Three-Dimensional Geometry
Geometry lifted into space — locating points by three coordinates, writing lines and planes as vector equations, and measuring the angles and distances between them, right up to the gap between two lines that never meet
- Locating points in space and the distance and section formulas in three dimensions.
- Direction cosines and direction ratios of a line.
- The vector and Cartesian equations of a line, and the angle between two lines.
- The equation of a plane in normal, Cartesian and intercept form.
- The angles between two planes and between a line and a plane.
- The distance of a point from a plane, and the shortest distance between two skew lines.
Coordinates & Distance in Space
A point in space needs three coordinates \((x,y,z)\), measured along three mutually perpendicular axes that divide space into eight octants. The distance between two points extends the plane formula with a third squared term, and the section formula carries over coordinate by coordinate.
A point dividing \(P_1P_2\) in the ratio \(m:n\) has coordinates \(\big(\tfrac{mx_2+nx_1}{m+n},\,\tfrac{my_2+ny_1}{m+n},\,\tfrac{mz_2+nz_1}{m+n}\big)\), and the midpoint is their average. Distance from the origin is simply \(\sqrt{x^2+y^2+z^2}\).
Direction Cosines & Ratios
A line's orientation in space is captured by the angles \(\alpha,\beta,\gamma\) it makes with the three axes. The cosines of these angles are the direction cosines; any triple proportional to them is a set of direction ratios, which are usually far easier to read off.
From direction ratios \(a,b,c\), recover the cosines by dividing by \(\sqrt{a^2+b^2+c^2}\). The line joining \(P_1\) and \(P_2\) has direction ratios \((x_2-x_1,\,y_2-y_1,\,z_2-z_1)\) — its components are exactly the displacement vector.
The Equation of a Line
A line is fixed by one point on it and a direction. In vector form it is the running position \(\vec a\) plus any multiple of the direction \(\vec b\); eliminating the parameter gives the symmetric Cartesian form.
The point is \((x_1,y_1,z_1)\) and the direction ratios are \((a,b,c)\). A line through two points \(\vec a,\vec b\) is \(\vec r=\vec a+\lambda(\vec b-\vec a)\), using the join as its direction.
Angle Between Two Lines
The angle between two lines is the angle between their direction vectors — and a single dot product delivers it. The same expression instantly tells you when the lines are parallel or perpendicular.
The lines are perpendicular when \(a_1a_2+b_1b_2+c_1c_2=0\), and parallel when the ratios match, \(\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}=\tfrac{c_1}{c_2}\). The absolute value ensures the acute angle is reported.
The Equation of a Plane
A plane is pinned down by a single point on it and a normal direction perpendicular to it. Every point \(\vec r\) of the plane makes the displacement \(\vec r-\vec a\) perpendicular to the normal, which is the whole equation in one line.
The coefficients \((a,b,c)\) are the components of the normal. The intercept form \(\tfrac xa+\tfrac yb+\tfrac zc=1\) reads the axis crossings directly, and a plane through three points takes its normal as \((\vec P_2-\vec P_1)\times(\vec P_3-\vec P_1)\).
Angles with Planes
Because a plane is represented by its normal, every angle question reduces to angles between vectors. Two planes meet at the angle between their normals; a line meets a plane at the complement of the angle between the line and the normal.
A line is parallel to a plane when \(\vec b\cdot\vec n=0\), and perpendicular to it when \(\vec b\parallel\vec n\). Two planes are parallel when their normals are parallel.
The angle between a line and a plane is measured to the plane itself, not to its normal — so it is the complement of the line–normal angle, and that is why the formula uses \(\sin\theta\) rather than \(\cos\theta\). Mixing the two up is the single most common mistake here; reach for sine only when one object is a line and the other a plane.
Distance of a Point from a Plane
The distance from a point to a plane is the length of the perpendicular dropped onto it. Substituting the point into the plane's equation, then normalising by the length of the normal, gives that distance directly.
Substitute the point \((x_1,y_1,z_1)\) into the left side of \(ax+by+cz+d_0=0\) and divide by the normal's length. The distance between two parallel planes follows by taking any point on one and applying the same formula, or simply \(\tfrac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}\).
Shortest Distance Between Skew Lines
Two lines in space that are neither parallel nor intersecting are skew. The shortest distance between them is measured along their common perpendicular — and the scalar triple product packages it into one clean formula.
Here \(\vec a_1,\vec a_2\) are points on the two lines and \(\vec b_1,\vec b_2\) their directions. The lines intersect (are coplanar) exactly when the numerator vanishes. For two parallel lines use \(d=\dfrac{|(\vec a_2-\vec a_1)\times\vec b|}{|\vec b|}\) instead.
Putting It to Work
Problem. Find the distance between \(P(1,-2,3)\) and \(Q(4,2,15)\).
Solution. Apply the distance formula to the differences \((3,4,12)\):
Problem. Find the direction cosines of the line joining \((1,2,3)\) and \((4,5,6)\).
Solution. The direction ratios are \((3,3,3)\propto(1,1,1)\); divide by \(\sqrt{3}\):
Problem. Write the vector and Cartesian equations of the line through \((1,2,-1)\) with direction \(2\hat i+3\hat j+4\hat k\).
Solution. Use \(\vec r=\vec a+\lambda\vec b\) and eliminate \(\lambda\):
Problem. Find the angle between the lines with direction ratios \((1,1,2)\) and \((2,1,1)\).
Solution. Both directions have length \(\sqrt6\), and their dot product is \(2+1+2=5\):
Problem. Find the plane through \((1,1,1)\) with normal \(2\hat i+3\hat j-\hat k\).
Solution. Use \((\vec r-\vec a)\cdot\vec n=0\):
Problem. Find the shortest distance between \(\vec r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)\) and \(\vec r=(2\hat i-\hat j-\hat k)+\mu(2\hat i+\hat j+2\hat k)\).
Solution. With \(\vec a_2-\vec a_1=(1,-3,-2)\) and \(\vec b_1\times\vec b_2=(-3,0,3)\), so \(|\vec b_1\times\vec b_2|=3\sqrt2\):
Chapter Summary
\(PQ=\sqrt{\sum(\Delta)^2}\); section formula coordinate-wise.
\(l^2+m^2+n^2=1\); ratios proportional to cosines.
\(\vec r=\vec a+\lambda\vec b\); symmetric Cartesian form.
\((\vec r-\vec a)\cdot\vec n=0\); coefficients are the normal.
Planes use \(\cos\) of normals; line–plane uses \(\sin\).
Point–plane \(\tfrac{|ax_1+\cdots|}{\sqrt{a^2+b^2+c^2}}\); skew lines via \(\vec b_1\times\vec b_2\).
Problems
Identify the objects first — points, lines, or planes — then pick the matching formula. Difficulty rises down the list.
- Find the distance between \((1,2,3)\) and \((-2,6,3)\).
- Find the coordinates of the point dividing \((1,2,3)\) and \((4,5,6)\) in the ratio \(2:1\).
- Find the direction cosines of a line with direction ratios \((1,2,2)\).
- Write the vector and Cartesian equations of the line through \((2,-1,3)\) parallel to \(3\hat i+2\hat j-\hat k\).
- Find the angle between the lines with direction ratios \((2,2,1)\) and \((4,1,8)\).
- Find the equation of the plane through \((2,1,-1)\) with normal \(\hat i+2\hat j+3\hat k\).
- Find the distance from \((1,2,3)\) to the plane \(2x-y+2z+3=0\).
- Find the angle between the planes \(x+y+z=1\) and \(x-y+z=5\).
- Find the angle between the line \(\tfrac{x-1}{2}=\tfrac{y}{3}=\tfrac{z+1}{6}\) and the plane \(2x+y-z=5\).
- Find the equation of the plane with intercepts \(2,3,4\) on the axes.
- Find the shortest distance between \(\vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+\hat k)\) and \(\vec r=(2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k)\).
- Show that the points \((-2,3,5),(1,2,3),(7,0,-1)\) are collinear.
- Find the distance between the parallel planes \(2x-2y+z+3=0\) and \(2x-2y+z+9=0\).