Hello! Find the equation of the projection of the line \(\frac{x}{4}=\frac{y-4}{3}=\frac{z+1}{-2}\) of the plane x-y+3z+8=0. So the line projects its self on the plane... First I find equation of line which passes throught the line and it has vector "a" parallel to the normal vector of the plane... The equation of that line is: \(\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}\) The equation of the line we need to find is probably: \(\frac{x-x_1}{4}=\frac{y-y_1}{3}=\frac{z-z_1}{-2}\) So we just need to find the point... That point is \(M(\frac{1}{11},4-\frac{1}{11},-1+3\frac{1}{11})\) But in my text book results I get tottaly different solution: \(\frac{x+9}{7}=\frac{y+1}{4}=\frac{z}{-1}\) I don't know what is the problem. Please help.
