Find the volume of the pyramid with base in the plane z 5 an

Solution

Area of the pyramid =1/3 *(area of base)*(height)... Here height is the perpendicular distance between the intersection point of y=0 , y-x=3 ,2x+y+z=3...and plane z=-5.... clearly intersection point becomes (-3,0,9).. perpendicular distance =(9+5)/sqrt(1) =14.. therefore height =14.. intersection points at base are for( y=0 and y-x=3) is (-3,0,-5) and for(y=0 and 2x+y+z=3) is (4,0,-5) because the plane is z=-5; intersection point of y-x=3 and 2x+y+z=3 is (5/3,14/3,-5) Area of the traingle with points (-3,0,-5),(4,0,-5),(5/3,14/3,-5). this is noyhing but the area of (-3,0),(4,0),(5/3,14/3). therefore clearly are is =1/2*(3+4)*14/3 =49/3.. So Volume of the pyramid is =1/3*14*49/3 =686/9.. Thank YOu.,,..