Find the area of the part of the plane x + z = 4 that lies above the square with vertices (0, 0), (1, 0), (0, 1), and (1, 1).
A)1
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D)2
E) $\sqrt { 5 }$
F) $\sqrt { 6 }$
G) $\frac { \sqrt { 5 } } { 2 }$
H) $\frac { \sqrt { 6 } } { 2 }$

Question

Find the area of the part of the plane x + z = 4 that lies above the square with vertices (0, 0), (1, 0), (0, 1), and (1, 1).
A)1
B) $\sqrt { 2 }$ 
C) $\sqrt { 3 }$
D)2
E) $\sqrt { 5 }$ 
F) $\sqrt { 6 }$  
G) $\frac { \sqrt { 5 } } { 2 }$ 
H) $\frac { \sqrt { 6 } } { 2 }$

Quizplus · Accepted Answer

The plane x + z = 4 intersects the xy-plane at z = 4 - x. The area of the part of the plane above the square is the area of the projection of the plane onto the xy-plane, which is a square with side length 1. The normal vector to the plane is (1, 0, 1), and its magnitude is $\sqrt{2}$. The area of the projection is scaled by the magnitude of the normal vector, giving an area of $\sqrt{2}$.

Find the Area of the Part of the Plane X 2\sqrt { 2 }2​

Find the Area of the Part of the Plane X $\sqrt { 2 }$