How can I represent a vector equation of a line segment in C++? -

i working computer graphics.

i represent line 2 end points, and, line2d class have method returns vector2d object.

suppose, have following classes:

struct point2d {     int x;     int y; }; 

then, can represent line segment using 2 points:

class linesegment2d { private:     point2d start;     point2d end; public:     ...     ... }; 

according definition, vector composed of magnitude , direction.

class vector2d { private:     point2d p; public:     double magnitude(void);     point component(void);     vector2d normal();     vector2d & add(vector & rhs);     vector2d & subtract(vector & rhs);     vector2d & multiply(int scalar);     int dotproduct(vector2d rhs);     vector2d & crossproduct(vector2d rhs); }; 

one object of point2d sufficient represent vector. example, magnitude of vector = sqrt(p.x*p.x + p.y*p.y);. and, p.x , p.y collectively represent direction.

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on other hand, know vector equation of line passing through (x0,y0,z0) is, r =r0 + tv where, r vector subject line. r0 position vector points direction of point (x0, y0, z0). since, r0 position vector, obviously, origin of r0 (0,0,0). t real numbered value, where, −∞<t<∞ –. v vector parallel our subject straight line.

vector equation of line segment between points p(1, 3, 2) , q(-4, 3, 0):

according above formula, vector equation of line pq either

r =<1,3,2> + tv  

or,

r =<-4,3,0> + tv 

the vector connects 2 points p , q is,

pq  = <(-4-1), (3-3), (0-2)>     = <-5, 0, -2> 

and, vector parallel our subject line sure.

so, can write,

r   =<1, 3, 2> + t <-5, 0, -2>     =<1, 3, 2>+<-5t, 0, -2t>     = <(1-5t), (3+0), (2-2t)>     =<1-5t, 3, 2-2t> 

according vector equation of line segment, think, vector class should following:

class linevector2d { private:     vector2d v;     double t; public:     .......... }; 

is correct representation?

if so, how can calculate/set/find value of t?

i think there's confusion because of following

according definition, vector composed of magnitude , direction.

there more 1 way of representing vector. think in question mean vector can represented magnitude (scalar) , unit vector indicating direction. vector can ordered triplet (for 3 dimensions) indicate magnitude (sqrt(x^2 + y^2 + z^2)) , direction origin.

i think answer question is, don't need calculate t. correct me if i'm mistaken, think you're interpreting t magnitude? can calculate v sqrt(x^2 + y^2 + z^2), v can hold both magnitude , direction ordered triplet itself.

edit:

template <typename t> struct point2d {     t x;     t y;      point2d operator + (const point2d<t>& rhs) const     {         return point2d<t>{x + rhs.x, y + rhs.y};     }     point2d operator - (const point2d<t>& rhs) const     {         return point2d<t>{x - rhs.x, y - rhs.y};     }     // ...      point2d operator * (const t value) const     {         return point2d<t>{x*value, y*value};     }     point2d operator / (const t value) const     {         return point2d<t>{x/value, y/value};     }     // ... };  template <typename t> class vector2d { private:     point2d<t> p; public:     vector2d(const point2d<t>& point) : p{point} {}      /*double magnitude();     point2d<t> component();     vector2d normal();     int dotproduct(vector2d rhs);     vector2d& crossproduct(vector2d rhs);*/      vector2d operator + (const vector2d<t>& rhs) const     {         return p + rhs.p;     }     vector2d operator - (const vector2d<t>& rhs) const     {         return p - rhs.p;     }     // ...      vector2d operator * (const t value) const     {         return p*value;     }     vector2d operator / (const t value) const     {         return p/value;     }     // ... };  template <typename t> class linevector2d { private:     point2d<t>  p;     vector2d<t> v;  public:     linevector2d() = default;     linevector2d(const point2d<t>& point, const vector2d<t>& direction) : p{point}, v{direction} {}      /// returns point on line time/value `t`     point2d<t> valueat(t t)     {         return p + v*t;     } };