We are given a plane x+y+z=7 (3D rectangular coordinates) and two points namely A(5,-1,4) and B(4,-1,3).we have to find projection of AB on the given plane.
My problem is that while first time i approached this problem, i found out the feet of perpendicular on the plane from both the points and then used distance formula between those feet. But im getting the wrong answer. Is there any lapse in my understanding?
actual answer is sqrt(2/3) but i am getting the answer as 1
the solution given in the book finds the projection by dotting the vector AB with the normal of the plane and ive understood that but still i dont find any problem in my method. Please help me pointing out my mistake
I believe the method you describe should give you the correct answer. Are you sure you've made no mistakes in your calculations? For the foot of the perpendicular to the plane from point A I get $$P_A=\frac{1}{3}\left(14,-4,11\right)\ ,$$ for the foot of the perpendicular to the plane from point B I get $$ P_B=\frac{1}{3}(13,-2,10)\ , $$ and the distance between $\ P_A\ $ and $\ P_B\ $ is $$ \frac{\sqrt{(14-13)^2+(-4+2)^2+(11-10)^2}}{3}=\frac{\sqrt{6}}{3}=\sqrt{\frac{2}{3}}\ . $$