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JEE Mains Maths · Formula sheet

Three Dimensional Geometry formulas

24 formulas and 24 common traps for JEE Mains Maths Three Dimensional Geometry, grouped by subtopic.

Full notes with worked examples

Direction Cosines and Equations of Lines

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Direction cosines and direction ratios

Direction cosines from direction ratios

(l,m,n)=(a,b,c)a2+b2+c2,l2+m2+n2=1(l,m,n)=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}},\qquad l^2+m^2+n^2=1

Two lines from a pair of equations in l, m, n

Angle between the two directions

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12 a22+b22+c22\cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}

The angle between two lines, perpendicular and parallel lines

Angle between two lines

cos⁡θ=∣d⃗1⋅d⃗2∣∣d⃗1∣ ∣d⃗2∣\cos\theta=\frac{|\vec d_1\cdot\vec d_2|}{|\vec d_1|\,|\vec d_2|}

Writing a line: through a point, two points, or perpendicular to two lines

Parametric point and a common perpendicular direction

(x1+aλ, y1+bλ, z1+cλ),d⃗=d⃗1×d⃗2(x_1+a\lambda,\ y_1+b\lambda,\ z_1+c\lambda),\qquad \vec d=\vec d_1\times\vec d_2

Triangles and tetrahedra with coordinates

Area and volume

[ABC]=12∣AB⃗×AC⃗∣,V=16∣[AB⃗ AC⃗ AD⃗]∣[ABC]=\tfrac12|\vec{AB}\times\vec{AC}|,\qquad V=\tfrac16\left|[\vec{AB}\ \vec{AC}\ \vec{AD}]\right|

Common traps

Ratios are not cosines until you divide

(2,3,6)(2,3,6) gives 4+9+36=494+9+36=49, not 11. Divide by 77 first; only then do the squares add to 11.

Remove a variable with the LINEAR equation first

Only after substituting from the linear equation is the quadratic homogeneous in two variables. Dividing the original quadratic by l2l^2 straight away leaves three unknowns.

Coefficient of xx, yy, zz must be +1+1

In 2−x3\frac{2-x}{3} the direction ratio for xx is −3-3, and in 3y−2k\frac{3y-2}{k} it is k3\frac{k}{3} for yy. Reading the denominators as they stand gives a wrong angle.

Step with the UNIT vector

A+kd⃗A+k\vec d moves k∣d⃗∣k|\vec d|, not kk. For a stated distance, divide d⃗\vec d by its length first.

Half for a triangle, a sixth for a tetrahedron

The cross product's length is a parallelogram's area and the triple product is a parallelepiped's volume. Forgetting the 12\frac12 or the 16\frac16 doubles or sextuples the answer.

Foot, Image and Distance from a Line

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The foot of the perpendicular on a line

Parameter of the foot

t=(P−A)⋅d⃗∣d⃗∣2,M=A+td⃗t=\frac{(P-A)\cdot\vec d}{|\vec d|^2},\qquad M=A+t\vec d

The image of a point in a line

Image from the foot

Q=2M−PQ=2M-P

Distance of a point from a line

Distance from a line

d=∣AP⃗×d⃗∣∣d⃗∣d=\frac{|\vec{AP}\times\vec d|}{|\vec d|}

Triangles with a side on the line

Half-chord and area

MQ=k2−PM2,[PQR]=12 QR⋅PMMQ=\sqrt{k^2-PM^2},\qquad [PQR]=\tfrac12\,QR\cdot PM

Common traps

Divide by ∣d⃗∣2|\vec d|^2, not ∣d⃗∣|\vec d|

tt multiplies d⃗\vec d itself, so the projection must be divided by the SQUARED length. Using ∣d⃗∣|\vec d| gives a point that is off the perpendicular.

2M−P2M-P, not M+PM+P

The foot is the MIDPOINT of PP and its image, so Q=2M−PQ=2M-P. Adding MM and PP gives a point nowhere near the line.

Divide by ∣d⃗∣|\vec d|

∣AP⃗×d⃗∣|\vec{AP}\times\vec d| is a parallelogram's AREA. Only after dividing by the base length ∣d⃗∣|\vec d| is it a distance.

The height is PMPM, not PQPQ

The height of the triangle is the perpendicular distance to the line. PQPQ is a slanted side; using it overstates the area.

Shortest Distance, Intersection and Coplanar Lines

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Shortest distance between skew lines

SD=∣(a⃗2−a⃗1)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣SD=\frac{\left|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\right|}{|\vec d_1\times\vec d_2|}

Distance between parallel lines

d=∣(a⃗2−a⃗1)×d⃗∣∣d⃗∣d=\frac{|(\vec a_2-\vec a_1)\times\vec d|}{|\vec d|}

The line of shortest distance: its feet on the two lines

Conditions for the feet

PQ⃗⋅d⃗1=0,PQ⃗⋅d⃗2=0\vec{PQ}\cdot\vec d_1=0,\qquad \vec{PQ}\cdot\vec d_2=0

Where two lines meet, and when they are coplanar

Condition for coplanar lines

(a⃗2−a⃗1)⋅(d⃗1×d⃗2)=0(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0

A line that meets two given lines

Condition for a given direction

Q−P=k v⃗Q-P=k\,\vec v

Common traps

Absolute value, and two answers

The formula has ∣…∣|\dots|. With an unknown inside, ∣expression∣=c|expression|=c gives two cases; questions asking for 'the sum of all values' need both.

Check for parallel directions first

(2,3,4)(2,3,4) and (4,6,8)(4,6,8) are parallel. Plugging them into the skew-line formula divides by zero; spot it before computing.

Perpendicular to BOTH lines

Making PQ⃗\vec{PQ} perpendicular to only one direction gives the foot from a point, not the common perpendicular. Both dot products must vanish.

Always check the third equation

Any two coordinates can be solved for λ\lambda and μ\mu. Only if the third coordinate also agrees do the lines actually meet.

Two parameters, not one

Using the same letter for the parameters on both lines forces the meeting points to correspond, which they do not. Give each line its own parameter.

Equation of a Plane

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A plane from a point and a normal, three points, or its intercepts

Point-normal form

a(x−x1)+b(y−y1)+c(z−z1)=0a(x-x_1)+b(y-y_1)+c(z-z_1)=0

The normal as a cross product: planes containing lines or perpendicular to planes

Normal from two in-plane directions

n⃗=u⃗×v⃗\vec n=\vec u\times\vec v

The family of planes through a line of intersection

Planes through a line of intersection

P1+λP2=0P_1+\lambda P_2=0

The angle between two planes; parallel and perpendicular planes

Angle between planes

cos⁡θ=∣n⃗1⋅n⃗2∣∣n⃗1∣ ∣n⃗2∣\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1|\,|\vec n_2|}

Common traps

Intercepts are not the normal

x2+y3+z6=1\frac x2+\frac y3+\frac z6=1 has normal (12,13,16)\left(\frac12,\frac13,\frac16\right), i.e. (3,2,1)(3,2,1), not (2,3,6)(2,3,6). Clear the fractions before reading the normal.

Perpendicular to a plane means CONTAINING its normal

A plane perpendicular to x+y+z=1x+y+z=1 has (1,1,1)(1,1,1) lying in it, so (1,1,1)(1,1,1) goes into the cross product. Using (1,1,1)(1,1,1) as the new normal gives a PARALLEL plane instead.

Parallel to a line: dot with its DIRECTION

A plane parallel to a line has the line's direction perpendicular to its normal: n⃗⋅d⃗=0\vec n\cdot\vec d=0. Setting n⃗\vec n parallel to d⃗\vec d instead makes the plane perpendicular to the line.

Take the absolute value for the acute angle

A negative dot product gives the obtuse angle between the normals. The angle between planes is conventionally the acute one, so use ∣n⃗1⋅n⃗2∣|\vec n_1\cdot\vec n_2|.

Distance, Foot and Image in a Plane

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Distance from a plane, parallel planes and sides of a plane

Distance of a point from a plane

d=∣ax1+by1+cz1+d∣a2+b2+c2d=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

The foot and the image of a point in a plane

Foot and image

F=P−k n⃗,Q=P−2k n⃗,k=ax1+by1+cz1+da2+b2+c2F=P-k\,\vec n,\quad Q=P-2k\,\vec n,\quad k=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}

Projections onto a plane

Length of a projection onto a plane

∣v⃗∣2−(v⃗⋅n^)2\sqrt{|\vec v|^2-(\vec v\cdot\hat n)^2}

Common traps

Match the normals before subtracting constants

2x+y−2z=12x+y-2z=1 and 4x+2y−4z=114x+2y-4z=11 are parallel, but ∣11−1∣/3|11-1|/3 is wrong: rescale one equation so both have the same a,b,ca,b,c first.

Keep the sign of the plane's value

kk can be negative, which moves the point along +n⃗+\vec n. Taking ∣k∣|k| sends the foot and image to the wrong side.

Subtract the normal component, not the whole vector

The projection keeps the part of v⃗\vec v that lies in the plane. Using v⃗⋅n^\vec v\cdot\hat n as the answer gives the part that was removed.

Lines Meeting Planes

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Where a line meets a plane, and the ratio a plane cuts

Ratio in which a plane divides AB

ACCB=−P(A)P(B)\frac{AC}{CB}=-\frac{P(A)}{P(B)}

The angle between a line and a plane; parallel and lying in

Angle between a line and a plane

sin⁡θ=∣d⃗⋅n⃗∣∣d⃗∣ ∣n⃗∣\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}

Distance measured parallel to a given line

Distance along a direction

distance=∣t∣ ∣d⃗∣,P+td⃗ on the target\text{distance}=|t|\,|\vec d|,\quad P+t\vec d\ \text{on the target}

Common traps

0⋅t=c0\cdot t=c means no meeting point

If tt drops out, the line is parallel to the plane: no solution when the constant is non-zero, every point when it is zero (the line lies in the plane).

SINE for a line and a plane

The dot product with the NORMAL gives the angle with the normal. The angle with the plane is its complement, so the formula has sin⁡θ\sin\theta. A question stating cos⁡θ\cos\theta of the line-plane angle needs converting first.

Multiply by ∣d⃗∣|\vec d| at the end

tt counts steps of d⃗\vec d, not units of length. Unless d⃗\vec d is a unit vector, the distance is ∣t∣ ∣d⃗∣|t|\,|\vec d|.

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