3D Geometry Formulas for JEE Main

Direction Cosine Relation

#1

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

💡 Direction cosines (l, m, n) are the cosines of the angles a line makes with the positive x, y, z axes. They always satisfy l^2 + m^2 + n^2 = 1.

⚠ Confusing direction cosines with direction ratios. DRs (a, b, c) do not satisfy a^2 + b^2 + c^2 = 1 in general.

Symmetric Form of a Line

#2

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

💡 Here (x1, y1, z1) is a point on the line and (a, b, c) are direction ratios of the line. If any DR is zero, the corresponding numerator must also be zero.

General Equation of a Plane

#3

ax + by + cz + d = 0

💡 Here (a, b, c) are the direction ratios of the normal to the plane. The normal vector is n = a i + b j + c k.

Angle Between Two Lines

#4

\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}}

💡 Use direction ratios (a1, b1, c1) and (a2, b2, c2) of the two lines. Take the absolute value of the numerator to get the acute angle.

⚠ Forgetting the absolute value in the numerator. Without it you may get the obtuse angle.

Angle Between a Line and a Plane

#5

\sin\theta = \frac{|al + bm + cn|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}

💡 theta is measured between the line and the plane (not the normal). The formula uses sin, not cos. Here (l, m, n) are DRs of the line and (a, b, c) are DRs of the normal to the plane.

⚠ Using cos instead of sin. The angle between a line and a plane is the complement of the angle between the line and the normal.

Distance from a Point to a Plane

#6

d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

💡 Substitute the point (x1, y1, z1) directly into ax + by + cz + d. The absolute value ensures a non-negative distance.

⚠ Dropping the absolute value sign and getting a negative distance, or forgetting the constant d in the numerator.

Shortest Distance Between Skew Lines

#7

d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}

💡 Lines: r = a1 + t*b1 and r = a2 + s*b2. Compute b1 x b2 first, then dot it with (a2 - a1). Take the absolute value and divide by |b1 x b2|.

⚠ Using (a1 - a2) instead of (a2 - a1) and then forgetting the absolute value, leading to a sign error.

Condition for Coplanarity of Two Lines

#8

(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0

💡 Two lines are coplanar (intersect or are parallel) if and only if the shortest distance between them is zero. This is equivalent to the scalar triple product being zero.

Foot of Perpendicular from Point to Plane

#9

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}

💡 The foot lies on the line through (x1, y1, z1) with DRs (a, b, c) (the normal direction). Substitute the parametric point into the plane equation to find the parameter value.

Image of a Point in a Plane

#10

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}

💡 The image is obtained by going twice the distance from the point to the plane along the normal. The parameter value is exactly double that of the foot of perpendicular.

⚠ Using the single distance formula instead of double. The image is at twice the perpendicular distance, not equal to it.