Passes through a point & parallel to a vector
Equation for a line that:
- passes through r0=⟨x0,y0,z0⟩
- is parallel to v=ai+bj+ck
Parametric equation
r=r0+tv;t∈R
Symmetric equation
ax−x0=by−y0=cz−z0
Passes through 2 points
Equation of a line passes through A=(x1,y1,z1), B=(x2,y2,z2).
rA and rB are the position vectors of A and B.
Parametric equation
r=(1−t)rA+trB;t∈R
Symmetric equation
x2−x1x−x1=y2−y1y−y1=z2−z1z−z1
Intersection
To show that two straight lines intersect in 3D space, existence of a point
which satisfies both lines must be proven.
It is not enough to show that the cross product of their parallel vectors is
non-zero.
Normal to 2 lines
Let α,β be two lines.
α:a1x−x1=b1y−y1=c1z−z1;β:a2x−x2=b2y−y2=c2z−z2
Here v1=⟨a1,b1,c1⟩, v2=⟨a2,b2,c2⟩ are
2 vectors parallel to α,β respectively.
Normal to both lines: v1×v2.
Unit normal
∣v1×v2∣v1×v2
Angle between 2 straight lines
Using the α,β lines mentioned above:
cosθ=∣v1∣⋅∣v2∣v1⋅v1=∣a1i+b1j+c1k∣⋅∣a2i+b2j+c2k∣(a1i+b1j+c1k)⋅(a2i+b2j+c2k)
Here v1,v2 are 2 vectors parallel to α,β respectively.
Shortest distance from a point
The distance can be calculated using Pythogoras’ theorem.
d2=r−p2−[∣n∣n⋅(r−p)]2
Here:
- P is the arbitrary point
- p is the position vector of P
- r is the position vector of a point on the line
- n is parallel to the line