PYQ (JEE Main & Advanced)

High-Yield JEE Main Concepts for 3D Geometry

1. Family of Planes

A very common JEE Main pattern is to find a plane passing through the line of intersection of two planes $\pi_1=0$ and $\pi_2=0$. The standard family is $$\pi_1 + \lambda \pi_2 = 0.$$ Then use the extra condition, such as passing through a point or being perpendicular to another plane, to determine $\lambda$.

2. Image of a Point in a Plane

If the image $(x',y',z')$ of a point $(x_1,y_1,z_1)$ in the plane $ax+by+cz+d=0$ is required, the direct formula is $$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}.$$ For the foot of the perpendicular, replace the factor $-2$ by $-1$.

3. Shortest Distance Between Lines

If the direction vectors are proportional, the lines are parallel and the formula is $$d = \frac{|(\vec{a}_2-\vec{a}_1)\times \vec{b}|}{|\vec{b}|}.$$ If they are not proportional, first check whether they intersect or are skew. For skew lines, $$d = \left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times \vec{b}_2)}{|\vec{b}_1\times \vec{b}_2|}\right|.$$ If the scalar triple product in the numerator is zero, the lines are coplanar, and if they are also non-parallel, the shortest distance is zero.

4. Line as Intersection of Two Planes

If a line is obtained as the intersection of two planes, $$a_1x+b_1y+c_1z+d_1=0, \qquad a_2x+b_2y+c_2z+d_2=0,$$ then its direction vector is the cross product of the normals: $$\vec{n}_1 \times \vec{n}_2.$$ To find one point on the line, set one variable conveniently, often $z=0$, and solve the remaining two equations.