JEE Mathematics Three Dimensional Geometry Notes Set 01

Planes: Equation Forms

Every first-degree equation \( ax+by+cz+d=0 \) (a, b, c not all zero) is a plane, and (a, b, c) are direction ratios of its normal. Through the point \( (x_1,y_1,z_1) \) with normal (a, b, c): \( a(x-x_1)+b(y-y_1)+c(z-z_1)=0 \). A plane parallel to a given one keeps the coefficients of x, y, z and changes only the constant. In intercept form \( \frac xa+\frac yb+\frac zc=1 \), the intercepts of \( ax+by+cz+d=0 \) are \( -\frac da,-\frac db,-\frac dc \).

Planes through Given Points or Directions

The plane through three non-collinear points is obtained from the determinant of the vectors \( (x-x_1,y-y_1,z-z_1) \), \( (x_2-x_1,\dots) \), \( (x_3-x_1,\dots) \) set equal to zero; four points are coplanar when the corresponding 3 x 3 determinant vanishes. A plane through a point and parallel to two given directions has normal equal to the cross product of those directions. If one variable is missing from the equation, the plane is parallel to that axis; if two are missing, it is parallel to a coordinate plane (for example, z = k is parallel to the XY-plane).

Distances and Reflections

The distance of \( (x_1,y_1,z_1) \) from \( ax+by+cz+d=0 \) is \( \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \). Two parallel planes with identical coefficients are \( d_1 \) and \( d_2 \) apart by \( \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \), and the plane midway has constant \( \frac{d_1+d_2}2 \). Let \( t=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2} \). The foot of the perpendicular from the point is \( (x_1-ta,\,y_1-tb,\,z_1-tc) \) and its image is \( (x_1-2ta,\,y_1-2tb,\,z_1-2tc) \). The plane bisecting the join of two points at right angles has the segment as normal and passes through the mid-point.

Sides of a Plane and the Ratio Formula

Two points lie on the same side of \( ax+by+cz+d=0 \) if the expression has the same sign at both, and on opposite sides if the signs differ. The plane divides the segment joining \( P_1,P_2 \) in the ratio \( -(ax_1+by_1+cz_1+d):(ax_2+by_2+cz_2+d) \); a negative ratio means external division.

Angle between Planes, Bisectors and Families

The angle between two planes equals the angle between their normals: \( \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \). Planes are parallel when the coefficients are proportional and perpendicular when \( a_1a_2+b_1b_2+c_1c_2=0 \). The angle between a line and a plane satisfies \( \sin\varphi=|\cos(\text{angle between line and normal})| \). The bisecting planes are \( \frac{a_1x+b_1y+c_1z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm\frac{a_2x+b_2y+c_2z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \). Make \( d_1,d_2 \) positive first; then the plus sign gives the bisector containing the origin, which bisects the acute angle when \( a_1a_2+b_1b_2+c_1c_2<0 \). Any plane through the line of intersection of \( P_1=0 \) and \( P_2=0 \) has the form \( P_1+\lambda P_2=0 \).

Useful Results for Variable Planes

If a plane meets the axes at A, B, C with the centroid of triangle ABC at (p, q, r), its equation is \( \frac xp+\frac yq+\frac zr=3 \). The area of the triangle cut off by \( \frac xa+\frac yb+\frac zc=1 \) is \( \frac12\sqrt{a^2b^2+b^2c^2+c^2a^2} \). If the plane stays at distance p from the origin, then \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=\frac1{p^2} \), and the centroid of triangle ABC satisfies \( x^{-2}+y^{-2}+z^{-2}=p^{-2} \), while the centroid of the tetrahedron OABC satisfies \( x^{-2}+y^{-2}+z^{-2}=16p^{-2} \).

Worked Examples

Example 1. Find the foot of the perpendicular from \( P(1,3,4) \) to the plane \( 2x-y+z+3=0 \).
Answer:
Here \( t=\frac{2-3+4+3}{4+1+1}=1 \).
Foot \( =(1-2,\,3+1,\,4-1)=(-1,4,3) \).

Example 2. Find the distance between the parallel planes \( 2x+2y-z+3=0 \) and \( 4x+4y-2z+5=0 \).
Answer:
Rewrite the second as \( 2x+2y-z+\frac52=0 \).
Distance \( =\frac{|3-\frac52|}{3}=\frac16 \).

Example 3. Find the plane midway between \( 2x-3y+6z+7=0 \) and \( 2x-3y+6z-7=0 \).
Answer:
The constants are 7 and -7, so the mean constant is 0.
The plane is \( 2x-3y+6z=0 \).

Example 4. Find the area of the triangle cut off from the plane \( 2x+3y+6z+9=0 \) by the Y-axis and Z-axis.
Answer:
The intercepts on Y and Z are -3 and -3/2.
Area \( =\frac12\cdot3\cdot\frac32=\frac94 \) square units.

Example 5. Find k if the planes \( x+2y+kz=0 \) and \( 2x+y-2z=0 \) are at right angles.
Answer:
Condition: \( 1(2)+2(1)+k(-2)=0 \).
So \( 4-2k=0 \) and k = 2.