JEE Mathematics Three Dimensional Geometry Notes Set 03

Straight Lines in Space: Equations

A line through \( (x_1,y_1,z_1) \) with direction ratios (a, b, c) has the symmetric form \( \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \). A zero denominator means that coordinate is constant along the line. Setting each ratio equal to t gives the parametric form \( (x_1+at,\,y_1+bt,\,z_1+ct) \), and the vector form is \( \vec r=\vec a+\lambda\vec b \). The line through two points uses their difference as direction. Lines parallel to the axes have directions (1, 0, 0), (0, 1, 0), (0, 0, 1), and a line equally inclined to the axes has direction (1, 1, 1).

Line Given as Two Planes

For the line \( a_1x+b_1y+c_1z+d_1=0=a_2x+b_2y+c_2z+d_2 \), the direction is the cross product of the two normals. A point on it is found by giving one variable a convenient value (usually 0) and solving the other two equations. A point at distance r from \( (x_1,y_1,z_1) \) along a line with direction cosines (l, m, n) is \( (x_1\pm lr,\,y_1\pm mr,\,z_1\pm nr) \).

Angles

The angle between lines with directions \( (a_1,b_1,c_1) \) and \( (a_2,b_2,c_2) \) satisfies \( \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}} \). Lines are parallel if the ratios are equal and perpendicular if the dot product is zero. The angle between a line (direction l, m, n) and a plane (normal a, b, c) satisfies \( \sin\theta=\frac{|al+bm+cn|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}} \). A line is parallel to a plane when \( al+bm+cn=0 \), and perpendicular to it when (l, m, n) is proportional to (a, b, c).

Coplanar and Skew Lines

Two lines are coplanar if they are parallel or intersect; otherwise they are skew. Lines through \( (x_1,y_1,z_1) \) and \( (x_2,y_2,z_2) \) with directions \( (a_1,b_1,c_1) \) and \( (a_2,b_2,c_2) \) are coplanar exactly when \( \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0 \). The plane containing two coplanar lines is given by the same determinant with (x, y, z) in place of the second point. A line lies in the plane \( ax+by+cz+d=0 \) when the point satisfies the plane and \( al+bm+cn=0 \).

Distances

For skew lines \( \vec r=\vec a_1+\lambda\vec b_1 \) and \( \vec r=\vec a_2+\mu\vec b_2 \), the shortest distance is \( \frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|} \), and it is zero when the lines are coplanar. For parallel lines with the same direction \( \vec b \), the distance is \( \frac{|\vec b\times(\vec a_2-\vec a_1)|}{|\vec b|} \). The distance of a point P from a line through A with direction \( \vec b \) is \( \frac{|\overrightarrow{AP}\times\vec b|}{|\vec b|} \). To find a foot of perpendicular, write a general point of the line and make the join vector perpendicular to the direction; the image is twice the foot minus the point.

Distance Measured Along or Parallel to a Direction

To measure the distance of a point P to a plane along a given direction, write the point \( P+t\,\vec d \), substitute it into the plane to find t, and the distance is \( |t|\,|\vec d| \). When the distance is measured along a line, the same method applies with that line's direction. For a point on a given line at a fixed distance from a given point, form the squared distance as a quadratic in t and solve; two points result, one on each side.

Worked Examples

Example 1. Find the Cartesian and vector equations of the line through \( (-1,0,2) \) and \( (3,4,6) \).
Answer:
Direction \( =(4,4,4)\propto(1,1,1) \).
Cartesian: \( \frac{x+1}{1}=\frac{y}{1}=\frac{z-2}{1} \). Vector: \( \vec r=(-\hat i+2\hat k)+\lambda(\hat i+\hat j+\hat k) \).

Example 2. Write \( 3x+2y-z-4=0,\ 4x+y-2z+3=0 \) in symmetric form.
Answer:
Direction \( =(3,2,-1)\times(4,1,-2)=(-3,2,-5) \).
Putting z = 0: \( 3x+2y=4,\ 4x+y=-3 \), so x = -2, y = 5.
The line is \( \frac{x+2}{-3}=\frac{y-5}{2}=\frac{z}{-5} \).

Example 3. Find the image of \( (1,6,3) \) in the line \( \frac x1=\frac{y-1}{2}=\frac{z-2}{3} \).
Answer:
Foot \( =(t,2t+1,3t+2) \); the join vector must satisfy \( (t-1)+2(2t-5)+3(3t-1)=0 \), so t = 1.
Foot (1, 3, 5), image \( =2(1,3,5)-(1,6,3)=(1,0,7) \).

Example 4. Find the shortest distance between \( \vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+\hat k) \) and \( \vec r=(2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k) \).
Answer:
\( \vec b_1\times\vec b_2=(3,-1,-7) \), magnitude \( \sqrt{59} \); \( \vec a_2-\vec a_1=(1,0,-1) \).
Triple product \( =3+7=10 \). Distance \( =\frac{10}{\sqrt{59}} \).