JEE Mathematics Three Dimensional Geometry Notes Set 02

Equation of a Plane

Every first-degree equation \( ax+by+cz+d=0 \) with \( a,b,c \) not all zero is a plane, and \( (a,b,c) \) are the direction ratios of its normal. The plane through \( (x_1,y_1,z_1) \) with normal ratios \( (a,b,c) \) is \( a(x-x_1)+b(y-y_1)+c(z-z_1)=0 \). A plane parallel to \( ax+by+cz+d=0 \) has the same \( a,b,c \) and a different constant. If one variable is missing, the plane is parallel to the axis of that variable; for instance \( by+cz+d=0 \) is parallel to the x-axis and perpendicular to the YZ-plane.

Planes Fixed by Other Conditions

Through three non-collinear points, the plane is the determinant \( \begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0 \), or equivalently use the cross product of two in-plane vectors as the normal. Through a point and parallel to two given directions, the normal is the cross product of the directions. Four points are coplanar when the \( 3\times3 \) determinant of their difference vectors is zero. The plane bisecting the join of A and B at right angles passes through the mid-point of AB and has AB as its normal.

Intercept Form and Normal Form

If a plane cuts the axes at \( (a,0,0),(0,b,0),(0,0,c) \), its equation is \( \frac xa+\frac yb+\frac zc=1 \). For \( ax+by+cz+d=0 \) the intercepts are \( -\frac da,-\frac db,-\frac dc \). The normal form is \( lx+my+nz=p \), where \( (l,m,n) \) are the direction cosines of the normal and \( p>0 \) is the distance from the origin; to obtain it divide by \( \sqrt{a^2+b^2+c^2} \) after making the constant positive on the right side. The triangle cut off by the plane has area \( \frac12\sqrt{(ab)^2+(bc)^2+(ca)^2} \) with the three axes, and \( \frac12|ab| \) with two of them. Its centroid is \( (\frac a3,\frac b3,\frac c3) \).

Distances and Positions

The distance of \( (x_1,y_1,z_1) \) from \( ax+by+cz+d=0 \) is \( \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \). The distance between parallel planes \( ax+by+cz+d_1=0 \) and \( ax+by+cz+d_2=0 \) is \( \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \), and the plane midway between them has constant \( \frac{d_1+d_2}2 \). Two points lie on the same side of a plane when the plane expression has the same sign at both, and on opposite sides when the signs differ. A plane divides the segment joining P and Q in the ratio \( -f(P):f(Q) \), where f is the plane expression; a negative ratio means external division.

Foot of Perpendicular and Image

For the plane \( ax+by+cz+d=0 \) and a point \( P(x_1,y_1,z_1) \), let \( f=ax_1+by_1+cz_1+d \) and \( N=a^2+b^2+c^2 \). The foot is \( P-\frac fN(a,b,c) \) and the image is \( P-\frac{2f}N(a,b,c) \). The foot is the mid-point of P and its image. The reflection of the plane \( a_1x+b_1y+c_1z+d_1=0 \) in the plane \( ax+by+cz+d=0 \) is \( 2(aa_1+bb_1+cc_1)(ax+by+cz+d)=(a^2+b^2+c^2)(a_1x+b_1y+c_1z+d_1) \).

Angle Between Planes and Bisectors

The angle between two planes is the angle between their normals: \( \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \). The planes are parallel when \( \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \) and perpendicular when \( a_1a_2+b_1b_2+c_1c_2=0 \); they coincide if the constant terms are in the same ratio too. 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 both constants positive; then the plus sign gives the bisector containing the origin, and it is the acute-angle bisector when \( a_1a_2+b_1b_2+c_1c_2<0 \), the obtuse one otherwise. The angle between a line and a plane is the complement of the angle between the line and the normal, so \( \sin\theta=\frac{|al+bm+cn|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}} \).

Family of Planes Through a Line of Intersection

Every plane through the line of intersection of \( P_1=0 \) and \( P_2=0 \) is \( P_1+\lambda P_2=0 \). Choose \( \lambda \) from one extra condition: a point on the plane, perpendicularity to a given plane (dot product of normals zero), parallelism to an axis, or a given distance from the origin. The direction of the line of intersection is the cross product of the two normals. Homogeneous equations such as \( 2x^2-6y^2-12z^2+18yz+2zx+xy=0 \) represent two planes through the origin, found by factorising.

Variable Plane Results

If a plane at constant distance p from the origin meets the axes at A, B, C, then \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=\frac1{p^2} \). The centroid of triangle ABC is \( (\frac a3,\frac b3,\frac c3) \), so its locus is \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac9{p^2} \); the centroid of the tetrahedron OABC is \( (\frac a4,\frac b4,\frac c4) \) and its locus is \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac{16}{p^2} \). The planes \( x=\pm a,\ y=\pm b,\ z=\pm c \) form a rectangular box, and planes through two opposite vertices parallel to the coordinate planes give a box whose diagonal is \( \sqrt{a^2+b^2+c^2} \).

Worked Examples

Worked Example 1. Find the distance between the planes \( 2x-2y+z+3=0 \) and \( 4x-4y+2z+5=0 \).
Answer:
Rewrite the second plane as \( 2x-2y+z+\frac52=0 \).
Distance \( =\frac{|3-\frac52|}{\sqrt{4+4+1}}=\frac{1/2}{3} \).
Distance = \( \frac16 \).

Worked Example 2. Find the image of the point \( (-1,3,4) \) in the plane \( x-2y=0 \).
Answer:
\( f=-1-6=-7 \) and \( N=1+4=5 \).
Image \( =(-1,3,4)-\frac{2(-7)}{5}(1,-2,0)=\left(\frac95,-\frac{13}5,4\right) \).
Image = \( \left(\frac95,-\frac{13}5,4\right) \).

Worked Example 3. Find the plane through the line of intersection of \( x-2y+3z-1=0 \) and \( 2x+y+z-2=0 \) and the point \( (1,2,3) \).
Answer:
Take \( (x-2y+3z-1)+k(2x+y+z-2)=0 \).
At \( (1,2,3) \): \( 5+5k=0 \), so \( k=-1 \).
The plane is \( x+3y-2z-1=0 \).

Worked Example 4. Find the area of the triangle cut off from the axes by the plane \( \frac x2+\frac y3+\frac z4=1 \).
Answer:
Here \( ab=6,\ bc=12,\ ca=8 \).
Area \( =\frac12\sqrt{36+144+64}=\frac12\sqrt{244} \).
Area = \( \sqrt{61} \) square units.