The general vector equation of a straight line is,(x,y,z)=a+λdwhereλis a parameter anddis the direction vector.Let the vector equation of the line be defined asL1.(x,y,z)=(1,2,3)+(2,1,−2)tThe vector equation of the line perpendicular toL1passing through lineP(x0,y0,z0)is given byax+by+cz=ax0+by0+cz0.....(1)Define the vector equation of the lineperpendicular to line passing throughPasL2.(a,b,c)in(1)is the direction vectorof the line passing throughL1,thus,(a,b,c)=(2,1,−2).∴2x+y−2z=ax0+by0+cz0.IfL2passes throughM(2,1,2),2x+y−2z=(2,1,−2)⋅(2,1,2)=4+1−4=1=1Thus,2x+y−2z=1passes through the point(2,1,2)and is perpendicular toL1.To find the point where the linesx(t),y(t)andz(t)intersects the plane,substitute the parametric equationsof the vector equation of the line2x+y−2z=12(1+2t)+(2+t)−2(3−2t)=12+4t+2+t−6+4t=1−2+9t=19t=3,t=31Thus, att=31,L1intersectsL2The coordinates of a point N onL1att=31aregiven by(35,37,37).Thus, the vector equation of the line passingthrough pointM&Nis the line passing through(2,1,2)and(35,37,37)which is given by(x,y,z)=(2,1,2)+λ((35,37,37)−(2,1,2))(x,y,z)=(2,1,2)+λ(3−1,34,31)......(2)Thus, equation(2)shows the relationshipbetween(2,1,2)and the vector equation of the lineL1.
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