If the line segment joining the points P(x1,y1) and Q(x2,y2) subtends an qngle @ at origin O

prove that OP.OQcos@ = x1x2 +y1y2

hey yar

op.oqcos@=>x1.x2cos0+x1.y1cos90+x2.y1cos90+x1.y2cos90+x2.y2cos90+y1.y2cos0

since (x1 and x2) and (y1 and y2)are in straight line

Approved

hey yar

by resolution of vectors

op.oqcos@=>x1.x2cos0+x1.y1cos90+x2.y1cos90+x1.y2cos90+x2.y2cos90+y1.y2cos0

since (x1 and x2) and (y1 and y2)are in straight line

therefore the answer becomes

OP.OQcos@=x1x2+y1y2

since cos0=1 and cos 90=0

thus prooved

 As the line OP and OQ subtends an angle @, so we can use cosine law...........

    2.OP.OQ.cos@ =(X1^2+Y1^2)+(X2^2+Y2^2)- {(X1-X2)^2+(Y1-Y2)^}

                                          =2X1X2+2Y1Y2

   SO OP.OQ.COS@= X1X2 + Y1Y2

Approved