Applications of Vector Algebra
Applications of Vector Algebra tests lines and planes in three dimensions: vector and Cartesian equations, angles, perpendicular distances, shortest distance between skew lines, and images of points. The exam usually supplies coordinates and asks for one distance or one equation.
Standard results
A line through a with direction b is r=a+λb. A plane with normal n through a point with position vector a is (r−a)·n=0, or r·n=d. In Cartesian form n1x+n2y+n3z=d. The angle between two planes is the angle between their normals. The distance from (x0,y0,z0) to ax+by+cz=d is |ax0+by0+cz0−d|/√(a²+b²+c²). The shortest distance between skew lines r=a1+λb1 and r=a2+μb2 is |(a2−a1)·(b1×b2)|/|b1×b2|. Two lines are coplanar when (a2−a1)·(b1×b2)=0.
Worked example
Find the distance from (1,1,1) to the plane 2x−3y+6z=14. Substitute: 2−3+6−14=−9, and √(4+9+36)=7. Distance = |−9|/7=9/7.
Traps
Use the normal vector for a plane, but the direction vector for a line; mixing them gives a wrong angle. Keep the constant on the same side before applying the distance formula. Distance is absolute, so a negative numerator is not a negative distance. For skew lines, the formula uses (a2−a1), not (a1−a2) without taking absolute value; the sign disappears anyway because of the modulus.