JEE Mathematics Three Dimensional Geometry Notes Set 04

Direction Cosines and Direction Ratios

If a line makes angles \( \alpha,\beta,\gamma \) with the positive axes, its direction cosines are \( l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma \) with \( l^2+m^2+n^2=1 \). Direction ratios a, b, c are any numbers proportional to l, m, n; divide by \( \sqrt{a^2+b^2+c^2} \) to get the direction cosines.
For the line joining \( (x_1,y_1,z_1) \) and \( (x_2,y_2,z_2) \), direction ratios are \( x_2-x_1,\ y_2-y_1,\ z_2-z_1 \).
The angle between lines with direction ratios \( a_1,b_1,c_1 \) and \( a_2,b_2,c_2 \) satisfies \( \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \). The lines are parallel if the ratios are proportional and perpendicular if the sum of products is zero.

Straight Line in Space

Symmetric form through \( (x_1,y_1,z_1) \) with direction ratios a, b, c: \( \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \). Vector form: \( \vec r=\vec a+\lambda\vec b \). Two-point form uses the ratios from the two points.
A line can also be the intersection of two planes; its direction is the cross product of the two normals.
Foot of the perpendicular from a point P to a line: take a general point \( Q \) of the line and impose \( \vec{PQ}\cdot\vec b=0 \). The image of P is the reflection through this foot.
Distance of P from the line through A with direction \( \vec b \): \( \frac{|\vec{AP}\times\vec b|}{|\vec b|} \).

Skew Lines and Shortest Distance

Two lines \( \vec r=\vec a_1+\lambda\vec b_1 \) and \( \vec r=\vec a_2+\mu\vec b_2 \) are coplanar if \( (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0 \); otherwise they are skew.
Shortest distance between skew lines: \( d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|} \). For parallel lines, \( d=\frac{|(\vec a_2-\vec a_1)\times\vec b|}{|\vec b|} \).
The common perpendicular has direction \( \vec b_1\times\vec b_2 \); its feet are found by writing general points on both lines and making the joining vector parallel to that cross product.

Equation of a Plane

General form \( ax+by+cz+d=0 \) with normal \( (a,b,c) \). Normal form: \( \vec r\cdot\hat n=p \) where p is the distance from the origin. Intercept form: \( \frac xp+\frac yq+\frac zr=1 \).
Plane through \( (x_1,y_1,z_1) \) with normal \( (a,b,c) \): \( a(x-x_1)+b(y-y_1)+c(z-z_1)=0 \). Through three points: determinant of the points with a row of ones set to zero (or normal from a cross product).
Plane through the line of intersection of two planes \( P_1=0 \) and \( P_2=0 \): \( P_1+\lambda P_2=0 \); choose lambda from an extra condition (a point on it, parallel to a direction, or passing through the origin).
The plane containing a line and a point, or two coplanar lines, has a normal obtained by a cross product of two direction vectors in it.

Distances and Angles

Distance of \( (x_1,y_1,z_1) \) from \( ax+by+cz+d=0 \): \( \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \). Distance between parallel planes with equal normals: \( \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \).
Foot and image of a point in a plane: move along the normal. The foot is \( (x_1,y_1,z_1)-t(a,b,c) \) with \( t=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2} \), and the image uses twice that shift.
Angle between planes = angle between normals: \( \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{|\vec n_1||\vec n_2|} \). Angle between a line (direction \( \vec d \)) and a plane (normal \( \vec n \)): \( \sin\phi=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|} \).
A line lies in a plane when one of its points satisfies the plane and its direction is perpendicular to the normal. The line is parallel to the plane when only the direction condition holds.
Two planes \( P_1=0,P_2=0 \) bisected by \( \frac{a_1x+b_1y+c_1z+d_1}{|\vec n_1|}=\pm\frac{a_2x+b_2y+c_2z+d_2}{|\vec n_2|} \); the bisector containing the origin is found after making both constants positive.

Worked Examples

Worked Example 1. Find the shortest distance between the lines through \( (3,5,7) \) with direction \( (1,2,1) \) and through \( (-1,-1,-1) \) with direction \( (7,6,1) \).
Answer:
Cross product of directions: \( (1,2,1)\times(7,6,1)=(-4,6,-8) \), magnitude \( 2\sqrt{29} \).
Joining vector \( (-4,-6,-8) \); dot product \( 16-36+64=44 \).
\( d=\frac{44}{2\sqrt{29}}=\frac{22}{\sqrt{29}} \).

Worked Example 2. Find the foot and the image of the point \( (1,2,3) \) in the plane \( x+y+z=3 \).
Answer:
\( x+y+z-3 \) at the point equals 3, so \( t=\frac33=1 \).
Foot \( =(1,2,3)-1(1,1,1)=(0,1,2) \); image \( =(1,2,3)-2(1,1,1)=(-1,0,1) \).

Worked Example 3. Find the plane through the intersection of \( x+y+z=6 \) and \( 2x+3y+4z+5=0 \) and through the point \( (1,1,1) \).
Answer:
Plane \( (x+y+z-6)+\lambda(2x+3y+4z+5)=0 \); at \( (1,1,1) \): \( -3+14\lambda=0 \), so \( \lambda=\frac3{14} \).
\( 14(x+y+z-6)+3(2x+3y+4z+5)=0 \), i.e. \( 20x+23y+26z=69 \).

Worked Example 4. Find the angle between the planes \( x+2y+2z=3 \) and \( 2x-y+2z=1 \).
Answer:
\( \cos\theta=\frac{2-2+4}{3\cdot3}=\frac49 \), so \( \theta=\cos^{-1}\frac49 \).

Worked Example 5. Show that the lines \( \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \) and \( \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} \) are coplanar.
Answer:
Joining vector \( (1,1,1) \), directions \( (2,3,4),(3,4,5) \): \( \begin{vmatrix}1&1&1\\2&3&4\\3&4&5\end{vmatrix}=0 \), so the lines are coplanar (they intersect at \( (-1,-1,-1) \)).