Vectors and Three-Dimensional Geometry
Vectors and Three-Dimensional Geometry
This domain tests vector algebra and its use in lines and planes: dot and cross products, angles, direction ratios, and shortest distances. ASSEB problems often mix a vector calculation with a coordinate-geometry equation.
Core rules
- If a=a1i+a2j+a3k and b=b1i+b2j+b3k, then a·b=a1b1+a2b2+a3b3=|a||b|cosθ.
- a×b is perpendicular to both; |a×b|=|a||b|sinθ. Its direction follows the right-hand rule, so a×b=−(b×a).
- Direction cosines of a are a1/|a|, a2/|a|, a3/|a|; their squares sum to 1.
- Line through point a with direction b: r=a+λb. Plane through point a with normal n: (r−a)·n=0, i.e. n·r=n·a.
- Distance from origin to plane ax+by+cz+d=0 is |d|/√(a^2+b^2+c^2).
Worked example
Find the plane through P(1,2,3) with normal n=2i−j+2k, and its distance from the origin. Using (r−a)·n=0: 2(x−1)−1(y−2)+2(z−3)=0. Expanding gives 2x−2−y+2+2z−6=0, so 2x−y+2z−6=0. The distance from O(0,0,0) is |−6|/√(2^2+(−1)^2+2^2)=6/√9=6/3=2. The unit normal is (2i−j+2k)/3, and the plane is 2 units from the origin.
Traps
Do not use cross product for a dot-product angle question. Reverse order in a×b changes the sign. In a plane equation, the normal is the coefficient vector (a,b,c), not a direction vector lying in the plane. When finding the distance, the formula requires the plane in general form ax+by+cz+d=0 and a unit normal.