

You only need the dot product of the connecting segment with the line's direction to vanish — that gives one equation in one unknown. Do not try to force the whole vector

\overrightarrow{PF}

to equal or be parallel to anything.

Use the symmetric form's parameter consistently

For

\frac{x+3}{5} = \frac{y+1}{2} = \frac{z+4}{3} = \lambda

, the point is

(5\lambda-3,\,2\lambda-1,\,3\lambda-4)

— the fixed point comes from setting each numerator to 0, the multipliers

(5,2,3)

are the direction. Mixing up which is which scrambles the sign of every coordinate (option B/C/D in these PYQs are exactly the sign-flipped traps).

Don't forget to substitute back

Finding

\lambda

is the middle of the problem, not the end. The answer is the coordinate

\vec{a} + \lambda\vec{d}

; a half-finished solution that reports

\lambda

itself is a guaranteed wrong option.

Carry the constant $d$ with its correct sign

Write the plane as

ax+by+cz+d=0

first. For

2x-3y+z = 11

that is

d = -11

, so

t = -\frac{(\cdots) - 11}{14}

. Dropping the sign of

d

is the single most common foot-on-plane arithmetic slip.

Divide by $a^2+b^2+c^2$ , not by $\sqrt{a^2+b^2+c^2}$

The parameter

t

uses

|\vec{n}|^2

in the denominator (the distance formula's

\sqrt{}

appears only for actual distances, not for

t

). For

\vec{n} = (2,-3,1)

use

14

, never

\sqrt{14}

Image is $2F - P$ , the foot is only halfway

A very common error is to report the FOOT as the answer to an image question. The foot is the midpoint; you must step the same distance again:

P' = 2F - P

. The foot's coordinates are usually one of the distractor options.

$2F - P$ , not $F - 2P$ or $2P - F$

From midpoint

F = \tfrac{P + P'}{2}

you solve

P' = 2F - P

— twice the foot minus the original point. Swapping the roles of

F

and

P

gives a reflection on the wrong side.

Backward problems use TWO conditions, not one

When the image is given and a coordinate is unknown, one equation (midpoint on line) is not enough if there are two unknowns. Add

\overrightarrow{PP'}\perp\vec{d}

(or a second midpoint-ratio equation) to pin both

a

and

b

Reflect in the LINE, not the line's fixed point

The mirror is the entire line, so the foot

F

is the nearest point on the line to

P

— found via perpendicularity — not the arbitrary point

\vec{a}

printed in the equation. Reflecting through

\vec{a}

gives the wrong image.

Preserve the direction ratios — don't negate them

For a line parallel to the plane, the image line keeps the SAME

(p,q,r)

. Distractors flip the signs of the direction (and shift the image point) — only the point moves under reflection, the direction stays put. Verify

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

before relying on this.

Reflect a point ON the line, then reattach the direction

Don't try to reflect the direction vector through the plane separately. Reflect a concrete point (the numerators give you one), get

P' = 2F - P

, and the image line is

P'

with the original direction.

Divide by $|\vec{d}|$ , not $|\vec{d}|^2$

The projection LENGTH uses

|\vec{d}|

(one power) in the denominator. Dividing by

|\vec{d}|^2

gives the scalar multiplier for the projection VECTOR, not its length — a frequent mix-up with vector projection.

$\overrightarrow{AB} = B - A$ — direction matters inside the dot product

Get the head-minus-tail right. The magnitude is unaffected by swapping

A

and

B

(the absolute value cleans up the sign), but a component mistake in

B - A

changes the dot product itself.

Plane projection SUBTRACTS the normal part; line projection KEEPS the direction part

Onto a line you want the component ALONG

\vec{d}

. Onto a plane you want what's LEFT after removing the component along

\vec{n}

. Using the line formula for a plane question (or vice-versa) is the headline trap — they are complementary pieces of the same vector.

Use the UNIT normal inside the square

The removed piece is

(\overrightarrow{AB}\cdot\hat{n})^2 = \left(\frac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2

. Forgetting to divide by

|\vec{n}|

over-counts the normal component. For

\vec{n}=(1,1,1)

, divide the dot product by

\sqrt{3}

before squaring.

Answer is $\sqrt{2/3}$ , not $2/3$

Take the final square root.

\sqrt{\frac{2}{3}}

and

\frac{2}{3}

are both offered as options in this PYQ — the un-rooted value is the classic distractor.

Intersection, Coplanarity, and Skew Lines

Learn this subtopic in the notes

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}

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

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

$(x_1{+}at,\,\ldots)$ 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

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

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

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

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

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

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}|}

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

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}

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

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}

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

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}

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

Common traps

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.

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.

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.

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.

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.

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.

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).

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.

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.

Tetrahedron Geometry — Centroid, Volume, and Vertices

Learn this subtopic in the notes

Centroid of a tetrahedron and a triangle

Centroid (tetrahedron and triangle)

G_{\text{tet}} = \dfrac{A + B + C + D}{4} \qquad G_{\triangle} = \dfrac{A + B + C}{3}

$A, B, C, D$ position vectors (or coordinate triples) of the vertices
$G$ centroid — the component-wise average of the vertices

Volume of a tetrahedron via the scalar triple product

Tetrahedron volume

V = \dfrac{1}{6}\left|\det\begin{bmatrix} \overrightarrow{AB} \\ \overrightarrow{AC} \\ \overrightarrow{AD} \end{bmatrix}\right| = \dfrac{1}{6}\big|\,\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})\,\big|

$\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$ the three edge vectors from a common vertex $A$
$[\,\cdot\ \cdot\ \cdot\,]$ scalar triple product = determinant of the edge rows
$\tfrac{1}{6}$ a tetrahedron is one-sixth of the spanning parallelepiped

Volume of OABC from a plane cutting the axes

Plane normal, intercepts, and OABC volume

\vec{n} = \vec{d_1}\times\vec{d_2} \qquad V_{OABC} = \dfrac{1}{6}\,|a\,b\,c| = \dfrac{k^3}{6}\ \ (\text{for } x+y+z=k)

$\vec{d_1}, \vec{d_2}$ direction ratios of the two parallel lines
$a, b, c$ the $x$ -, $y$ -, $z$ -intercepts of the plane
$k$ the constant in $x+y+z=k$ ; each intercept when $\vec{n}\propto(1,1,1)$

Common traps

Divide by 4 for a tetrahedron, by 3 for a triangle

The single most common slip: a four-vertex solid averages over 4, a three-vertex triangle over 3. Read the figure named in the stem — "tetrahedron" means

/4

, "triangle" means

/3

. Using the wrong divisor lands you on a tempting distractor every time.

Inverse problems: rearrange, don't re-guess

When the centroid is given and a vertex is unknown, multiply through by the divisor first:

x_A + x_B + x_C = 3x_G

, then subtract the knowns. Skipping the multiply-through step is where sign errors creep in.

Watch which coordinate the puzzle reuses

Some stems reuse a letter across the centroid AND a vertex (e.g. centroid

(r, q, 1)

with a vertex coordinate

q

). Match coordinates by POSITION, solve the chain in order, and don't conflate a centroid component with a same-named vertex component.

It's $\tfrac{1}{6}$ for a tetrahedron, not $\tfrac{1}{3}$ or 1

The parallelepiped volume is

|\det|

; the tetrahedron is one-SIXTH of it, not one-third. (

\tfrac{1}{3}

belongs to the pyramid-volume formula

\tfrac{1}{3}\times\text{base}\times\text{height}

, a different setup.) Forgetting the

\tfrac{1}{6}

gives an answer six times too big.

Build edges from ONE common vertex

Use

\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}

— all three leaving the same vertex

A

(head minus the SAME tail). Mixing tails (e.g.

\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{AD}

) breaks the triple-product meaning and the volume is wrong.

Set $|\det| = 6V$ , not $|\det| = V$

In a solve-for-

x

problem, multiply the target volume by 6 BEFORE equating to the determinant:

\dfrac{1}{6}|\det| = \tfrac{11}{6}

means

|\det| = 11

, not

\tfrac{11}{6}

. Dropping the

\times 6

corrupts the linear equation for

x

Normal = cross product, then the constant comes from the POINT

Two steps, in order: first

\vec{n} = \vec{d_1}\times\vec{d_2}

gives the coefficients, THEN plug the given point into

\vec{n}\cdot\vec{r}=k

to fix

k

. Skipping the point and assuming

k

is wrong; the point is what pins the plane down.

Volume of OABC is $\tfrac{1}{6}\,abc$ , not $\tfrac{1}{6}k$ or $abc$

It's one-sixth the PRODUCT of the three intercepts. For

x+y+z=k

that's

\tfrac{1}{6}k^3

(since all three intercepts equal

k

) — e.g.

k=3

gives

\tfrac{27}{6}=\tfrac{9}{2}

, not

\tfrac{3}{6}

. Don't forget to cube

k

before dividing by 6.

Simplify the cross-product normal before reading intercepts

A normal like

(4,4,4)

is parallel to

(1,1,1)

— reduce it so the plane reads cleanly as

x+y+z=k

. Carrying the un-reduced

(4,4,4)

into the intercept step muddies the constant and the intercepts.

Line and Plane formulas

Line — Equation, Direction Cosines, and Vector Form

Direction ratios, direction cosines, and the l² + m² + n² = 1 identity

Symmetric Cartesian form and vector form of a line

Line through two points

Normalising a non-standard Cartesian equation

Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes

Unit vector perpendicular to two lines

Angle a line makes with an axis; equal inclination

Direction cosines from a linear + quadratic constraint pair

Common traps

Plane — Equation, Normal, and Construction

Equation of a plane and its normal

Direction cosines of the normal

Planes parallel to a coordinate plane or to a given plane

Plane from the foot of the perpendicular from the origin

Plane through a point with normal fixed by axis angles

Plane perpendicular to two given planes

Plane through a point parallel to two lines

Plane through three points

Perpendicular bisector plane of a segment

Family of planes through a line of intersection (lambda engine)

Intercept form, intercept triangle area and centroid

Recovering a plane from a point and its mirror image

Common traps

Angles — Line, Plane, and Direction Conditions

Direction ratios, direction cosines, and the dot/cross toolkit

Angle between two planes (and solving for an unknown coefficient)

Angle between a line and a plane (the solve-for-lambda variant)

Parallel and perpendicular conditions (lines, and line-parallel-to-plane)

Line lies in a plane (two conditions, solve the unknowns)

Direction-cosine systems and equal-angle lines

Line of intersection of two planes, and its angle with an axis

Common traps

Distances in 3-D

Distance of a point from the axes and the origin

Distance of a point from a plane

Equidistant points and the gap between parallel planes

Distance of a point from a line

Distance between two parallel lines

Shortest distance between skew lines (and solving backwards for a parameter)

Build a plane from conditions, then take a distance

Where a line meets a plane, and distance measured along a line

Common traps

Foot of Perpendicular, Image, and Projection

Foot of the perpendicular from a point to a line

Foot of the perpendicular from a point to a plane

Mirror image of a point in a plane

Mirror image of a point in a line

Image of a line in a plane (and planes through an image)

Projection of a segment onto a line

Projection of a segment onto a plane

Common traps

Intersection, Coplanarity, and Skew Lines

A general point on a line

Point where a line meets a plane

Point of intersection of two lines

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

Shortest distance between skew lines

Direction of the line of intersection of two planes

Transversal intersecting two given lines

Condition for a line to lie in a plane

Common traps

Tetrahedron Geometry — Centroid, Volume, and Vertices

Centroid of a tetrahedron and a triangle

Volume of a tetrahedron via the scalar triple product

Volume of OABC from a plane cutting the axes

Common traps

More MHT-CET Maths formula sheets